<r
ft'^
\^
KEY
TO
MR J. B. LOCK'S
ELEMENTARY TRIGONOMETRY
MACMILLAN AND CO., Limited
LONDON • BOMBAY • CALCUTTA
MELBOURNE
THE MACMILLAN COMPANY
NEW YORK • BOSTON • CHICAGO
ATLANTA • SAN FRANCISCO
THE MACMILLAN CO. OF CANADA, Ltd.
TORONTO
KEY
TO
MR J. B. LOCK'S
ELEMENTARY TRIGONOMETRY
BY
HENEY CARR, B.A.
OF THE GBAMMAB SCHOOL, LAGOS, WEST AFBICA.
MACMILLAN AND CO., LIMITED
ST MARTIN'S STREET, LONDON
1910
\f) '
Q.
Firs^ Edition 1889.
Reprinted 1893, 1896, 1900, 1904, 1910.
PKEFACE.
The solutions of the elementary questions are given by-
Mr Oarr with considerable detail in the hope that the book
may be useful to students who are studying the subject
without the help of a Teacher.
I have myself read the proofs and have given solutions
of many of the questions, particularly of those towards the
end of the book.
Any notice of misprints or inaccuracies of any kind will
be gratefully acknowledged either by the Publishers or
myself.
J. B. LOCK.
358351
KEY
TO
ELEMENTARY TRIGONOMETRY.
EXAMPLES. I. Page 2.
1. 1 mile = 5280 feet
^280 __ ^
= -TiT.- X 66 feet ;
DO
.-. the measure required is — - -- = 80.
^V 66
2. The area of the square which is the unit is 22 x 22 square yards,
i. e. 484 square yards.
Now 1 acre = 4840 square yards
""ZftJ ^ ^^* y ' •** ^ measure required is j~ = 10.
3. 1 ton = 2240 lbs.
2240 2240 2240
= -— ,- X 14 lbs. = — -— X 1 stone = —. — - ^ x 10 stones ;
14 14 14 X 10
2240
.*. the required measure is — — -— = 16.
14 X 10
2300
4. Length of Atlantic cable is 2300 miles =-—,- x 21 miles,
2300
i.e. — — - X length of cable from England to France;
^1
2800
.*. the required measure is . =109^{.
5. Area of the field is 22 x 1100 square yards,
. 22x1100 ._.. ,
1. e. — TA-r^ X 4840 square yards
4840
22x1100 , r , ^
= — r?7T?. — X 1 acre = 5x1 acre = 5 acres.
4840
L. T. K. e 1
2 MEASUREMENT. I.
6. a miles = a x 1760 yards
a X 1760 ^ ^ ^^ . , . 1760a
= ^7 — X ^ yards ; .-. the measure required is — , — .
7. The given distance is ac feet = -— x 3 feet
ac ^ ^ ac -
= y X 1 yard =^ yards.
8. The first measure is 24 and the first unit is half a sovereign ;
.'. the amount is 24 x 10 shillings = 240 shillings.
To find the second unit proceed thus : —
the given amount is 240 shillings = 240 x 1 shilling ;
.*. the second unit is one shilling.
To find the third unit proceed thus : —
240
the given amount is 240 shillings = 960 x ^^ shillings ;
.^ . 240 ..... 240x12 , . ^,
.-. the unit IS ^^ shillings = — ^^^r — d., i. e. three pence.
Or we may adopt the following solution : —
The student will observe that for a given quantity if the U7iit be
increased or decreased the measure is decreased or increased proportionally ;
i. e. for a given quantity the measure and unit vary inversely.
24
Therefore the second unit =r^ of 10 shillings = Is.,
24 1
the third unit = ,r^ of 10 shillings = 7«.
yoU 4
EXAMPLES. II. Pages 4, 5.
Area of second field _ 1
•*■' "Area of fiFst field^ ~ 20 *
Area of second field = ^t: x area of first field
11 „ 1 , 3,097,600
^20 "" 2 ^^^^^^ ^^^^ = 40 ^^* ^40 — ^^* ^
= 77440 square yards.
Height of first person 9
Height of second person ~ 8 *
9
Height of first person = - x height of second person
o
=|x5ft. 4in.=|x64in. = 72in,=:6ft.
2.
RATIO. II.
3. Area of the field = 65 x 3 acres = 195 acres = 195 x 1 acre ;
.'. ratio to an acre =195.
4. The first field = 6| x 2^ acres ;
.*. the measure of the first field in terms of second is 6j;
the ratio is * , = -7- ;
2J 4
the required fraction is ^ = — .
5.
Since
3-125 tons = 3-125xl ton = 3-125 x 20cwts.
62*5
= 62*5 cwts. = "2— X 4 cwts. ;
62*5
.*. the required measure is —r- =15*625.
o^«^x 62-5 , ,
3-125 tons = -J- X 4 cwts.,
o i^^x J. ' 4 X 62-5 ,. 625 ,^^
3-125 tons contain 4 cwts. —7— times = -77^- = 154 :
4 40 ^'
62-5
X 4 cwts.
,. . 3-125 tons
the ratio is — . — = ■
4 cwts.
4
4 cwts.
62-5 ,
'°' 125
"■- 8 '
,, . J r x- • 3-125 tons
the required fraction is — :
4 cwts.
4 ^^
4
125
The sum of money 22
3 guineas ~ 7
00 f!fJ
The sum of money = ^x3gs. = -— xl guinea
66 ^^ ..„. 66x21 , , .„. 66x21 ^^
= y x 21 shillings = — - — X 1 shilhng = - — — x 20s.
66x2 1
' 7 x 20'
X 1 pound ;
,, . , . 66 X 21 198 ^ „
.-. the measure required is — — — - = — -- = 9i^^.
7 X Zy) 2\)
22 S
The sum of money = — x 3 guineas = 22 x - guinea;
3 3 X 21
.*. when the measure is 22, the unit is - guinea = — = — 8,=9s.
1—2
RATIO. II.
7. a miles = 1760a yards =—^-— x 22 yards
1760a , , . 1760a ^ , .
= -^^r— X 1 chain = ^^, x h chains ;
22 2>«o
,^ . ^ . •1760a 80a
.-. the required measure is = -^— .
a miles = 1760a yards = — ^x- x 22 yards
1760a , , . 1760a , .
= -TT^- X 1 chain = -7^77— x c chains ;
22 22c
.-. the required number of times is -^r- — = .
22c c
Similarly it may be shewn that a miles =—7- x d chains ;
,, . -, ,. a miles 80a
.'. the required ratio
It may be also shewn that a miles = j; x k chains;
.-. the required fraction is
d chains d
< k chains ;
a miles 80a
k chains k
8. £20 = 400s. X 1 shilling = ^ x 21 shillings ;
.*. the required unit is 21 shillings.
£25 = 500s. =: 500 X 1 shilling = ^ x 21 shillings ;
.*. the required sum is 21 shillings.
£30 600
a certain sum in pounds 21 '
,600 shillings 600
a certain sum in shillings 21 '
.'. the required sum is 21 shillings.
Again, £10 or 200 shillings = —— x a certain sum in shillings;
.'. the required sum is 21 shillings.
EXAMPLES. III. Pages 7, 8.
1. Let the hypotenuse be x feet,
then x^ square feet = 36 square feet + 64 square feet (Eucl. I. 47),
.-, a:2 = 36 + 64,
.-. a;2 = 100,
.*. X =10;
.». the length of the hypotenuse is 10 feet.
LINEAR MEASUREMENT. III. 5
2. Let X yds. be the required length.
1002 sq. yds. = 60^ sq. yds. + x^ sq. yds.,
.-. a:2^ 1002 -602 = (100 + 60) (100 -60) = 160x 40 = 6400,
.-. a: = 80; .-. the required length is 80 yards.
3. In the figure of E. T. p. 6, let AD represent the pole, AC the rope,
then the required length will be represented by DC.
^D = 48feet; AC==52{eet, CD = xteet,
Now (Eucl. L 47) Aa^ = AD^+CD^
.-. 522 = 482 + a;2, .-. a:2=:52-'-482=(52 + 48)(52-48) = 100x4 = 400,
.*. a; = 20 ; .*. required distance is 20 feet.
4. Let AD represent the height of the houses, DC the width of the
street, A C will represent the length of the ladder,
^D = 40ft., CZ) = 30ft., ^C = a;ft.
Now by (Eucl. I. 47) AC^ = AD^+ CD\
.'. a;2 = 402 + 302=3 2500, ... x = oQ\
.-. the required length of ladder is 50 ft.
5. Let AD represent height of wall, CD width of the moat, and AC
length of scaling ladder,
AD = 11 feet, CD = 54 feet, AC = x feet.
Since A (72 = AD'^ + CD^, x'^ = 722 ^ 542 ^ siOO,
.*. .r = 90; .-. required length is 90 feet.
6. Let AD represent the length of the field, and DC the width, AC will
represent the length of the path.
AD = \ of a mile = J x 1760 yards = 440 yards.
DC = tV of a mile = T% x 1760 yards = 330 yards.
AC = x feet.
Since AC^ = AD^-{-DC\ .-. a;2 = 4402 + 3302 = 302500, ... ^^ = 550;
.♦. length of gravel path = 550 yards (nearly).
Width of gravel path = 2 ft. =| yd.
Depth of gravel path = 2 inches = | ft. = ^^ yd.
.'. volume of gravel path = (550 x | x ^V) cubic yards
= \Y cubic yards = 20|f cubic yards (nearly) .
7. Let AD represent the length of the field = 4a ft.,
DC represent the width of the field = 3a ft.,
AC represent the diameter of the field = a; ft.
Since AC^ = AD^ + DC^, .-. a;2 = (4a)2+ (3a)2 = 16a2 + 9a2 = 25a2,
.'. x=5a; .*. required diameter is 5a feet.
6 LINEAR MEASUREMENT. III.
8. In the figure ABC (E. T. p. 6). Let AB, AG be the equal sides,
BC the base, AD the perpendicular from A on BG,
AD bisects BG, .-. BD = IBG,
AB = AC=lSa yards, BD = ^BG = 5a yards, ^D=a; yards.
Since AB^ = AD^ + BD^, .-. (lSaf = x^-\-{5a)^
x^ = 169a^ - 25^2= 144^2, /. x = 12a ; .*. the required length is 12a yards.
9. In the figure E. T. p. 7, let AD, AB represent the equal sides of the
isosceles right-angled triangle, DB be the base ; from the vertex A to base
DB draw perpendicular AK, then DK=KB = \DB,
Let DB be x feet.
Then the square on DB = sum of squares on DA and AB.
a;2sq. ft.=a2sq. ft. + a2sq. ft., .*. ic2 = 2a2, :. x=J2.a = DB\
.-. DK=\DB = lJ2.,aii.=^KB.
From Eucl. III. 25 it may be shewn that K is the centre of the circle
circumscribing ADB ; .* . KA = KD = KB =z^^2 .a feet ;
.*. the length of required perpendicular is J;^2 . a feet.
Or proceed thus: — Let AK=x feet. Since AK is perpendicular to DB,
.-. AD^=AK^ + KD'\ a2 = a:2 + (J^2.a)2r=a:2 + Ja2;
2_a2_2a2 _^2a
- ^-2"-"r' •*• ^-"x*
10. In the figure E. T. p. 7, let DB = a feet, AD = DB = x feet.
Since DB^ = AD^ + AB^, .\ a^ sq. ft. = a;2 sq. ft . + a;2 sq. ft. ,
a2^2a;2, .-. 4a;2 = 2a2, .-. 2a; = a^2; .-. x = l^2a.
11. In the equilateral triangle ABG, let AD be drawn perpendicular on
BG, ,\BD=DG = iBG; .-. AB = ait., BD = iBG = iatt, AD = xn.
Now AB^=AD^ + BD\
:. a2 sq. ft. =a;2 sq. ft. + (^j sq. ft.,
a2 = a;2 + _; ... a;2 = a2 - — = —-; ,-. x=^.a;
4 4 4 2
.•. length of perpendicular is J^/S . a ft.
12. In the equilateral triangle of (11) let AD=a, AB=^x feet, and
5D==iJ5C=Ja;feet. Now ^£2=^1)2 + 51)2,
.-. a;2sq. ft.=a2sq. ft. + ^lV sq. ft.; .-. a;2=a2 + ^;
2^^_ 2 g_4a2 4.3.a2
.-. ^ -a ; .-. a: _ ^ - ^ ;
.•• required length of side is i/J^ , a.
2 ^'_ 2 3a;2_ 2 ,_4a2 4.3.a2 _2V3
x^-^a; ... — _a; ...a:---- ^ ; .'.x-~^.
LINEAR MEASUllI:MENT. III. 7
13. It is obvious that the diameter of the circle coincides with the
diameter of the inscribed square.
Let a? = diameter of circle = diameter of inscribed square,
2/ = side of inscribed square ; .*. x^ = y^ + y^ = 2y^; .*. x — y^2;
X ^2
The side of inscribed square y J2 1 ^, • -, x- . -. ,^
rp, ,. 7 ^-----— . -— =?.=y = ' ... the required ratio is 1 : ^2.
The diameter of the circle x 2 ^2 ^
14. Let AB he the given chord, the centre of the circle, let the per-
pendicular from on AB bisect AB in Cy then
OC=a;feet, 0.4 = 0^ = 10 feet, ^C=Ci^ = i^B = 4 feet.
Now OA^ = AG^+CO\
102 sq. ft. =42 sq. ft. +a;2 sq. ft. ;
.-. 0^2 = 102-42 = 100- 16 = 84;
/. X = ^y^ ; .". the required distance is ,^84 ft.
15. With the notation of (14) 0^ = ayds. =Sa feet, 0(7=6 feet,
AB = x feet and AC=ix feet. Now OA^ = AC^+CO^,
:. (3a)2sq. ft.^d") sq.ft. + 62 sq.ft., or 9^2 = ^ + 62.
.-. a:2 = .36a2-462 = 4(9a2-62); ... x^2 J^a^-h^\
.-. the required length is 2 ^9a^~ b^ feet.
16. Let X be the common difference of the a. p.; then since the hypote-
nuse is the greatest side, the other two sides are 5a -x and 5a - 2x.
Eucl. L 47, {5a-x)^+(5a~2xf=={5a)^;
.-. 50a2-30aa; + 5x2 = 25a2; .-. x^-Qax^-oa^
From this quadratic a; = a or 5a,
(1) when a; = a, 5a-a: = 5a- a = 4<x,
and 5a-2x = 5a-2a = Sa;
(2) when a; = 5a, 5a- x = 5a- 5a = 0,
and 5a-2x^5a- 10a = - 5a.
In the first case the other two sides are 4a and 3a. In the second case
the other two sides are and - 5a, i. e. the triangle becomes a straight line;
.*. only the first case is admissible.
EXAMPLES. IV. Page 13.
1, Let X be the number of feet in the diameter of the square.
Then x^sq. ft.=:72sq. ft. + 72sq. ft., or a;2 = 72 + 7^
.-. a;2 = 2x72; .-. a: = 7V2 = 7 x 1-4142... =9-899... ft. ;
,\ the length of the diameter is 9-899... ft.
INCOMMENSURABLE QUANTITIES. IV.
2. N.B. A yard is less than the thousandth part of a mile.
Let X be the number of miles in the diameter of the square.
Then a^^sq. mi. = 12sq. mi. + 1^ sq. mi., .-. a;^ sq. mi. = 2 sq. mi.,
x = j2 = l-4:14:2..,] .'. the length of the diameter is 1*4142... miles
= 1-4142... X 1760 yards = 2489 yards.
3. Let X be the number of inches in the hypotenuse.
Then x^ sq. inches = (42 J)^ sq. in. + (40)^ sq. in. ;
■=(S)'.(«,=
7225 6400 13625
+ -
4 ' 4 4 '
.-. a; = iVl^25 = 1x116-726 = 58-36;
.-. the length of the hypotenuse is 58-36 inches.
4. Let x^ be the number of square inches in the square,
X ,, ,, of inches in each of the sides;
.-. a;2 = 1000x 9x144 = 1296000; .-. a; = 1138.
5. Let X be the number of feet in each of the sides,
.-. x^ sq. ft. =the area of the square and the length of the diameter is x ^2,
.^2= (4840 x 9 X 10) sq. ft. = 435600 sq. ft. ;
.-. a;= 7^35600= 660,
length of diameter = a; ;^2 ft. = 660 x 1-4142 ft. = 933-372 feet;
6. Let x^ be the number of square inches in the square,
X ,, ,, inches in each of the sides,
x^ sq. inches = 1 acre = (l x 43560 x 144) sq. in., or {c2 =43560 x 144,
.-. i»= ^43560 X 144 = 12 ^^43560=12 x 208-7 = 2504;
.-. the length of side is 2504 inches.
7. Let ABG be the equilateral triangle, AD perpendicular on J5C,
therefore BI) = DG = \BC.
^D = a;feet, ^5 = 10ft., BD = 5ft.,
lO^sq. ft. = a:2 sq.ft. +52 sq.ft.; or ^2 = 102-5^ = 75 ;
.*. a;=/y75 = 8-66 ft. ; .•. height of the equilateral triangle is 8-66 ft.
8. x^ sq. ft. = (5-32)2 sq. ft. _ (2-66)2 gq. ft. .
.-. a;2=(5-32)2 .- (2-66)2=7-98 x 2-66 = 21-2268;
.-. a;=V21-2268 = 4-607;
.-. height of the equilateral triangle is 4-607 ft.
9. a;2sq. in. =(24-6)2 sq. in. + (41-3)2 sq. in.,
x'' = (24-6)2 + (41-3)2 = 605-16 + 1705-69 = 2310-85 ;
.-. a; = ^2310-85 = 48-07... ; .'.the length of diameter is 48 inches.
INCOMMENSURABLE QUANTITIES. IV. 9
10. x^ sq. ft. = 782 gq. ft, + 352 gq. ft. ,
x^ = 782 ^ 362 = 6084 + 1296 = 7380 ;
.-. ar= ^7380 = 85-9; .'. the length of diameter is 85-9 ft. =85 ft. 11 in.
EXAMPLES. V. Pages 21, 22.
1. The circumference = tt x diameter
22 22
= — X 1 yard=— yds. =3f yds.
2. The circumference = tt x diameter = tt x 2 x radius
=:7r X 8 ft. =^ X 8 ft. = 4^ ft. =254 ft.
7 7 ^
3. The circumference =7r x diameter
= 7rx 48m. = — x48m. = — =— m.=:150f m.
4. IT X diameter = circumference = 10 ft. ;
.-. diameter = ?^ = ,1 x 10 f t. = ^ ft. = S^\ ft.
5. The circumference = 1 revolution = ' = — ^^ — - = 8 yds. ;
.•. TT X diameter = 8 yds. ,
,. , 8yds. 7 ^ , 56 , 168., _^ _
diameter = -^ = 22 "" ^^^'"^22 ^"^^-""22 ^^'^^^^ ^^'
6. Here the diameter is 36 inches.
n- , ;.• . 22 _. 792. 792 „, 6Q ..
Circumference = tt x diameter = — x 36 m.=:-— - in.= — — ft. =:-=-it.
, . .' ' M Iml. 5280x7 o^ _ ^nr,
The no. of revolutions in a mile = -^ ^ = — jr:; — =80 x 7 = 5bO.
circumierence 00
7. The diameter of first wheel is 50 inches.
Now circumference = tt x diameter ;
22 1100
.'. circumference =— x 50 inches =—^ inches.
The diameter of second wheel is 52 inches.
22 1144
Circumference = tt x diameter = — x 52 in. = — =— inches.
7 7
The no. of revolutions in a mile made by first wheel
_ 1ml. 52 80 X 7
~~ circumference " 1100 '
• -1 1 i. 1 X. 1. :. u , 5280x7x12
similarly no. of revolutions by second wheel = TTJl •
5280x7x12 5280x7x12
Difference of no. of times = ^^v^r z-^rr.
1100 1144
KOQA n io/1144-1100\ 5280x7x12x44 1008 _^^ _, .
=''''^^^^Hn44^oo) =-Ti44^m^^^^^^
10 CmCUMFiERENCE OF A CIRCLE. V.
22 110
8. Here diameter is 5 feet. Circumf . = tt x dia. = -— x 5 ft. = -=- ft.
7 7
^, , , ,. . -i^n -1 100 mis. 100 mis. X 7
No. of revolutions m 100 miles = - . = — ttt^-fz
circumierence 110 it.
52800x7 „„„.„
= ^;j^- = 33600.
9. In 1 sec. the engine makes 3 revolutions,
.-. in 60 X 60 sees. „ ,, 3 x 60 x 60 revolutions ;
i. e. in 1 hour the engine makes 10800 revolutions.
From (8) engine makes 33600 revolutions in 100 miles,
.*. engine makes 336 ,, ,, 1 mile.
.*. engine makes 1 revolution in ^^ of a mile;
.*. engine makes 10800 revolutions in miles;
i.e. engine „ 10800 ,, in 32f miles;
but engine makes 10800 revolutions in an hour;
.'. the engine runs 32f miles per hour.
10. In 1 second the engine makes 4 revolutions,
in 60 X 60 seconds ,, ,, ,, 4 x 60 x 60 revolutions;
i.e. in 1 hour ,, ,, „ 14400 revolutions.
But in 1 hour the engine goes 60 miles ;
.'. the engine makes 14400 revolutions in 60 miles,
the engine makes 1 revolution in rrrr^ miles, i.e. in — tttf^tt- feet = 22 feet;
14400 14400
therefore the circumference of the engine is 22 ft. ;
TT X diameter = circumference =22 ft.;
.-. the diameter =-- = ^ x 22 ft. = 7 ft.
11. The extremity of the hand of the clock travels round the face of the
clock once in an hour, and therefore it makes 24 revolutions in a day.
The diameter here is 2 x 11 ft. = 22 ft.
22 484
Circumference =7r x diameter = — x 22 ft.=-^ ft. = length of one revolution;
o^ 1 4.- 484 _, „^ 484x24 ^ ^„3, ,
24 revolutions = -;;- x 24 ft. = -— — — yds. = 553f yds.;
7 7x3
.'. the extremity of the large hand travels 553f yds. per day.
In 1 minute the extremity of the large hand makes ^ of a revolution,
1 -484,^ 484x12, 484, ,^ ^ .
i.e. g^of — ft.=^^^^m. = — m.=13.8...m.
CIRCUMFERENCE OF A CIRCLE. V. 11
22 2376
12. Circumference = tt x diameter = -«- x 108 ft. = — ^— = 339 ^ ft.
13. It is shewn in Euc. IV. 17 that radius of a circle = side of the
inscribed regular hexagon ;
.*. diameter of the circle = 2 x 3 ft. = 6 ft.
22 132 132 X 12
Circumference = tt x diameter = -=- x 6 ft. = -=r- ft. = in. = 2263 in.
7 7 7
Perimeter of hexagon = 6 x 3 ft. = 18 ft. = 18 x 12 in. = 216 in.;
.'. total length of wire =(226*3 + 216) in. = 442-3 in.,
hence 443 in. are required.
14. The radius of the circle = side of inscribed hexagon ;
IT X diameter = circumference,
,. ^ circumference 113x10^, 226^,
diameter = = — --- — ft. =-=t- f*-;
TT 355 71
113 X 5 113
.-. radius = ~okk— ^*- = 71" ^*- Perimeter of hexagon = Vi^ ft. x 6 = ^^ ft. ;
total length of wire required = 10 ft. + -V/ ft.
.*. no. of inches necessary = 120 in. + 115 in. = 235 in.
15. The diameter of the circle coincides with the diameter of the
inscribed square. Let a; = diameter of square = diameter of circle;
then, x^ sq. in. = 242 gq i^^ ^ 242 sq. in. or x^ = 242 + 242 = 2 x 242 .
.-. a; = 24^^2 = 24x1-414 = 33-9 in.
355
Circumference = TT x diameter = -— x 33-9 in. = 107 in.
J.xO
Perimeter of square = 4 x 24 in. = 96 in.;
.-. total length of wire required = 96 in. + 107 in. = 203 in.
16. The diameter of the circle coincides with diameter of the inscribed
square. Also tt x diameter = circumference;
^. , circumference 113 ^^ ^^ 113x144. 16272. ,^ ^ .
diameter = = 355 ^ 12 ft- =-355- ^^'=^^ m. = 45-8... m.
Let X = side of inscribed square,
then a; ,^2 = diameter of inscribed square = diameter of circle = 46 in.;
.-. a: = ^.^'in. = ^5%^in. = 23-9... in. x 1-4142... =32-4... in.;
.*. perimeter of square = 4 x 32-4... in. = 130 in. nearly;
.-. total length of wire = 144 in. +130 in. =274 in. nearly.
22 5 55
17. Circumference = tt x diameter = -— x t in. = :.-t in.
7 4 14
/55 \ 330
Area of surface of handle of bat= ( — x 12 ) sq. in. =— — sq. in.
T -. 4, ^ . area of surface of handle 330 1 330 x 40 13200
Length of string = —-—. = — ; — ;- 77; = ^ — = — ^r"
diameter of string 7 40 7 7
= 1885-7... in., .-. 1886 in. are required.
12
ANGLES.
'0
I
EXAMPLES. VI. Page 25.
1. If OV starts from the position OH and turns about O
in the direction contrary to that of the hands of the clock to
the position OV (so that the angle ROF is a right angle) it
will describe the angle equal to + 3 right angles.
2. If OF turning in the direction contrary to that of the
hands of the clock starts from OH^ makes about the point
one complete revolution (i.e. four right angles) and then turns
in the same direction to the position 0I\ so that the angle
BOF is equal to a right angle, it will describe + 5 right
angles.
3. If OF turning in the direction contrary to that of
the hands of the clock starts from OF, makes about the
point O one complete revolution (i.e. four right angles) and
then turns in the same direction to the position OF so that
the angle FOF is equal to half a right angle (i.e. 45<^), it
will describe + 4 J right angles.
4. If OF turning in the direction contrary to that of
the hands of the clock starts from OF, makes about the
point one complete revolution, and then turns in the
same direction to the position OF so as to make the angle
FOF equal to four right angles less | of a right angle, OF
will describe + 7 J right angles.
5. In the Figure of 1, if OF turning in the same direction as that of the
hands of the clock starts from OF and turns about the point to the
position OF so that the angle i2 OP is equal to a right angle, OF will describe
- 1 right angle.
6. If OF turning in the direction contrary to that of
the hands of the clock starts from OF, makes two com-
plete revolutions (i.e. eight right angles) about the point 0,
and then turns in the same direction to the position OF
so that the angle FOF is equal to + 2 right angles and
60 degrees, OF will describe + 10| right angles.
7. If OF turning in the same direction as that
of the hands of the clock starts from OF, makes
about the point two complete revolutions and
then turns in the negative direction to the position
OF so that the angle FOF is equal to - 2100, OF
will describe - \^\ right angles.
8. If OF turning in the direction contrary to that of the hands of the
clock starts from OF makes one complete revolution about the point 0, so as
to be again in the position OF from which it started, it will describe + 4 right
angles.
9. If OF turning in the same direction as that of the hands of the clock,
starts from OF and makes about the point one complete revolution, so as
to be again in the position OF from which it started, it will describe - 4 right
angles.
/?
ANGLES. VI. 13
10. n represents any whole number positive or negative. 4n will always
be divisible by 4, and 4?i right angles will represent n complete revolutions.
If therefore OP turning in the positive or negative direction, starts from the
position OR and makes about the point 0, n complete revolutions, so as to
be again in the position OR from which it started, it will describe 4n right
angles.
11. From 10, when OP has described 4n right angles it will be in the
position OR from which it started. From this
position let OP turn about the point in the p O E
positive direction to the position OP so that the
angle ROP is equal to + 2 right angles; OP will thus have described (4;i + 2)
right angles.
12. -(4w + 4) right angles = -47i right angles + ( — 4) of
a right angle. When OP has described - in right angles it
will again be in the position OR from which it started (10).
From this position let OP turn about the point in the
negative direction to the position OP^ so that the angle ROP
is equal to half a right angle (45^); OP will thus describe
- (471 + ^) right angles.
EXAMPLES. VII. Page 28.
1. 63* 2V 18-- = 106 + 10000 + 1000000 °^ ^ "^^* ^"^le
: -632118 of a right angle.
,_104 26 99-1
~ 100 "^ 10000 "^ lOOOOOC
= 1-0426991 of a right angle.
2 18 27
00 "^ 10000 "^ 1000000
•021827 of a right angle.
._J_ 29 4894^
"100"^ 10000 "^ioooooo
= -03294894 of a right angle.
;2 41
OOO "^ 1000000 ^* ^ ^^^*
= •006241 of a right angle.
,1000 8 12
~ 100 "^ 10000 "^ 1000
= 10-000812 of a right angle.
S2^ 4 52
^ 100 "^ 10000 "^ 10000000
= •3204052 of a right angle.
104 26 QQ-1
2. m.26« 99-r = j^ + j^+ j^^^of arightangle
2 18 27
3. 2« 18- 27 = j^ + ^-^^ + 3^j^^^ of a right angle
4. 3. 29> 48-94" = 4 + ^+ ^-^-^^^-^ of a right angle
62 41
^^' ^^ =10006 + 1000000 °^ " "S^* ^"^Sle
6. 1000« 8- 12"=^ + _J_^ + _1L_ of a right angle
32« 4- 5-2" = ^ + loUoO + lOOO^ °^ * "^ht angle
14 DECIMALS OF A RIGHT ANGLE. VII.
8. 1^2^3-4-=~+^^ + j^Jl^ofarightangle
= -0102034 of a right angle.
9. 69«0'7-r = ^«| + ^^^ofarighta„gle
= •6900071 of a right angle.
10. 119. 3- 0-45" = Jg + jJ^+ -^^. of a right angle
= 1-19030045 of a right angle.
11. 1006« 18^ r = ^^^ + j^+ j^^ of a right angle
= 10-061801 of a right angle.
2 26 48
12. 2« 26^ 4-8-= j^ + j^^ +10000000^^ ^ ''^^^ ^^^^^
= -0226048 of a right angle.
13. '36 of a right angle = 36 grades,
•00,78,, „ „ =78 minutes,
•00,00,91 „ „ „ =91 seconds;
.-. the angle is 36« 78' 9V\
14. 1 right angle =100 grades,
•04 of a right angle = 4 grades,
•00,30 „ „ „ = 30 minutes,
-00,00,21 „ „ „ = 21 seconds;
.-. the angle is 1048 30' 2r\
15. *01 of a right angle = 1 grade,
•00,20,, „ „ =20 minutes,
-00,00,03 „ „ „ = 3 seconds;
.*. the angle is 1^ 20' 3'\
16. -00,10 of a right angle = 10 minutes,
•00,00,2 „ „ „ = 2 seconds;
.*. the angle is 10' 2".
17. '06 of a right angle = 6 grades,
•00,25 „ „ „ =25 minutes;
.*. the angle is 6s 25'.
18. 3 right angles = 300 grades,
•02 of a right angle = 2 grades,
-00,12 „ „ „ = 12 minutes;
00,00,50 „ „ „ = 50 seconds;
/, th© angle is 302« 12' 50'\
DECIMALS OF A RIGHT ANGLE. VII. 16
19. 1 right angle ^100 grades,
•00,10 of a right angle = 10 minutes;
/. the angle is 100k 10\
20. '01 of a right angle = 1 grade,
•00,01 „ „ ,, =1 minute,
•00,00,001,, „ „ =-1 second;
.-. the angle is 1^ 1' 01'\
21. 6 right angles = 600 grades,
•45 of a right angles 45 grades,
•00,10,, „ „ = 10 minutes;
/. the angle is 6458 10\
22. *02 of a right angle = 2 grades,
•00,30,, „ „ =30 minutes;
.-. the angle is 2? 30\
23. -00,01 of a right angle = 1 minute,
•00,00,10,, „ „ =10 seconds;
/. the angle is 1' 10'\
24. -00,00,10 of a right angle = 10 seconds;
.-. the angle is 10'\
EXAMPLES. VIII. Page 30.
1. 60 ) 27 seconds 2. 60 ) 30 seconds
60 ) 15-4 5 minutes 60 ) 4-5 minutes
90 y8-2575 degrees 90 ) 6-075 degrees
•09,17,5 of a right angle -06,75 of a right angl©
= 9^17^ 50". =6«75".
3. 60 ) 15 seconds 4. 60 ) 19 seconds
60 y 5-25 minutes 60 ) 14^31666666... minutes
90 j 97-0875 degrees 90 ) 16-2 3861111... degrees
1^07,87,5 right angle ^8, 04,29012345679 of a right angle
= 1078 87' 50". =18«4^ 29\..
5. 60 )_6 minute^ 6, 90 ) 4 9 degree s
90 ) 132-1 degrees -54 of a right angle
1^467 right angle ^54^ 44' 44^4'\,,,
=:146«7r 77-7"...,
16
DECIMALS OF A RIGHT ANGLE. VII.
7, IK 37' 50'' = -01375 of a right angle 8. 8* 75' = -0875 of a right angle
90
90
1-23750 degrees
60
7*8750 degrees
60
14-2500 minutes
15-600 seconds
the result is 1° 14' 15".
52-5000 minutes
60
30-0000 seconds
the result is 7° 52' 30".
9. 170K 45' 35'' = 1-704535 of a right angle
90 10. 24k 0' 25" = -240025 of a right angle
153-408156 degrees
60
24-489000 minutes
60
" 29-340000 seconds
.-. the result is 153° 24' 29-34".
11. 18S 1' 15" = -180115 of a right angle
90
16-210350 degrees
60
12-621000 minutes
60
37-260006 seconds
the result is 16° 12' 37-26"
90
21-602250 degrees
60
36-135000 minutes
60
'8-10000 seconds
the result is 21° 36' 8-1".
12. 358= -35 of a right angle
90
31-5 degrees
60
30 minutes
.-. the result is 31° 30'.
1, (1) TT radians =7rx
EXAMPLES. IX. Page 35.
2 right angles
37r
(2) -7- radians = -7- x
4 4
= 2 right angles = 2 x 90° = 180° ;
/I
.-. the angle contains 180°.
Stt 2 right angles 6
of a right angle
3 3
= jr of a right angle = - x 90° = 135° ; .-. the angle contains 135°.
(3) iradian=^i:^i^^^^Hil!?=l?2! = 57-2957...;
TT TT
.-. the angle contains 57°* 2957...
(4) 3 radians = 3 x — = - right angles ;
.-. the angle contains - right angles.
CIRCULAR MEASURE. IX. 17
(5) 3-14159265 etc. radians
= IT radians = tt x — — = 2 right angles = 180° ;
TT
.*. the angle contains 180°.
... 2 ,. 2 2 right angles 4 . , , ,
(6) - radians = - x — ~ — = -s right angles ;
TT TT IT ir^
4
.". the angle contains —^ right angles.
/,,^ ., T ^2 right angles 26 . , ^ ,
(7) radians = 6 x -— — - — = — right angles;
TT TT
26
.'. the angle contains — right angles.
TT
(8) -00314159 etc. radians
= mo "^^^^'''^I^ ^ 2right^angles ^ _^|_ ^.^^^ ^^gies = -002 right angles.
(9) IOtt radians = IOtt x ^ ^^g^^ ^^g^^s ^ ^q ^^^-^t angles ;
TT
.*. the angle contains 20 right angles.
2, Let a be the circular measure of the given angle: then
_c 180°
(1) — = :;t:7T7.=1; .'. a^=ir^: .*. tt is the required circular measure.
^ ' 7r« 180"
(2) — — r57ro=2; .'. a*'=27r«; .*. 2ir is the required circular measure.
(^) ~ — TofTo = o ' • • ^'^ = a ' •'• Q ^^ *^® required circular measure.
,,^ a« 224° 45° 1 7r« TT. ^, • ^ • ,
^^ ^^ "^ 185^ ^ 360^ ^ 8 ' •'• 8 ' •*• 8 ^^ required circular measure.
^ ^ 7r<^ 180° 180' •• " 180'
•"• Ton ^^ *^^ required circular measure.
a« _ 57-295° etc . _ 2 right angles ^ _ 180^ _1_
^^ 7r« ~ 180° ~ TT ^ 180° ~ TT ^ 180°
1 180° 1 . 7r« ,. -. . .t, • 1 ..
= - X TTTT^ = - ; •*• a«= — = l'^ ; /. 1 IS the circular measure required.
TT 180° TT TT
^^ T«~"i80°~i80' •*• ^"""180'
.*. r--^ TT*' is the circular measure required.
loO
L. T. K. 2
18 CIRCULAR MEASURE. IX.
^^^ 7r«" TT ''l80°~27r' ''•"'-2^-2-^"
.'. ^ is the circular measure required.
^^ 7r« ~ 180° ~ 180 ' •*• " ~ 180 '
' * Tftn ^^ *^® circular measure required.
3. (1) 338 33' 33-3" = 33-333333,.. grades = 33-3 grades.
Let a be the required circular measure,
a« _ 33'3g _ 33-3 _ 100 J^_l. ^_7r^.
7r« ~ 200e ~ 200 ~ 3 ^ 200 ~ 6 ' •'*"*'" 6 '
.-. ^ is the circular measure required.
,^. a« 50« 1 ^ 7r« TT . ^, . ,
(2) — = 20^ = t; .*. a«=— ; /. ^ is the circular measure required.
g^ _ 16-68 _ 16-6 _ 50 J__jl^. . «_^
^ ^ 7r«~ 2008 ~ 200 ~ 3 ^ 200" 12' *'• °''~12'
. Y^ is the circular measure required.
^^> 5^2^8 = 250' ••^^=^' •• ^ is the circular measure required.
(5)
(6)
1' 1
a^ = z
20000' 20000 ' • • "- 20000 '
/. ^A^^ is the circular measure required.
a« 10''
7r« 2000000" 200000' " 200000'
' * 900000 ^^ *^® circular measure required.
^^ 7r« ~ 2008 ~ 200 ' '**"""200'
" 200 ^^ *^^ circular measure required.
,^. a« 2008 1 1 , 7r« ,
^^^ ^^ = -^-^2008=;^' ''--'=^ = ^"^
.'. 1 is the circular measure required.
(9) ^"2008 ~^' * * "'''=^'^*'J •'• ^^ is the circular measure required.
CIRCTTLAR MEASURE. IX. 19
4. (1) 180° = .o; .-. l° = f^; .-.4.5-^^x45 = 1;
,^^ StT TT StT TT 4 1 ,, . . 1
.-. 45° -j — 7- = -r -^ /- = T X n— = o ; •*• the required ratio is - .
4 4 4 4 OTT D o
(2) If D and (r be the number of degrees and grades respectively in any
angle,
180 = 200^ •••180-0 = 2-00 = 3' .•- ^ -3- the number of grades in 60° ;
,^, ,^ 200« ,^ 200 10 ,, . ^ .. . 10
60°-^608=— ^ — '-QO«=^^r--^ = -^ ; .-. the required ratio is — .
/o^ D 258 1 ^ 180° 45 ,. ,, „ , . . _ ,
(^) 180 = 200i=8= ••• ^ = -^ =y(i-e-tlie no. of degrees in 25«);
258^ 22° 30' = 25«-T-224° = — — f- -— = 1 ; /. the required ratio is 1.
,AK (r a 24 a 247r ,. ^, i_ i. j. . ^. .
^^^ 200 = ,r' •'•200 = ^' .-.a^^OO^'-^'*^^''"^^^"'^^^^^^^'^''^^^'^'
^^'-^^ =200-^^ = 200 ^2 = 50' ••• the required ratio IS ^.
(5)
J) _ l-75'^ _ 175
180 ~ 7r« "IOOtt'
175 X 180
' ■ ^~~~T00 ^^•^* *^^ iiumber of degrees in 1*75«) ;
1 .750^^° 1'^^ ^ 1^^° 1^^° 175 X 180 X TT
"" OA»
IOOtt * TT IOOttxIOO "20
.*. the required ratio is ~ .
a. \ TT
(^) ~ = T3a ' •'• "=" ToA (^•®* number of radians in 1°) ;
^°-^^'=iio^^ = iio*' ••• the required ratio is j^.
EXAMPLES. X. Pages 37—39.
1 . The angle = — ^. — x (a radian) = -~ radians = -7^X7^^ radians ~ li°.
radius ^ ^25 2 25 ^
2. The angle=^^ X (a radian) = f^ x ' "^^^ ""^^^ -^ = % x ^Jg!=90°.
radius ^ '10 tt 10 tt
3. The angle = — -^, — x (a radian) = - „ . -', — x (a radian)
radius ' 3^^ inches ^ '
24 2 right angles 24 11 7 " . , ^ , 168 . ,^
^3A "" ^ = T ^ 35 "^ 22 ^ ^ "^^* ^^^^^'= 35" ''^^* ^^^^''
= 4f right angles.
2—2
20 CIRCULAR MEASURE. X.
- ^, , 1 in. , _. . 1 20000 French minutes
4. The angle = g-jj-^^ x (a radian) =^^ x
1 7 700
— Tfu\ ^ oo ^ 20000 French minutes = —r- French minutes,
5. The angle = — -,-. — x (a radian). Let x = length of arc ;
. . ^ radianszr^ ; /. a;=| x 25 = ^=112J ft.
6. The angle = — r^ — x (a radian),
radius ^
. ^„ X 180° X 7 -^^o 315j; 80°x22 ,,^
,.e.80°=5X — =jx-xl80-^^; .-. ^= -^^^-=5^;
.'. length of arc required is 5fJ ft.
7. Let X be the length of the arc.
The angle= -^ x (a radian), i.e. 60« = ^x— = £^x^x200« = '^;
radius ^ ' 10 w 10 22 11 '
fifi
.*. x='=- = 9f; .*. length of arc required is 9f ft.
8. The diameter of the sun can be here represented as approaching the
magnitude of the arc subtended by an angle of 32' in the centre of the circle
of radius 90,000,000 miles.
Let X represent length of arc.
Then "^ 2 right angles _ 7x 180x60 _
90,000,000 IT " ' 90,000,000 ^ 22 ""^^ '
21a; =17, 600, 000; /. a; = 831,095 miles; .-. length of arc is 838,095 miles.
But the diameter of the sun coincides with this arc to three significant
figures; hence the diameter of the sun is equal to about 838,000 miles.
9. The train in 1 hour travels 20 miles,
,. ,» M 1 second „ .,, ^. miles,
60x60
20x10 ... 1 . .,
., „ „ 10 seconds „ _ - ,^ miles, i.e. zr^o{& mile.
bi) X dU io
Now number of radians = —3^ — = — x 2 = - .
radius 18 9
.-. the angle = - of a radian;
^, , 1 2 right angles 7x180° 70° .^„
/. the angle = ^ x \ = -^^^=. ^=6^°.
CIRCULAR MEASURE. X. 21
10, The train in 1 hour travels 60 miles,
„ „ „ 1 minute ,, 1 mile,
„ »» n T minute ,, j of a mile.
Now — 5^ — = number of radians, t x k = 7^',
radms '428
,, . . 3 180° 3xl80°x7 945° ^,,,^
.-. the angle ib g x — = "^-^y- = ^=21^1°.
11, The angle = --^. - x (a radian)
1 180° 1 180x60x60
= 7?^?^ ^ = JT^Fi ^ seconds
4000 IT 4000 TT
7 X 180 X 60 X 60 1134
= 4000x22— ^^^^-^ ^ secs. = olA".
12, The angle = ~-^^ — x (a radian). Let x represent length of arc,
i-e. l" = 77^x — x60x60;
4000 TT
4000x22 ., 11 ., 11x1760 ^ 19360 ^
••• ^=180 X-60V60 ^^^"^==567 "^^^^^ = -56^" ^^^'^ ^67" ^^''
= 33*95 yds., i.e. 34 yds. nearly.
13. Let TT be the required ratio.
^ radms tt ' 12-5 tt '
.*. T= -_ — =3*1416; .*. the required ratio is 3*1416.
o X l.Z'o
1^ m. 1 arc 180° . ^.^ 1-309 180°
14. The angle = --TT— x , i.e. 7i°= ^r^- x ;
radius tt ^ 10 ir
1*309x12 15*708 ^^,^^
/. TT = = — - — = 3*1416.
o 5
15. Let TT be the required ratio.
Then the angle = — ^. - x a radian), i.e. 22^° = —;^ x *,
radius ^ '' ^ 80 tt
. 377 xl80x2 _377_ ^
•• ''-"12x80x45 - 12 -''^"''-
22 CIRCULAR MEASURE. X.
16. The diameter of the moon approximates in magnitude to the arc
subtended by an angle of 30' at the centre of a circle of which the radius is
equal to the distance of the eye of the observer from the centre of the moon ;
similarly the diameter of the sun approximates in magnitude to the arc sub-
tended by an angle of 32' at the centre of a circle of which the radius is the
distance of the eye of the observer from the centre of the sun.
The ratio of these arcs will be then equal nearly to the ratio of the
diameters.
' Let x^ be the length of arc subtended by angle of 30',
iCg n >> " '» »» " *^^ >
Ti be the distance of the observer's eye and the centre of the moon,
375ri „ „ „ „ „ » ., » sun.
. angle j.
Then, since arc=^r— r-Ti^ — t— x tt x radms,
' 2 right angles
a^i _ 30' jr\_ _ 1
•*• ^2 ~ ^ ^ 375ri ~ 400 '
.-. the ratio of the arcs is -zr^ and so is the ratio of the diameters.
400
17. 180° = TT radians, 180 x 60 x 60 seconds = tt radians ;
■■■Isecond^^-^^^P^g^ofaradian;
.-. 10 second8 = jg^i^^ of a radian = ^ ^3- Jg^g^^^ of a radian
= 0000484... radian.
7322400
18. The two places evidently lie on a great circle of the globe.
The angle contained by the difference of their latitudes
arc of great circle , ,. .
= ^^^ X (a radian) ;
radius
1^° = — - — X — - ; :. r= -^ = 49 ^x inches.
19. It is evident that the centre of the smaller circle is in the middle
point of the perpendicular of the isosceles triangle ; and its radius is half of
the perpendicular. The radius of the greater circle is the perpendicular.
Now let d be the number of radians in the vertical angle ; then
arc of greater circle within the triangle =2/i^, when 2h is the height of
the triangle ;
arc of smaller circle without the triangle = /i (27r - 26).
But the two arcs are equal ;
.-. 2he = h(2w- 2d) ; .'. 2e = ir; :. the vertical angle is ^ .
CIRCULAR MEASURE. X. 23
20. Each of the angles in the triangle ABC (fig. Eucl. I. 1) is equal
to 60°.
mi- 1 T>^/^ arcjBC ^. _. radius ^5 _._
The angle BA G = — t-. t^ x a radian ; .-. radian = =-^^— x 60°.
^ radius AB arc BG
Let r represent AB; BG = AB = r. Let d represent the difference between
arc BG and BG, so that arc BG = r-\-d; for the shortest distance between two
points is the straight line joining them ; hence the arc BG is greater than the
straight line BG,
V
:. radian = , x 60°
r + d '
i.e. the radian (unit of circular measure) is less than 60°.
21. The magnitude of the diameter of the sun is constant; the change
is therefore due to the approach of the sun to the observer; i.e. to the varia-
tion of the radius of the circle of which the centre is the eye of the observer,
and on the circumference of which is the diameter of the sun ; this diameter
may be regarded as being nearly equal in magnitude to the arc it cuts off.
Let the two radii be r^ , r.^ , and the arc or diameter of the sun be d ;
-^^x (a radian) = 32' 36", /. - x (a radian) = 32' 36",
raQius T-t
dx (a radian) ^. ... dx (a radian)
•■• '■i = " 32' 36" • ^'""'^^'y '■^ =— 3r32^ '
••• '•» ■• '•^=32W = SlW-'ii = i^2 = ^«92 : 1956=473 : 489.
22. From Art. 71 it is seen that - is the measure of the angle where the
unit is a radian; for this angle to be represented by ~ , k must depend
solely on the unit.
(i) In order that — ^— may be the measure when the unit is a radian,
since the angle - x (a radian), /. /c = 1.
.... a , ,. , a 180° a 180 , , ,
(ii) - X (a radian) = - x = - x x (a degree).
In order that — ^ may be the measure of this angle when a degree is unit,
K,a a 180 . 180
= -x — , I.e. K= — -.
r r IT IT
24 ANGULAR MEASURE. XI.
EXAMPLES. XI. Pages 40—43.
1. Let the angle contain x right angles.
In X right angles there are 90^ degrees,
., ,, ,, ,, ,, 100:r grades ;
.-. 90a; + 100a; = 38, /. 190a;=38;
.-. X — -, The angle is - of a right angle, i-e. ^ x - = r^r .
2. Let the angles contain a; and i/ right angles respectively, then their
difference is 100 {x-y) grades; their sum is 90 {x-\-y) degrees;
100 (a; -2/) = 20, 90 (a; + i^) = 48 ;
i.e. a;-2/ = -; aj4-i/ = — . By addition 2a; = -— ;
o lo 15
.♦. a; = ~ of a right angle = — x 90° = 33°.
Now a; + 2/=Y^ ;
15
**• ^'^IS'^'^IS ~ SO'^e^^^^^^^*^"^^^' •*• 2/ = gX90° = 15°;
.*. the angles are 33° and 15°.
3. Let the 1st angle contain 2a; right angles, i.e. 180a; degrees,
„ ,, 2nd ,, „ x ,, ,, i.e. 100a; grades;
.-. 180a; + 100a; = 140;
.-. a; = ^ of a right angle = 45°, .*. 2a; = 2 x 45° = 90° ;
.*. the angles are 90° and 45°.
4. Let the angles contain 4a; and 5a; right angles respectively,
i. e. 400a; grades and 450a; degrees respectively ;
.-. 450a;- 400a; -2i; /. 50a; = 2J;
* * ^^20 ^^ ^ ^^^^* angle,
4a; = ^ of a right angle = 18°, hx = - of a right angle = 22^° ;
.-. the angles are 18° and 22^°.
5. Let the angles contain x and y right angles respectively,
i.e. — and -~ radians respectively, and 90a; and 902/ degrees respectively;
Tra; wy _Tr
T~T~9''
.-. 9(a;-2/) = 2, .-. 90 (a;-2/) = 20. But 90 (a; + y) = 56 ;
.-. by addition 180a; = 76, a; = — - of a right angle = :p_^ x 90° = 38°;
180 180
18
90x + 901/ = 56, 902/ = 56 - 38 = 18, y = -^oi^ right angle = 18° ;
.% the angles are 38° and 18°,
ANGULAR MEASURE. XI. 25
6. Let the angles contain ic - 1/, XyX + y right angles respectively; they
are in A. P.
Their sum is 3a; right angles ; but since they are the angles of a triangle
their sum is 2 right angles ;
2
/. 3ic = 2, X = - of a right angle = 60°.
o
7. Ijet the angles contain x-y, x, x + y right angles respectively ; they
are in A. P.
From 6 it is seen that x—QO°.
The number of grades in [x - y) right angles is 100 (x - y) ;
/. 100{x-y) : 60 :: 5 : 6;
600 {x-y) = 300, x-y = \ of a right angle ;
.-. a:-i/ = 45°, 60°-2/ = 45°; /. 2/ = 15°;
/. a; + 2/ = 60° + 15° = 75°;
.-. the angles are 75°, 60°, 45° respectively.
8. Let the angles contain x-y, x, x + y right angles respectively,
i.e. 90 (a; -2/) degrees, 90a; degrees, 100 (x + y) grades, respectively;
100(x + y) : 90x + 90(x-y) :: 10 : 11,
i.e. 1100 (x + i/)^ 1800a; -900^/;
7
•*• 7x = 20y, .-. 2/ = 2o^'
But from Example 7, a; = 60°;
.-. y=J-x60° = 21°, a;-2/ = 60°-21° = 39°, a; + 2/ == 60° + 21° = 81° ;
.'. the three angles are 39°, 60°, 81° respectively.
9. Let the angles contain x-y, x, x + y right angles respectively,
i.e. 100 {x-y) grades, 90a; degrees, and {x + y) ~ radians, respectively;
100{x-y) : (x + y)'^ :: 200 : 37r;
••• 3(a;-2/) = a; + 2/, x = 2y, y = 2'
2
But from Example 7, a; = - of a right angle;
.: y=(-x ^ = -A oi& right angle.
a;-y=(o-M = 3)ofa right angle = - x 100k=33-3s,
2 2
a; = - of a right angle = - x 100^ = 66-6k,
o o
a: + 2/=(o + o) of a right angle = 100« ;
/. the angles are 33-3K, 66*68, 100^, respectively.
26 ANGULAR MEASURE. XI.
n ^ »» ,T 9>=6 27 50,.
11- 9(b^=ioo^aoo= ■•^^^=iolao™=5o'"' ■■•27^='"=
.-. g3/-^M=m, 2M-~M=m, 2M-m=~M.
12. PromArt.70(El.Trig.),^ = 4 = ^.
_, ,. .. ace a- c e
By proportion if 5 = ^ =/ ' ''• 6^-^ =/ '
• 200-180 TT
13. Let a given angle contain M English minutes, and m French minutes ;
M . m ^^9x6 54 ,, ^,
i¥=,-^^— ,-^m = .-_-m, iif= -54^;
••90x60 100x100' 10x10 100
that is, by multiplying French minutes by '54 we have English minutes.
14. Let one part contain x right angles and the other part y right angles;
.*. both parts contain 90 (x + y) degrees ;
90 (x f y) = 383!,^°, .-. 900 (a; + 2/) = 33 1 .
In X right angles there are (a; x 90 x 60 x 60) English seconds,
In 2/ „ „ „ (y X 100 X 100 X 100) French seconds;
a; x 90 X 60 x 60=i/ x 100 x 100 x 100,
81
that is, 81a;= 250?/ ; .'. y = ^^ x.
Substitute this value of y in the equation 900 (x-{-y) = SSl\
5 5
/. a; = — of a right angle = — x 90° = 25°,
18 18
.-. the two parts are 25° and 8° 6'.
15. 12« 50' = -125 of a right angle.
60 ) 2 7 minutes
90" p-45 degrees _405_105_21
•105 of a right angle ~ 025 ~ 125 ~ 25 '
16. Let a be the number of radians ;
TT 180x60' 180x60 10800*
ANGULAR MEASURE. XL 27
17. It appears from Euclid I. 32, that all the interior angles of any
rectilineal figure, together with four right angles are equal to twice as many
right angles as the figure has sides.
Hence if n be the number of sides of any rectilineal figure, we have the
sum of its n angles + 90° x 4 = ^0° x2n; or the sum of its n angles
= (2n - 4) X 90°= {n - 2) x 180°.
If the figure be a regular polygon, its n angles are all equal and con-
sequently each of them is x 180°.
n
(i) In the regular hexagon n=6;
:. each angles — X 180° = 120°= ~X7r = ?^
b 6 3
6-2 400«
= -^x200«=--- = 133^33^ 33 •3\
D 6
(ii) In the regular octagon w=8 ;
.-. each angle = ^-^ x 180° = 135° = ^~ x tt = ^ = ?^ x 200^ = 150«.
o 8 4 8
(iii) In the regular quindecagon w = 15 ;
I, 1 1^-2 .ono ir^o 15-2 137r 15-2 ^^^ 520^
.-. each angle = -— —-X 180° = 156°=-— —X7r = -T^ = — r-^-x 2008 = -^r-
lo 15 lo 15 o
= 173-3K.
18. In the regular decagon n= 10;
10 — 2
.*. each angle = x 180^=144°.
In the regular pentagon w=5;
5 — 2
.-. each angle = — — - x 200° = 120s.
^ ,. 144 6
^"*^^ = i20=5-
19. From (18) the number of grades in an angle of the regular pentagon
is 120; from (17, i) the number of degrees in an angle of the regular hexagon
is 120 ; .-. they are equal.
20. Each angle of a regular polygon of 48 sides
= (—^ X 180 X 60 j minutes (English) = 10350 English minutes;
2 right angles = 108 x 60 English minutes = 10800.
Difference =10800 -10350 = 450 English minutes.
21. From (17, iii) the angle of a regular quindecagon is 156°. But the
exterior angle of the quindecagon together with the adjacent angle are
together equal to 180°; .-. the exterior angle=180°- 156°=24°.
Let X be the required measure ;
.-. (i) 90° = a:x24°, .-. ^ = || = 3|;
r'\ 180° o.o 180 15
(ii) =a;x24°, .-. x = ^--= — .
IT 2iir 27r
28 ANGULAR MEASURE. XI.
22. Let X be the measure of 1 degree, y the unit in degrees ;
.'.l=x.y, :.^=^^y; .••- + 10 = ^' •••^ = 19-
Butr = a;.2/, /. l° = g.2/, ••2/=io'
19°
.*. the required unit is — r.
23. Let a; be the measure of 1 degree, 2/ the unit in degrees;
Butl° = ;E.^°, .-. l°=l.y°, .-. 2,° = 90°,
.'. the required unit is 90°.
24. Let l° = a; . y°, /. c°=ca;. y° and ^'^^^ • 2/°,
, ^ 96x 10a lOac ^ ^
but ^.^ =a, .-. x= -^rz-. :. ca:=-^;^—= measure of c°.
iU \jb 9o
25. Let \° = x. y°, .: b° = bx. y°, as = ^-"| . ^°,
9aic - 10c
10 ' " 9a + 106*
^ .10 o o 1° 9a + 106
xu • ^ •*• 9a + 106_
.*. the required unit is — r^r — degrees.
26. Let the angle contain x right angles, i.e. 100 x grades or 90x degrees,
18 9
100a; -90x = — , .-. x = -^— of a right angle.
TT OTT
9
Let y be the measure of a right angle when — - of a right angle is the unit.
OTT
Q ^_
1 right angle = y x p— of a right angle, .-. 2/ = -q .
/. the required measure is — - .
27. Let X be the number. The three angles are a;°, a;8, x° + a;« respec-
tively, I.e. x , -r-, -——respectively.
9a;° 19a;°
Since they are the angles of a triangle, ^° + ^177 + ~T?r~ = l^^*
1800° 1800
x^ radians =.§-;
19a;° bir 19 tt
10 ~ 19 ^ 10 ~ 2 •
•• '^~ 38 ~ 38
9x° Stt 9 97r
10 ~ 19 ^ 10 ~ 38 *
,, , StT 97r TT
the angles are jg, -, ^^
ANGULAR MEASURE. XI. 29
28. Let AjBjG he the angles of the triangle, and u the measure ; so that
9m 9m
A=u degrees = uxl degree, B = u grades = ^ degrees = — x 1 degree,
180w ^ 180u ^ ^
C=u radians = degrees = x 1 degree.
TT TT
Since they are the angles of a triangle they are equal to two right angles
9w 180m ,^^ 18007r
180°= 180x1 degree; .-. M+-r + = 180, .\ u=-
10 TT ' •• 197r + 1800
29. Let the angles of the triangle be x degrees, 10a: grades and lOOac
radians respectively; i.e. x degrees,
10a; X 9 ^ ^ , 100a; x 180 . , . ,
— ;,- — or 9a; degrees, and degrees respectively.
10 TT
Of these x degrees is the smallest ; and the sum of the three is equal to
two right angles ;
^ 18000a; ,^^ o 1807r ,
., ^ + 9a. + -^ =180; .'. a;- j^^^^^^^ degrees
~Vl07r + 18000^180y
X ^J radians = j^^^-^^^g^ radians.
30. Let n be the number of tides of the polygon, and .-. the number of
terms of the a. p. is n ;
the sum of the series is - {2 x 120°+ (w - 1) 5}.
But from (17) the sum of the angles of the polygon
= (w-2)xl80°, .-. ^{2xl20 + (n-l)6}=(7i-2)xl80°.
From this quadratic n = 16 or 9.
Both these values are admissible upon trial.
EXAMPLES XII. Pages 60—52.
1. (i) Angle of reference being ^J5D:
DA is the perpendicular j for it is opposite to ABD and is perpendicular
to BD.
BD is the base^ for it is adjacent to the angle ABD and to the right angle,
(ii) Angle of reference being BAD :
DB is the perpendicular ^ for it is opposite to BAD and is perpendicular
to^D.
AD is the hase^ for it is adjacent to the angle BAD and to the right angle,
(iii) Angle of reference being A CD :
DA is the perpendicular, for it is opposite to ACD and is perpendicular
to CD.
CD is the base, for it is adjacent to the angle ACD and to the right angle,
(iv) Angle of reference bein^ DAG:
DC is the perpendicular, for it is opposite to DAC and is perpendicular
to AD,
AD is the base, for it is adjacent to the angle DAG and to the right angle.
30 TRIGONOMETRICAL RATIOS. XII.
o ;;\ M-.,KJ7> perpendicular DB base CD
2. (1) sm BAD = -v-—— ; = TB ' (") cos A CD = , ■ = ,
hypotenuse AB ^ ^ hypotenuse CA'
n;;\ +or,nji^ perpendicular DC r\ ■ a nr^ perpendicular DA
(m) tsinDAC= v — — — ^ft-Tj (iv) sin^/?i) = i--— ^^ = — .
base DA' ' ^ hypotenuse BA*
(V) tan^^Z)=:?"^Pf^^?^=?^, (vi) sinD^C=P?^:P^Hdi£^r^5^
base D^ ' ^ ^ hypotenuse AC'
/.r\i\ ^^c T\n 4 ^^^^ ^^ I •••X 4. r^n M perpcndicular DC
(vn) cos DCA = t- r = ^-- , (vni) tan DCA= ^—^ = — ,
hypotenuse CA ^ ' base DA
(ix) cos^5D = ^^^^-^-=^^, (X) sin^CD=P^^?^^i^^^ = ^.
hypotenuse BA ^ hypotenuse CA
3. It is seen from Euc. VI. 8 that the triangles ABC, BDC and BDA are
similar to one another.
(i) In the right-angled triangle ABC, sin ACbJ^-^^^^^^^^^^=:?A
hypotenuse CA '
In the right-angled triangle BDC, sin J C^ = ^f ^"^^^^^^^^ = gg .
hypotenuse CB '
1 BA . DB
:. two values are — — and — - .
CA CB
(ii) In the right-angled triangle ABC, cos ^0^ = .-- J^ = ^
hypotenuse CA
In the right-angled triangle BDC, cos ABC=^, ^^'5? = — •
hypotenuse CB '
. 1 CB ^ CD
.'. two values are -— and — .
CA CB
(iii) In the right-angled triangle ABC, t^n AC B = ?^^''^^^'^ = ^
base CB '
In the right-angled triangle BDC, t^n A CB = ^-^^'^^^^ = ^1 .
base CD '
. , BA ^ DB
.'. two values are — — and -— - .
Cx5 CD
(iv) In the right-angled triangle ABC, sin BAC=^~^^^^^^!^ = ^.
hypotenuse 'A '
In the right-angled triangle BDA, sin BAC =^'^^^^^^' = ^ ■
hypotenuse AB '
1 DB ^ BC
.'. two values are — - and — — - .
AB AC
(v) In the right-angled triangle ABC, cos BAC=^~^^ = — .
hypotenuse AC
In the right-angled triangle BDA, cos BAC=, — ^^ = '^^- ;
hypotenuse AB
.-. two values are —r- and —7^ .
AB AC
TRIGONOMETRICAL RATIOS. XII. 31
BC
(vi) In the right-angled triangle ABC, tan BAG= — ,
In the right-angled triangle BBA, i2LnBAC = —j.\
DB .BC
.'. two values are —^ and -— ^ .
AD AB
DA
4. (i) sin BDA = ^-^ for DBA is a right angle.
BA
(ii) sin BE A = -^ for ABE is a right angle
AG
= -— - for GAE is a right angle.
EC
(iii) sin GBD = ^ . (iv) cos BAE = 4^ .
x)C AE
(V) COS B^B =^ = ^ • (vi) cos CBD =|g.
(vu) tan BOZ) = ^-^ = ^ = — . (vni) tan i)B^ = ~ .
(ix) tanB^4 = g = g. (X) tanCBi)=|g .
(xi) sinD^B = 5| = J§ (xii) sinB4£ = ||.
5. In the triangle ABC CB is perpendicular ^ AG is base and AB hypo-
tenuse when A is the angle of reference ;
, perpendicular CB 3 ft. 3
.*. sm^ — ^ ^ — — -
.*. cos^ =
hypotenuse AB 5 ft. 5'
base AC 4 ft. 4
hypotenuse AB 5 ft. 5
perpendicular _ 05 _ 3 ft. _ 3
.;. tan A - ^-^ " Ic " Ift". " 4 '
When B is the angle of reference, CA is perpendicular^ BC is base and
BA is hypotenuse;
perpendicular _ CA _ 4 ft. _ 4
.*. sin/i=:
.'. 0O8J5 =
.-. tan 5 =
hypotenuse BA 5 ft. 5*
base BC 3 ft. 3
hypotenuse BA 5 ft. 5
perpendicular _ C'^ _ 4 ft. _ 4
base ~ BG~ Bit. " 3 '
32 TRIGONOMETRICAL RATIOS. XII.
/^ -o ,^, . , 05 a ^ AC b ^ ^ CB a
6. From (5) sm A = -— i =: - , cos A= -,-=-, tan A = — -^ = - .
^ ' AB c AB c AC b
From Euc. I. 47 AC^+ CB^ = AB^,
AC^ CB^ fACy /(75Y_
•'' ^B2"^^J52--^' •'• \ab) '^[abJ "•^'
•• iff'^iiy^^' *'• sinM + cos2J=l.
(i) Since - = sin^, /. a = cQmA.
(ii) „ r=—, = ~ = 8UiB. .\b = csinB,
^ BA c
-nri „
(iii) „ — — =-=cosJ5, :. a — cco^B.
r \ AC b , ^ .
(IV) „ — ^ = -=COSii, /. 6=:CCOS^.
' AB c
(v) sin ^ = — 5 = COS B, (vi) cos A = -— = sin B.
AB AB
OB
(vii) tan^ = — - =cot jB.
A C
7. Let ABC be the right-angled triangle; C the right angle.
. , BC 5 , AC 12 ^ ^ BC 5
. ^ AC U ^ BC 5 ^ ^ AC 12
''''^=AB = i3' ''''^=AB = rr *^"^=^=5~-
8. Let ABC be the right-angled triangle; C the right angle.
. ^ BC 1 . AC J3 ^ ^ BC 1
. „ ^C ^3 ^ BG 1 , ^ ^O ,„
9. In (7) sinM + cos^^ = (A)%(gy = ^H.
In (8) sinM + cosM = Qy + (^y = l + J = l.
144
169" ■
25 ,
+ i69='-
RATIOS OF ONE ANGLE. 33
EXAMPLES. XIII. Page 60.
1. 2.sinD.cosD = 2. -— . -y^ = l = sin90° = sin ^.
2. 2 . sin C. cos = 2 . ^ . "^^ = -^ = 8in 60° = sin B.
3. cos^i? - sin^i? =^- -\=-\^ 1 - 2sin25=:l -?= - \ ;
4 4 2 2 2
/. cos2jR - sin^^ = 1-2 sin^B.
4. sin jB cos C + sin C cos 5= ^ . ^ + ^ . ^ = l = sin90°=sin ^.
5. cos2D-sin2D = /^-iy- /"-l y=0=cos90° = cos^.
7. sin-^B + cos»B = (f )' + (y'= ? + J = l.
10. sinB.cosC-sin(7cos£ = ^. ^ --. - = - = sin30 = sin (7.
11. 2(cos2^.cosD + sin5..sinZ))2 = 2. Q . A + ^Z? . ^ V
12. 2 (sin D . cos C- sin G . cosD)2 = 2 ('A . ^^ - ^ . A V
13. sin 30°= i = -5.
14. sin46o=^ = >Z?^iil|?lM = .7071068.
15. sin60o=f = lj!~:i-=.866025...
L. T. K. S
34 RATIOS OF ONE ANGLE. XIIL
16. tan 60° =V3= 1-7320508...
17. tan30o = -L=f = l:Z5?^- = .57735...
18. Bin .s^^Jtll^^-^^^^O- - 1 ^ ljgggggg:.^.3090170...
EXAMPLES. XIV. Pages 63—65.
1. Take fig. E.T. p. 54. Then, 0M= 179 feet, POJtf =45°, PM=a;; then
^=.jfg = ta„450;...^g = l;... . = 179 feet.
2.
.-. a; = 200^3 = 200x1-732 = 346 ft.
3. Take fig. E. T. p. 55.
Let be top vertical cliff ; OFM angle of depression of point P, PM= 150
feet; OM=a;feet; then
JJ=tan30o = -^-., .■.^, = },; .•..=^°=50V3 = 60xl.732 = 86-6ft.
4. Take fig. E. T. p. 55. Let be the top of tower, OPiV/= anj^le of
depression, OM height of tower above top of house=(117 -37) ft. = 80ft.,
PiH=j; ft.
PM r
^=cot30° = V3; .-. ^ = v/3, .*. a; = 80 ^3 = 80 x 1-732 = 138-5 ft.
5. Let ED be the lamp post so that DBG is horizontal and EAC is a
straight line, then j-^ = tan DCE = r=— , .
DC B(j
DE BA
ED
" DCBC " 4| + 19 19'
- ^D = T^(4i+19)-Ax^=V=7ift.
PRACTICAL EXAMPLES. XIV. 35
6. With the diagram of (5) if AB represent height of lamp post and BG
length of its shadow, C angle of elevation of the sun,
tan = 1^ = g^- = ^3 = tan 60° ; .-. C = 60°.
Let h ft. represent height of tower ;
/. -^ = tan60°=V3; /. /i= 100 x ^3 = 100 x 1-732 = 173-2 feet.
7. The breadth of the river is represented by PQ = x yds.;
.-. ^ = tanPi^g=tan32°17', .^ = -6317667, a; = 63-17yds.
QK luU
8. Take fig. E. T. p. 62, Ex. 2.
Let PQ be height of flagstaff, i.e. 25 ft.,
MQ house, i.e. icft.
Angle POM angle of elevation of the top of flagstaff, i. e. 60°,
QOM bottom i.e. 45°.
Let OM=y feet.
Then ^ = tan 60°; .-. ^±^^ = tan60°, ^ = tan 45°; .'. - = tan45°.
OM y OM y
By division —^-=^3; .-.0;=;^—^ = 34-15;
.*. height of house is 34-15 feet.
9. In the diagram referred to in (8) let be the point of the cliff ; P
and Q the two ships. OQlf=45°, OPiIf=30°, Oi¥= 100 ft., MQ = xieQt,
PQ = y feet;
.-. ^=tanOQM; .'. — -=tan45° = l; .-. a;=100:
MQ X ^
.-. ^ = tanOP3f; /. — = tan 30° = -^ ; .'. a: + ?/ = 100 . /3.
MP x + y ^6 ^ ^
Since x-\-y = 100^3;
:, 2/ = 100V3-a;--=100V3-100 = 100(V3-l) = 100x-732 = 73-2;
.-. the required distance is 73-2 feet.
10. In the diagram referred to in (8) let PQ be the tower 100 ft. high,
POiH=75°, (3OM=60°, OM=yn., QM=xU.;
.•.^ = tan60°; .-. -=^3, -^-j, = tan 75° ; .-. --- = 2+^3.
^ ,. . . a; + 100 2 + jS , 100 , 2 100 2
By division = — ^; .-. 1 + — =1+-—; .-. — =-7w;
.-. a; = 50^3 = 50x1-732 = 86-6;
.-. the height of the cliff is 86-6 feet,
3—2
36
PRACTICAL EXAMPLES. XIV.
11. In the diagram referred to in (8) let be the position of the house,
PM the direction of the road, P and Q the two consecutive milestones;
OPiV/= 30°, angle first observed; OQlf = 60° angle next observed; OM
( = 1/ miles) the required distance of house from the road. MQ — x miles,
MP=x+l miles.
^= tan OPM;
MP '
OM
By division
MQ
x + 1
= ta.nOQM;
y
x+l
y
= tan 30° = -
/8'
tan 60° = ^3.
= 3;
1+1 = 3, .-. -=2, /. X=:\.
XX 2
Since |=\/3; .*. y = x>jS = ^
V3 _ 1-732
= -866 of a mile
= -866 X 1760 = 1524 yards ;
.*. the required distance is 1524 yards.
12. The angle ^C makes
with AB is CAB ; and BC
makes with BA the angle
CBA . (The student should ^ - ' "
take notice of the direc- ^^^
tions AB and BA.) Draw ^^^
CD perpendicular to AB, ^^^
AB = A00 yards, BD = x ^--"
yards ;AD = (400 - x) yards, ^^"^30^^
CD = y yards (the breadth a 400 yds.
of the river),
^^P
60/ \
CD
BD
CD
AD
= tanOPD; /. ^ = tan60° = v/3 (i),
X
= ta.n CAD:
IO^^ = *^«^'^°=VB <">•
Divide (i) by (ii), .*.
400-
= 3, 150^4, ... .^100.
From (i) ?/ = a; ^3 = 100 x 1-732 = 173-2;
.-. the required breadth is 173-2 yards.
13. Draw the figure as indicated in the problem; then, since angle
^JB^ = 45°andangleP^^^ = 90°; .-. angle .4PJ5; = 45° and .-. AE = AB.
D
a yds. £
PRACTICAL EXAMPLES. XIV. 37
Because AB = AD (sides of the square), /. angle ^^Z> = angle ADB=45°,
But angle BE A =45°; .\ EA = AD. Let a; = the side of the square, then
ED = 2x and BD = x^2,
= — = tan-i v^2 = ^2; /. 2x = aj2; .: XsJ2 = a = BD.
14. In fig. E. T. p. 62, Ex. 2, let be the top of hill, MOQ and MOP
angles of depression of Q and P respectively the top and bottom of flagstaff,
PQ = 2o feet = height of flagstaff, QM — x ft. = height of hill above flagstaft,
height of hill = (25 + x) ft. , OJl/=?/ = distance of foot of hill from flagstaff.
^^= tan 3/0(3; .'. - = tanilfO(3 = tan45°13',
PM ic 4- 2 ^
— r.=tanilfOP; .*. -^^^—= tan MOP = tan 47° 12'.
OM y
^ .. . . a; + 25 tan 47° 12' . , 25 1-0799018 , -07231
By division — - =-,^:^^^^^,. i.e. 1 + - = ^-^^^^^^ = 1 + ^-^^^^^^^^^^
X 1-0075918 100759 2518975 ^^^
•• 25= "-O^Z^ST = "mr^ •• ^ = -7231- = ^^^ ^"^'^^'
.-. a; + 25r=348 + 25 = 873;
.-. the required height of hill is 373 feet.
15. Let A be the first station and B the second so that AB = 1 mile.
Let G be the place vertically under the balloon K; then since CA has a N.W.
bearing, i.e. CA. is inclined westward 45° to the line AB, .*. angle O^P = 45°;
similarly OP^ =45°, .-. CA = CB, and angle PO^ = 90°,
AC 1/2 j4C
^ = sin 45° = -— = ^-=-7071; .-. "^ = -7071, .-. ^C=J5(7--7071 mileij;
also, since the altitude of the balloon K at A is 45°, KC = AC;
.'. the height of the balloon is -7071 miles = -7071 x 5280 = 3733 ft.
16. Let AB be the height of the balloon, C the station due south; then
angle ACB = (jO°; draw ClJ perpendicular to PC and west of it, make CD
represent a mile, join BD and AD, then angle ADB = 4:5°.
Let AB the height of balloon = x miles,
.-. PO = ^Pcot^O/? = a;cot60°=^.
Because angle ADB = 4:6° and angle ^jBD = 90°;
.-. angle BAD = 45°, and BD = AB = x miles.
From Eucl. i. (47) BD^=BC'^ + CD^', .-. x''=(-^\\l;
.-. 2a:2 = 3; .'. a; = J^6 miles = J x 2-4495 x 5280 ft. =6468 ft.;
.-. the required height is 6468 feet.
17. Since the altitude of the sun is 30°, the length of the shadow ot
the height of the triangle will be to the height of the triangle as cot 30° : 1 :
.-. length of shadow is bjS. The base of the triangular shadow coincides
38 PRACTICAL EXAMPLES. XIV.
with the base of the triangle (2a); .*. the shadow is an isosceles triangle
with base 2a and height 6/^3. The height bisects the base and the vertical
angle;
/. tangent of half the vertical angle = - — — = "^ ,
18. Let AB represent the stick and BD the length of its shadow ;
Let GB be the position of AB when the length of its shadow is again equal
to BD, The shadow of GE^ perpendicular on BD, will in that position then
be equal to ED. And the angle GDE will be equal to the angle ADE. The
two triangles will therefore be similar (Eucl. vi. 2).
:. ~— = -r-- ; .'. A G and GD are in one and the same straight line.
ED BD
Because angle ADB^SO°; .\ angle BAD = 60", and since AB = BG,
.', angle AGB = QfO^, the remaining angle ABG is also = 60°.
Now (75E = 90°-^J5C = C0°-G0° = 30°.
.'. the angle of inclination of stick to horizon is 30°.
19. In fig. E. T. p. 79, let P be the position of the person, then NP is
radius of circle described by the person in consequence of the earth's rota-
tion. LetPOJ^ = 60°. Then A^P= 01/= OP cos 60°= OP cos 60° = ^V*-niiles;
.*. the diameter of the circle described by the person is 4000 miles.
Circumference = 4000 x 3-1416 = 12566-4 mis.
.'. the distance passed in Ihr. =^ of 12566-4 mis. ='523-6 miles.
EXAMPLES. XV. Page 7L
1. cos^ . tan^ . =cos^ . T=sin^.
cos^
... . cos ^ sin ^ ^
2, cot A . tan A = -.— ^ . — -— = 1.
sm A cos A
3, cos^ . .
. cos^ = sm^.-^ — ^ = sm^,cot^.
sm^
A . X . 1 COS^ 1 .
4, seCu4.cot^= , ,—. — - =^ - — ^=cosec^.
cos^ sm^ sm^
EASY IDENTITIES. XV. 39
, ^ J 1 sinA 1 .
5. cosec A . tan A = — — - . -. = -. = sec ^.
sin^ cos^ cos^
^ ,. . . .. . . J /sin A cos^\ .
6. (tan A + GoiA)smA . cos A = + --. — . sin A . cos A
" ^ ' \cos^ smAJ
sin2^ + cos2yi . • 2 I , 2^1
= : — -- . sinyi . cos yl=sm^^ + cos^4 = l.
cos^ . sin^
r. /. . . .V . . . fsinA cos^N . , .
7. (tan A - cot A) sin A . cos A = -, -. — r sin ^ . cos A
^ ' VcosJ' sm^/
(sin^ A - cos^ A ) sin A . cos A . „ . „ ,
= > , — — -^ = sin2 A - cos2 A .
sm A . cos A
8. cos2^-sin2^ = l-sinM-sinM = l-2sin2^.
9. (sin^ + cos^)2 = sin2^+cos2^ + 2sin^ . cos ^ = 1 + 2 sin ^ cos ^.
10. (sin A - cos ^y-* = sin2 A + cos^ ^ - 2 sin ^ . cos ^ = 1 - 2 sin ^ . cos A,
11. cos^ B - sin'* B = (cos^ B + sin2 B) (cos2 B - sin^ B) = cos^ B - sin^ B
= cos^B- (1 - cos2 5) = 2 cos2 jB - 1.
12. (sin2B + cos2 5)2 =12=1.
13. (sin2 B - cos2 5)2 = (1 - cos2 B - aos- Bf = (1 - 2 cos2 Bf
= l-4cos2J5 + 4cos4 5.
14. 1 - tan* J5 = (l + tan2 5) (1 - tan2 5) = sec2 J5 {1 - (sec2 5 - 1)}
= sec2 5 (2 - sec2 B) = 2 sec2 B - sec* J5.
15. (sec B - tan B) (sec B + tan B) = sec2 B - tan2 B
= l + tan25-tan25 = l.
16. (cosec d - cot d) (cosec ^ + cot 6) — cosec^ ^ - cot2 $
=:l + COt2(9_COt2^=l.
17. sin3 d + cos3 (9 = (sin 6 + cos 6) (sin2 ^ + cos2 ^ - sin ^ cos 6)
= (sin + cos ^) (1 - sin cos ^) .
18. cos3 - sin3 ^ = (cos ~ sin 0) (cos2 ^ + sin2 ^ + cos ^ sin 0)
= (cos - sin ^) (1 + sin cos ^).
19. sin^ + cos^ = (sin2 ^ + cos*- 0) (sin* ^ + cos* - sin2 ^ cos2 0)
= sin* ^ + cos* - sin2 cos2 ^
= sin* + cos* 0-\-2 sin2 ^ cos2 ^ - 3 sin2 cos2 ^
= (sin2 + cos2 ^)2 - 3 sin2 cos^ ^ = 1 - 3 sin2 cos3 ^.
20. sin« - cos^ = (sin2 ^ - cos2 0) (sin* ^ + sin2 cos2 ^ + cos* 0)
= (2 sin2 ^ - 1) {sin* + sin2 0(1- sJn2 (9) + (1 - sin2 ^)2}
= (2 sin2 ^ - 1) (1 - sin2 + sin* 0).
40 EASY IDENTITIES. XV.
tan ^ + tan B tan ^+ tan 5
21- cowT^^=-l — l^-^'-io^.]
tani
= tan A . tan B,
tan A tan B
tan A + tan B
tan A -r tan B
tan A . tan B
1 i o 1 + tan a . tan 8
+ tan B ^
__ cota + tanjS tana ^ tana tan/3 . .
22. I r o = -. - -. X X o = I — ^^cotatanS.
tan a + cot p , 1 1 + tan a . tan p tan a '^
tan a -f- r — -— —
tan p tan /3
23.
24.
1-sin^ _ ( l-sin^)(l-sin.-^) _ (l-sin^^ _ (1 - sin A)^
1 + sin ^ " (I + sin ^) (1 - sin J^) ~ 1 -sin^J^ ~ '~ cos^A
\ cos^ J \coaA cos A/ ^ '
1 + co s J. _ (1 + co s^) (1 + cos^) _ ( 1 + cos J)^ _ (l + cos^)2
1 - cos ^ "~ (1 - cos ^) (1 + cos A)'~ 1 - cosM ~ smM
/1 + C0S^\2 / 1 C0S^\2 , , ,^
= V^nT-j ^Vii^ + si^j =(cosec^+cot^)
25. 2 versin B - versin^ ^ = 2 (1 - cos ^) - (1 - cos 6^
= 2 (1 - cos <9) - (1 - 2 cos ^ + cos2^)
= l-cos2^ = sin2(?.
26. versin ^ (1 + cos <9) = (1 - cos B) (1 + cos ^) = 1 - cos^ ^ = sin^ B.
EXAMPLES. XVI. Page 74.
1. In fig. E. T. p. 72 let the measure of OP be 1, and let c be the
measure of Oilf , .*. c = cos A . Let x be the measure of MB ;
sin^=^J = ^^= VT^^^^, tan^4^.^^E£!.Vi^^_,
OP 1 ^ 01/ c cos^
, , OM c coaA , OP 1 1
cot A = --= = ; = z=r , sec ^ = -— - = - = ,
^^ Vl-c2 Jl-coB^A OM c COS A'
, OP 1 1
cosec A = -— - =
MP Ji-c^ J1-co8^a'
2. In fig. E. T. p. 73 let POikf be the given angle, t the measure of
OM and 1 the measure of PM;
.'. cot A = ~- = - or t = cot A,
RATIOS OF ONE ANGLE. XVI. 41
Let X be the measure of OP^
.-. x^- = l + t^ .'. x = Jl + t\
. , MP 1 1 , OM t cot A
Sm A = T-rr = — T-JT^zizr: = — , - , COS A = -
OP Jl^'t^ Vl + COt^^' OP ^1 + ^2 7l+cot2^'
MP_i__i^ OP _ViT?_/^/iTcoF^
^^''^-OM-t ~cot^' ^^''^~0i/~ ~T~~ cot^ '
cosec A = ^rr=, = ^^^ — = x/l + cot^ i4.
MP 1 ^
3. In the figure E. T. p. 73 let c be the measure of OP and let 1 be the
OP
measure of Oilf, /. sec^ = — r^^c.
It can be shewn that the measure of MP is ^c^ - 1,
. , MP Jc^^l JseG'A-1 ^ OM 1 1
sm A = --- = ^ = ^ .— , cos ^ = 7r^ = - = J ,
OP c sec A OP c sec A
^ , MP sl'^^^ , — 3-r— r . , OM
OM 1 '^ ' MP ^-^ZTi J^ec^-Zl'
^ OP OP c sec^
cosec A — TT^- = - - =
OM MP Jc^_i Jsec^A^'
4. In the figure E. T. p. 73 let OP have c for measure, let 1 be the
measure of MP, /. c — cosec A and Jc^ - 1 is the measure of OM,
. , MP 1 1 . ^^ Jc2 - 1 Jcosec22"^l
sm A = TT^ = - = 7 , cos A = ;-^ = > = y— .
OP c cosec ^ OP c cosec ^
MP 1 1
tan A = 77=-, =
OM Jc'^ _ 1 ^cosec2 ^ - 1 '
cot A = ~vv, = ^-^i = V cosec- A-1,
MP 1
OP c _ cosec ^
~OM"";^/p31~^cosec2^-l*
5. cos2 A + sin2 ^ ^ 1 (Art. 107) ;
.*. cos2 ^ - 1 - sin^ A ; .*. cos A ~ J'l - sin^^ ,
^ sin A sin A
001^="?^ = ^^^ (Art. 108),
sm^ sm^
sec ^ = =z . _^^z:rr=r ; coscc A ^ -. — 7 (Art. 102).
cos^ ^yi-sin^^ smJ
42 KATIOS OF ONE ANGLE. XVI.
1 1
6. sin A -
cosec^ ;^yi + cot2^
1 tan^ ..
(Arts. 105, 111.),
y('-iii) ^'*
tanM
cos^=-^, = . ^ - (Art. 105, ii.), cot^ = --^ (Art. 102),
sec^ ^l + tan2^ tan^ '
.. 1 Jl + tan2^
sec ^ = Wl + tan^ A (Art. 105, n.), cosec A = - — - = -^^— r- — -, *
^ ^ ' /» smA tan A
EXAMPLES. XVII. Page 75.
1. Draw a figure similar to that of E. T. p. 74. Let P03I be the give
angle A, take P so that the measure of OP is 5, then since
sin^=f, .-. MP = S,
Let X be the measure of OM, .*. a;2 = 5^-3^ = 16, .'. a; = 4,
. , MP 3 ^ OP 5
*^^^ = 0M = 4- ^"^^^^=iifP=3-
2. Fig. as in (1). Let 0P=3 and 01/= 1 for cos ^ fi- ^- ^ 1 = 3 5
.-. Jfp-^ = 32-l = 8; .-. MP = sjS = 2^2,
. ^ MP 2^2 ,j, 031 1
^^^^=op=-T-^^^*^=i^=2;r2-
r. MPi 4
3. Fig.asin(l). LetMP = 4; .-. 01/= 3 for tan ^ 1. e. -- I = - ;
.-. 0P2= 42 + 32=25; .-. OP =5,
. , MP 4: ^ OP 6
smA = ^=-^. sec^ = — =3.
4. Fig. as in (1). Let 0P = 4 and 0M=1; for sec 6 [=^^A =^'
.-. l/P2=OP2- 01/2 = 42 -1=15; .-. MP=JT5,
,^ OM 1 . ^ MP jr5
Qot6=-p=z= -==1, sin^=-r^=^ .
MP Ji^' OP 4
.'. 0P2 = 1/P2+ 01/2 = 3 + 1 = 4; .-. 0P = 2;
. ^ MP v/s ^ OM 1
'^^^=0P=^' ^^^^=0P=2-
RATIOS OF ONE ANGLE. XVIl. 43
6. Fig. as in (1). Let OM =2 and MP = ^5; for cot \^ = ^pj = 75 ;
. 0P2 = iHP2+ 0^2^:5 + 22 = 9; .-. 0P=3,
, ^ MP J5 . OP 3
7. Fig. as in (1). Let MP = h and OP = c; for sin ^ I = ^^ J = - ;
.-. OM'^=OP^-MP^ = c^-b^; /. OM=sJ^^^,
MP h
tan ^ = TTrr^ — ,_ .
OM Jc2_l2
r MP~[ a
8. Fig. as in (1). Let 3IP = a, OM=b; for tan 6 |_=^^J = p
.-. OP^=OM^ + MP^ = a^ + h^; .-. OP=Ja^ + b'\
. ^ J/P a ^ OM b
sin(9 = ^r^= ._- — - ; cos(9 = -
■OP 7^2:1^2' OP ^^2:^762*
9. Fig. as in (1). Let OP = a and OiV/^l ; for cos^ [= — | = -;
.-. iV/p2=OP2-OM2=:a2_i. ,'. MP=J^^^,
. . l/P V^^^ ^ ., OM 1
Sm ^ = T^pr = , cot ^ = YTTi = / .
OP a MP Ja^-1
N.B. The required ratios in each of the above examples may be found
without referring to the figure. If we combine the formulae of Art. 106, we
may shew that
sin e=zjr^^8H, and cot (? = -^J2l£= (see Ex. XVI. 1).
Vl-cos2 6/
If cos ^ = - as in (9), we have
10. Since sin ^ = a, .'. 1 -a2=:l - sin2 ^^cos^ ^, (Art. 105, i.)
„ tan^ = 6, .-. l + 62=:l + tan2(? = sec2^; (Art. 105, ii.)
.♦. (1 -a^) (l + i>2)^co62 e . sec2 ^ = cos2 . — 2- = l (Art. 102).
11. cos0 = h. tan^ = ;.= k: :. ain = k cos = hk,
cos^
From Art. 105, i., sin2<9 + cos2(9 = l; .-. h^ + hV=l; .'. h^ (l-\-k^) = h
44 TRACING THE CHANGES
EXAMPLES. XVIII. Page 77.
1. With the same construction and figure as in E. T. p. 76,
. OM
When the angle A is 0°, OP coincides with OR and then OM is equal to
OP; when A is equal to 90*^, OP coincides with OM and then OM vanishes ;
and as A continuously increases from 0° to 90°, OM continuously decreases
from OP to ; and OP is always equal to OR,
OM OP
Therefore as A approaches 0°, the fraction -^ approaches -r^ , that is 1 ;
when A — 90° the fraction ^^ is equal to — — , that is ; and as A con-
tinuously increases from 0° to 90°, the numerator of the fraction -^ con-
tinuously decreases from OP to while the denominator is constant, and
therefore the fraction -^ , which is cos A^ decreases continuously from 1 to 0.
2. With the same construction and figure as in E. T. p. 76,
'''' = 0M'
The changes in the secant may be traced by means of the figure in the
same way as those of the cosine (1) ; or we may use the formula sec ^=: ;
cos 6
as the angle continuously increases from to Jtt the cosine continuously
decreases from 1 to 0, and therefore the secant continuously increases from
1 to 'infinity.'
3. With the same construction and figure as in E. T. p. 76,
• . MP
When the angle A is 90°, MP is equal to 0P\ and when A is 0°, MP is
zero; as A continuously decreases from 90° to 0°, MP continuously decreases
from OP to zero ; and OP is always equal to OR,
MP OP
Therefore when A = 90°, the fraction — — is equal to -^ , that is 1 ; when
MP
J[ = 0° the fraction — ^ is equal to -r^, that is 0; and as A continuously
MP
decreases from 90° to 0° the numerator of the fraction ~-~ continuously
decreases from OP to zero, while the denominator is unchanged and there-
MP
fore i\ve fraction -—^ , which is sin A, decreases continuously from 1 to 0.
IN THE VALUES OF THE RATIOS. XVIII. 45
4. With the construction and figure of E. T. p. 76, cot ^= r— .
When the angle 6 is 0, OM is equal to OP^ and when the angle 6 is \ tt,
OM vanishes; and as the angle continuously increases from to Jtt, OM
continuously diminishes from OP to zero.
When the angle d is 0, MP is equal to zero ; and when ^ is J tt, MP is equal
to OP; :. as d continuously increases from to Jtt, 3IP continuously
increases from zero to OP.
OUT DP
Therefore when is 0, the fraction — - is equal to - , that is * infinity';
when ^ is J TT the fraction ^jj^ is equal to — = , that is zero ; and as con-
tinuously increases from to Jtt, the numerator continuously diminishes
from OP to zero, while the denominator continuously increases from zero to
OM
OP ; so that the fraction ~y ^ , that is cot 0, continuously decreases from a
number greater than any assir/nable numerical quantity until it is zero.
EXAMPLES. XIX. Page 82.
1. With a fig. similar to E. T. p. 79.
In OU, take ON so that the measure of ON is J. Draw NP parallel to
OR cutting the quadrant in P. Join OP, and draw PM perpendicular to OR.
Then jR OP is the angle required.
. __„ MP ON 1 ^ 1
For smROP=^ = -=-^l = -.
Therefore an angle POR has been drawn whose sine is ^.
2. Since the sine of an angle is the reciprocal of its cosecant, therefore
the sine is J of the angle of which the cosecant is 2. The question may
therefore be put into the form, 'Draw an angle whose sine is J;' which
is (1).
3. With the construction of Example 3, p. 80.
Let 0M= 1 and MP = 2. Then POM is the angle required.
MP 2
For tanPO^/=— =j = 2.
4. Yes. Since the tangent of an angle continuously increases from to
'infinity' as the angle continuously increases from 0° to 90°, there is there-
fore a value between 0° and 90° when the tangent of the angle is 431.
5. No. The cosine of an angle is never numerically greater than unity;
therefore no angle can be drawn whose cosine is .|.
6. Yes. The secant of an angle is numerically between 1 and infinity.
46 COMPLEMENTS. XIX.
7. (i) The complement of 30° = 90° - 30° = 60°.
(ii) „ „ 190° = 90°-190°=-100°.
(iii) „ „ 90° =90°- 90° = 0°.
(iv) „ „ 350° = 90° -350° =-260°.
(v) „ „ -25° = 90°- (-25°) = 90° + 25° =115°.
(vi) „ „ -320° = 90° -(-320°) = 90° + 320° = 410°.
, ... Sir IT Sir IT
M ,. .. T = 2-T==-i-
. .... IT TT f 7r\ TT TT 27r
(^"^^ " " -6=2 -(-6J =2-^6 =T-
8. sin^ = cos(90°~^), Art. 118; /. sin 70° = cos (90° - 70°) = cos 20°.
9. cos (90° -42° 44') = sin 42° 44'; /. cos 47° 16' = sin 42° 44'.
10. sin 79° = cos (90° - 79°) = cos 11°, sin 11° = cos (90° - 11°) = cos 79° ;
sin 79° cos 11° ^ r,no x..o
•'• ^^?;h=^— TTo; ••• tan 79° = cot 11°.
cos 79° sm 11°
11. sin 54° = cos (90° - 54°) = cos 36° ;
•'• - • g^o ^^ » i- e. cosec 54°= sec 36°.
sm54° cos 36°
12. In the triangle POM (E. T. p. 46), let angle POM=^ then angle
OPM= 90°-^.
/•\ • r^T^AT perpendicular OM . , /* i. r,- •• x
(i) sm OFM = ~~~T = — p- = cosme A ; (Art. 75, n.
' hypotenuse OP
.-. sin(90°-^) = cosJ[.
(ii) cot 0PM = ^^^^ , = ^ = tan ^ ; (Art. 75, iii.)
perpendicular OM
,\ cot(90°-^)=tan^.
(iii) cosec OPM^A yP^^^enuse ^OP^^^^ (Art. 78, v.)
perpendicular OM '
.', cosec (90° ~^) = sec vl.
pendicular OM , ,
I, TFT. = cot A
base MP
.-. tan(90°-^)=cot.d.
(iv) tanOPif=P^^^^^^^=.^ = cot^ ; (Art. 78, vi.)
^ ' base MP \ » /
EXAMPLES. XX. Page 84.
1, sin ^=-7^ , sin 45° = -75 ; /. ^ = 45° is a solution of the equation.
2. 4 sin ^ = cosec ^ ; .*. 4sin^ = -; — -;
sm ^ '
.-. igin^^:^!] ,-, 2^1n^ = l; ,-, sin^ = i; /, ^ = 30° is a solution.
EQUATIONS. XX. 47
3. .*. 2cos^ = Bec^; .'. 2co8^= -; .-. 2cos2^=l; .-. cos^=— ^.
cos^ ^2
.'. 45° is a solution.
3
4. .-. 4 sin 61-3 00860^ = 0; /. 48in(? = -^— ,; .*. 4sin2^ = 3;
sin e
.-. sin ^ = 4,^3 ; /. ^ = 60° is a solution.
3
5. 4 cos ^-3 sec ^ = 0; .-. 4 cos (9 = -; .'. 4cos2^ = 3;
cos 6
.'. cos^ = 4;^3; .-, ^ = 30° is a solution.
6. 3tan^ = cot^; .-. 3tan<? = 7 ^; .'.Stan^^r^l; .-. tan(9 = -T-;
tan u fjo
.-. ^ = 30° is a solution.
7. 3sin^-2cos2(9 = 0; .'. 3 sin ^- 2 (1 - sin^ (9)r=0;
.-. 2sin2^ + 3sin6> = 2.
From this quadratic sin 0= - 2 or J. The value - 2 is inadmissible; for
there is no angle whose sine is numerically greater than 1 (Art. 115) ;
.'. sin d = i.
But sin 30° = i; .'. ^ = 30° is a solution.
8. v/2 sin (? = tan <9 : /. J2 sin = ^ ; .\ either sin /9 = or J2 = .
If sin ^ = 0, ^ = is a solution of the equation.
If ^2= .; .'. cos0=-j~;
^ cos sJ2
.'. = 45° is a solution of the equation ; .-. ^ = 0° or 45°.
9. 2cos<?:..J3cot(9; .-. 2 cos 0= JS .^?^^;
^ ^ sm ^
.'. either cos^ = 0, or 2 sin = ^%.
If cos ^ = ; ,'. ^ = 90° is a solution of the equation.
If 2 sin (9 = ^3; .-. sin^ = i^3.
But sin 60° = i ^3; .-. 60° is a solution of the equation ; /. <9 = 90°or60°.
3
10. tan^=3cot6>; .-. tan(9 = -— -,; .-. tan2^ = 3; .'. tan<9 = ^3.
tan
But tan 60° = ^3 ; .-. ^ = 60° is a solution of the equation.
3
11. tan^-f3cot^ = 4; .-. tan^ + , ;, = 4; /. tan^ (9 -4 tan ^ + 3 =
tan
From this quadratic tan ^ = 3 or 1.
Both values are admissible.
If tan ^ = 1 ; .-. ^ = 45° is a solution of the equation.
48 EQUATIONS. XX.
12. tan^ + cot^=2; /. tan^+-— - = 2; .-. tan2(?-2 tan ^ + 1 = 0.
tan
From this quadratic tan ^ = 1 ; :, 6 = 45° is a solution of the equation.
13. 2sin2^ + ;^2cos^ = 2, 2 (1 -cos^ (?)+^2 cos (9 = 2 ;
.-. ^2cos^-2cos2^ = 0; /. either cos^ = or V2-2cos^ = 0.
If cos ^ = 0, ^ = 90° is a solution of the equation.
If ^2-2co8^ = 0; /. 008 6^ = -^.
But cos45°= y-;
.-. ^=45° is a solution of the equation ; .*. ^ = 90° or 45°.
14. 4sin2^ + 2sin^=l, sin^ ^ + 4 . sinl9 = J, sin^^ + i . sin ^ + i^ = t\;
.-. sin(9=J(±V5-l).
But sin 18° = J (/^5 - 1); .*. 6= 18° is a solution of the equation.
15. 3tan2^-4sin2^=l; .-. 3 ?^- 4 sin^ 6^ = 1 ;
" cos^^
.-. 3sin2|?-4sin2|?(l-sin2^) = l-sin2^; .-. 4sin4^ = l; .-. sin^==:-— .
\/2
But sin 45° =-7^ ; .*. ^ = 45° is a solution of the equation.
16. 2sin2(? + ^2sin^=2; .*. sin2^ + -^ sin ^ = 1 ;
The value - fJ2 is inadmissible, for there is no angle of which the sine is
numerically greater than 1 ; .*. sin ^=-^ .
But sin 45° = -7^^ ; .*. ^ = 45° is a solution of the equation.
17. cos2(?-;^3cos(9 + i = 0; .'. GO^e = U^.
But cos 30°= Jj^3 ; /. ^ = 30° is a solution of the equation.
18. cos2^ + 2sin2<9-f sin(9=:0;
.-. l + sin2(?-f sin^=0; .*. sin^ = 2or4.
The value 2 is inadmissible for there is no angle of which the sine is
greater than 1 ; .*. sin ^ = J.
But sin 30°= J; .*. ^=30° is a solution of the equation.
MISCELLANEOUS EXAMPLES. 49
EXAMPLES. XXI. Page 85.
1. 3 sin 60°- 4 sin360° = 3 . ^ - 4 . f^i^ ^ = 3 . i^ - 3 . ^ = 0,
2 \ 2 y 2 2
4cos3 30°-3 cos 30° = 4.(^^y-3. ^ = 5.^3 2.^3 = 0;
.-. 3 sin 60° - 4 sin^ 60° = 4 cos^ 60° - 3 cos 60°.
2. tan30°(l + cos30° + cos60°) = -i (^1 + ^ + 1) = ^^^i = ^-^^-^ ,
sin30° + sin60°=:i + i\/3-4(l + V3);
.-. tan 30° (1 + cos 30° + cos 60°) = sin 30° + sin 60°.
3. 2 cos2 d-7 cos ^+3 = 0, putting x for cos we have
.-. a; = ^±|=3 or i; .*. cos ^ = 3 or J.
The value 3 is inadmissible, for there is no angle of which the cosine is
greater than 1.
4. 8cos2^-8cos^ + l = putting x for cos 6;
/. a; = i±JV2 = J(2±^2); .-. cos ^ = i (2±^2).
Both values are admissible as they are neither greater than 1.
5. 8sin2^-10sin^ + 3 = 0, put a; for sin^;
.-. 8a;2-10a: + 3 = 0; /. a:^-i . ic + (|)2 = |f --| = ^V ; ••• a^-|=±i;
••• a;=|±i = ior|; .-. sin6> = |ori;
/. one value of sin ^ is 77 .
b
6. 12tan2(9--13 tan ^ + 3 = put a; for tan 6>;
.-. 12a:2-13a^ + 3 = 0; .-. a:2-i3^ + (if)2 = (i3)2_3^^2^5^. ., ^_i3_±^5^.
.-. a;=^f±^^ = | or J; /. tan ^ = J or J.
7. 3cos2^ + 2.^3 .cos<9 = 5Jput cos^ = a;; /. 3a;2 + 2 . ^3 . a; = ^ ;
•'• ^'^V'3*''"^W3; -12"^3~-12' •'•^ + v'~3~ W3'
^523 7 V3 7^3
2^3 2^3 ~ 2^3 2^3 2 6 *
7 /3
The value —-■ is inadmissible for there can be no angle of which the
cosine is numerically greater than 1 ; .*. cos = ^^/3.
But cos 30°- J ^3 ; /. ^ = 30° is a solution of the equation
i.e. ^ = i7r „ „ „ ,.
L. T. K. 4
50 MISCELLANEOUS EXAMPLES. XXI.
8. sin4<9 + cos4(9 + 2sin2^cos2^ = (sin2^ + cos2^)2 = l.
9. cos^ A +2 sin2 A . cos^ A = cos^ A (cos^ A + 2 sin^ A)
= cos2 A (cos2 A + sin2 ^ + sin^ A) = cos^ ^ (1 + sin^ A)
= (1 - sin2 ^) (1 + sin^ A) = l- sin^ A.
10. sin^ ^ + cos^ A = (sin2 ^ + cos^ A) (sin* ^ - sin^ A cos^ ^ + cos^ A)
= sin* A - sin2 ^ cos^ A + cos* ^ == sin* A - sin^ ^ (1 - sin^ A)-\-{l- sin^ ^)2
= sin*^ - sin2 A + sin*^ + 1-2 sin^ J. + sin*^ = 1-3 sin^^ + 3 sin* A.
,, . X ^^ -, sin*^ , (l-cos2(9)2 ^ l-2cos2(9 + co8*^
11. l + tan*^ = l + — ^^ = 1+ r^r-^ = l + tt.
cos*^ cos*(? cos*(?
_ l-2cos2^ + 2cos*(9
~ cos*d '
_ cos id + cos 5 sin ^ + sin B
sin J. - sin B cos ^ - cos B
_ (cos ^ + COS .B) (cos A - cos B) + (sin ^ + sin B) (sin ^1 - sin B)
"" (sin uil - sin B) (cos ^ - cos B)
_ cos2 ^ - cos2 B + sin2 A - sin2 .B _ (cos2 ^ + sin2 ^ ) - (cos2 ^ + sin2 B)
"" (sin^ -sin£) (cos^ -cos JB) ~~ (sin ^ - sin jB) (cos ^ - cos jB)
1-1
"" (sin ^ - sin ^) (cos ^ - cos 5) ~
,« / A . ^xo / 1 sin^\2 /l-sin^\2
13. see^-tan^2^ . r = t' )
' \cos^ cos^y \ cos^ J
- (^ - s in ^Y _ (l-sin.^) 2 _ 1 - sin .i
~~ cos2 A ~~ 1 - sin2 A ~ 1 + sin ^ *
14. We may proceed by figure after the manner of Examples XVIII., or
since cosec 6 = — — - , and as d continuously increases from to J tt, sin d con-
tinuously increases from zero to 1 ; .*. cosec 6 continuously diminishes from
to 1, that is from an infinitely large quantity to 1.
15. Since cot ^=7 and as d continuously decreases from Jtt to 0,
tan
t&nS continuously decreases from infinity to zero; /. cot d continuously in-
creases from . ^ . -- to , that is from zero to infinity.
mfinity zero
16. sin (^ + 0) = 1^3.
But sin60° = J;^3; .-. (9 + = 60° is one solution,
cos(^-0)r=4^3.
But cos 30°= i ^3 ; /. 6* - = 30° is one solution.
+ 0=60° and 6>-0 = 3O°; /. 2^ = 90° and ^ = 45° = j7r,
20 = 30°; .-. = 15°=:iV'r.
MISCELLANEOUS EXAMPLES. XXI. 51
Or, since sin (^ + 0) = J ^3 and cos (^ - 0) = J ^^3 ;
.-. sin(^ + 0) = cos(^-<^).
But sin (<9 + 0) = cos {90° -((9 + 0)}, (Art. 118); .-. ^-0 = 9O°-^-0;
.-. 2^ = 90°; .-. ^ = 45° = 4 TT, and since ^ + = 60°; .-. = 15° = TV7r.
EXAMPLES. XXII. Page 89.
Let Af B, Gy Df E be points in LR, such that the measures of ABy BG,
CDy DE are 1, 2, 3, 4 respectively.
1. AB + BG+GD = l + 2 + S=+6.
2. AB + BG + GA = l+2-(2 + l) = 0.
3. J5C+ CD + D^ + £C=+ 2 + 3 + 4 -(4 + 3) =+2.
4. ^D-(7D = (l + 2 + 3)-3=+3.
5. JD + i)B + J5^ = (l + 2 + 3)-(3 + 2) + 2 + 3 + 4=+10.
6. i?(7-^C + ^i)-^D = 2-(l + 2) + (l + 2 + 3)-(2 + 3) = 0.
7. CZ) + Z)5 + 5^ = 3-(3 + 2) + (2 + 3 + 4)=+7.
8. CD-i?D + 5^+^C+CJ5; = 3-(2 + 3)-l + (l + 2) + (3 + 4)=+7.
EXAMPLES. XXIII. Page 91.
1. The angle of 270° is the angle described by OP turning about from
the position ORj in the positive direction, to the position OD thus describing
3 right angles. [The angle ROD.]
2. 370° = 360° + 10°, i. e. the angle described by OP turning about O from
the position OR, in the positive direction, making one complete revolution
and then turning in the positive direction through the angle of 10°.
3. 425° = 360° + 65°, i.e. the angle described by OP turning about O
from the position OR, in the positive direction, making one complete revolu-
tion and then turning in the positive direction through the angle of 65°.
4. 590° = 360° + 180° + 50°, i. e. the angle described by OP turning about
from the position OR, in the positive direction, making one complete
revolution, then turning in the positive direction until it has described an
angle of 180° (i.e. being in the same straight line with OR), and from that
position turning round about in the positive direction, through the angle
of 50°. [Fig. III. E. T. p. 96.]
5. The angle described by OP turning about 0, from the position OR in
the negative direction through the angle of 30°. [ROP, in fig. ii. E. T.
p. 110.]
6. - 330°= - 360° + 30°, i. e. the angle described by OP turning about O
from the position OR, in the negative direction, making one complete revolu-
tion (when it has turned again to the position OR) and then turning back
in the positive direction through an angle of 30°. [ROP, fig. i. E. T. p. 110.]
4—2
52 ANGLES UNLIMITED IN MAGNITUDE. XXIII.
7. -480°= -360° -180° + 60°, i.e. the angle described by OP turning
about O from the position OR, in the negative direction, making one com-
plete revolution, then turning again in the negative direction until it has
described an angle of 180° (thus being in the same straight line with OB)^
and from that position turning back in the positive direction through an
angle of 60°. [Fig. ii. E. T. p. 96.]
8. - 750°= - 720° - 30°, i. e. the angle described by OP turning about
from the position OR in the negative direction, making two complete re-
volutions (when it has turned twice again to the position OR) and then
turning in the negative direction through an angle of - 30°. [ROP^, fig. ii.
E. T. p. 110.]
9. i?^7r = 67r + |7r = 6xl80° + 135° = 3x360° + 180°-45°, i.e. the angle
described by OP turning about from the position OR in the positive direc-
tion, making three complete revolutions, (when it has turned thrice again to
the position OJR), then turning again in the positive direction through an
angle of 180° (being thus in the same straight line with OR) and from that
position turning back in the negative direction through an angle of -45°.
[Fig. E. T. p. 94.]
10. 27i7r-f J7r = nx360°-f 30°, i. e. the angle described by OP turning
about from the position OR in the positive direction, making n complete
revolutions (when it has turned n times again to the position OR) and then
turning again in the positive direction through the angle of +30^. [ROP. fig*.
E. T. p. 104.]
Note. The case considered above is when n is positive ; if n is negative
we may retain the same figure and in the explanation write 'negative' for
^positive\ vid. Examples VI. (10).
11. (2n + l)7r + j7r = 27i7r + 7r-f j7r = nx360° + 180°+60°. When OP
has described the angle n x 360° it is n times again in the position OR ; from
this position let OP turning about O, in the positive direction, describe
the angle 180° (i.e. is in the same straight line with OR), for this position
turning again about in the positive direction let OP describe + 60°, vid.
Note to (10). [Fig. iii. E. T. p. 96.]
12. Fig. E. T. p. 94.
(27i-fl)7r-j7r = 2n7r + i7r = wx360°+135°=wx360° + 180°-45°.
When OP has described the angle n x 360° it is n times again in the
position OR. We may proceed for the rest as in (9).
13. 2n7r- j7r=wx360°-90°, i.e. the angle described by OP turning
about O from the position OR, making n complete revolutions (thus being
n times again in the position OR) and then turning back in the negative
direction through an angle of - 90°. [The angle ROD.]
14. (2n + l)7r-i7r = 2n7r-h47r = nx360° + 90°. The angle described by
OP turning about from the position OR, making n complete revolutions
(thus being n times again in the position OR) and then turning in the posi-
tive direction through an angle of +90°, vid. Note to (10). [The angle ROU.]
QUADRANTS. 53
EXAMPLES. XXIV. Page 94.
I. 120°= 180° - 60° represents an angle in the second quadrant.
2 340° = 360° -- 20° = 2 X 180° - 20° is an angle in the fourth quadrant.
(Art. 130, ii.)
3 490° = 540° - 50° = 3 X 180° - 50° is an angle in the second quadrant.
(Art. 130, iii.)
4. - 100°= - 180° + 80° is an angle in the third quadrant. (Art. 130, iv.)
5. -380°= -360° -20°= -2x180° -20° is an angle in the fotirth
quadrant. (Art. 30, ii.)
6. - 1000° = - 6 X 180° + 80°, i. e. an angle in the first quadrant.
(Art. 130, i.)
7. |7r = 7r-i7r, i.e. an angle in the second quadrant. (Art. 130, iii.)
8. 10 TT + J TT = an angle in the ^rs^ quadrant. (Art. 130, i.)
9. 97r-|7r = 87r + Jtt, i.e. an angle in first quadrant. (Art. 130, i.)
10. 27i7r- J7r = an angle in the fourth quadrant. (Art. 130, ii.)
II. (2n + l)7r + |7r = (2?i + l)7r +7r- Jtt = 2m7r - J7r = an angle in the
fourth quadrant. (Art. 130, ii.)
12. nw + ^Tr. If 71 be even = 2/71, then n7r + ^7r = 2wi7r + Jtt, i.e. an angle
in the first quadrant. (Art. 130, i.)
If n he odd = 2rM + l, then 7i7r + Jir = (27;i + 1) tt H^^tt, i.e. an angle in the
third quadrant. (Art. 130, iv.)
EXAMPLES. XXV. Page 98.
1. The angle of 60° is in the first quadrant ;
,-. the sine is positive,
the cosine ,, ,,
the tangent ,, ,, (Art. 133, i.),
i. e. the signs of the sine, cosine and tangent are + , + , + .
2. The angle of 135° is in the second quadrant ;
.-. the sine is positive,
the cosine is negative,
the tangent is negative (Art. 133, ii.),
i. e. the signs of the sine, cosine and tangent are +, - , - .
3. The angle of 265° is in the third quadrant ;
.'. the sine is negative,
the cosine is negative,
the tangent is poaitive (Art. 133, iii.),
i.e. the signs of the sine, cosine and tangent are - , - , +.
54 QUADRANTS. XXV.
4. The angle of 275° is in the fourth quadrant ;
.*. the sine is negative,
the cosine is positive,
the tangent is negative (Art. 133, iv.),
i. e. the signs of the sine, cosine and tangent are - , + , - .
5. The angle of - 10° is in the fourth quadrant ;
.*. the sign is negative,
the cosine is positive,
the tangent is negative (Art. 133, iv.),
i.e. the signs of the sine, cosine and tangent are - , + , - .
6. The angle of - 91° is in the third quadrant ;
.'. the sine is negative,
the cosine is negative,
the tangent is positive (Art. 133, iii.),
i. e. the signs of the sine, cosine and tangent are - , - , + .
7. The angle of - 193° is in the second quadrant ;
.*. the sine is positive,
the cosine is negative,
the tangent is negative (Art. 133, II.),
i. e. the signs of the sine, cosine, and tangent are + , - , - .
8. The angle of - 350° is in the first quadrant ;
.*. the sine is positive,
the cosine is positive,
the tangent is positive (Art. 133, I.),
i. e. the signs of the sine, cosine and tangent are + , + , + .
9. The angle of - 1000° is in the first quadrant [Examples XXIV. (6)] ;
.*. the sine is positive,
the cosine is positive,
the tangent is positive (Art. 133, I.),
i. e. the signs of the sine, cosine, and tangent are + , + , + .
10. The angle of 2mr+ tt J is in the first quadrant ;
.*. the sine is positive,
the cosine is positive,
the tangent is positive (Art. 133, I.),
i. e. the signs of the sine, cosine and tangent are + , + , + .
11. The angle of 2mr + Jtt is in the second quadrant ;
.*. the sine is positive,
the cosine is negative,
the tangent is negative (Art. 133, II.),
i. e. the signs of the sine, cosine, and tangent are +, - , - .
QUADRANTS. XXV. 55
12, The angle of 2mr - J tt is in the fourth quadrant ;
.*. the sine is negative,
the cosine is positive,
the tangent is negative (Art. 133, IV.),
i. e. the signs of the sine, cosine, and tangent are - , + , - .
EXAMPLES. XXVI. Page 100.
1. 150° is an angle in the second quadrant.
Let the angle ROP be 150° (fig. ii. E. T. p. 96).
Then the angle POL = 180° - 150° = 30°.
Therefore the Trigonometrical Ratios of 150° = those of 30° numerically j
and in the second quadrant the sine is positive and the cosine and tangent
are each negative;
.-. sinl50° = i; cos 150°= -J ^3 ; tan 150°= - J^^-
2. 135° is an angle in the second quadrant.
Let the angle BOP be 135° (fig. ii. E. T. p. 96).
Then the angle POL = 180° - 135° = 45°.
Therefore the Trigonometrical Ratios of 135° = those of 45° numerically;
and in the second quadrant the sine is positive^ and the cosine and the tan-
gent are negative;
:. sinl35° = J^2; cos 135°= -1^2 tan 135°= - 1.
3. - 240° is an angle in the second quadrant.
Let the angle BOP (fig. ii. E. T. p. 96) described by OP starting from the
position OB and turning about O in the negative direction be - 240°.
Here angle POL = - 240° - ( - 180°) = - 60° ;
.-. the Trigonometrical Ratios of - 240° = those of - 60° numerically ;
i. e. = those of 60° numerically ; and in the second quadrant the sine is posi-
tive and the cosine and tangent are negative;
,\ sin-240°=+W3; cos-240°=-i, tan - 240° = - ^3.
4. 330° is an angle in the fourth quadrant.
Let the angle EOP= 330° (fig. iv. E. T. p. 96).
Then the angle POjR = 330°-360° = -30°, i.e. an angle in the fourth
quadrant.
.'. the Trigonometrical Ratios of 330^ = those of 30° numerically.
Therefore also the Trigonometrical Ratios of 330° = those of 30° numeri-
cally ; and in the fourth quadrant the cosine is positive and the sine and
tangent negative ;
.-. sin 330°= -i; cos330°= +i>/3 ; tan330°= - 4^3.
66 SIGNS OF RATIOS. XXVI.
5. - 45° is an angle in the fourth quadrant.
The Trigonometrical Batios of ~ 45° = those of 45° numerically ; and in
the fourth quadrant the cosine is positive and the sine and tangent negative ;
.-. sin-45°= -i^2; cos-45°=+W2; tan 45°= -1.
6. - 300° is an angle in the first quadrant.
Let the angle ROP (fig. i. E. T. p. 96) described by OP revolving about
for the negative direction, from the position, be the angle of — 300° ;
.\ P0i2 = 360° -300° =60°;
.*. the Trigonometrical Ratios of -300° = those of ^0° numerically ; and
in the first quadrant, the sine, cosine and tangent are each positive ;
.-. sin~300°=+iv/3, cos - 300° = + i ; tan-300°=V3.
7. 225° is an angle in the third quadrant.
Let the angle BOP be 225° (fig. iii. E. T. p. 96).
Here the angle POL = 225° - 180° = 45° ;
.*. the Trigonometrical Ratios of 225° = those of 45° numerically ; and in
the third quadrant the sine and cosine are each negative and the tangent is
positive; .'. sin225°= -^^2 ; cos 225°= - J^2 ; tan225°=+l.
8. - 135° is an angle in the third quadrant.
Let the angle BOP be - 135° (fig. iii. E. T. p. 96).
Here the angle POL = 180° - 135° = 45° ;
.-. the Trigonometrical Ratios of- 135° = those of 45° numerically ; and
in the third quadrant the sine and cosine are each negative and the tangent
is positive ;
.-. sin-135°=-iV2; cos- 135°= -^^2 ; tan-135°=+L
9. 390° is an angle in the first quadrant.
Let the angle JROP = 390° (fig. i. E. T. p. 96) ;
.-. the angle POB = 390° - 360° = 30° ;
.*. the Trigonometrical Ratios of 390° = those of 30° numerically ; and in
the first quadrant the sine, cosine, and tangent are each positive ;
.-. sin 390°= +J; cos 390° = + i ^^3, tan 390°= + J V^.
10. 750° is an angle in the first quadrant.
Let the angle BOP = 750° (fig. i. E. T. p. 96).
Here the angle POjR = 750° - 720° = 30°;
.-. the Trigonometrical Ratios of 750° = those of 30° numerically ; and in
the first quadrant the sine, cosine, and tangent are each positive ;
.-. sin 750°= +4; cos750°=+W3; tan 750° = J V^.
11. - 840° is an angle in the third quadrant.
Let the angle BOP= -840° (fig. in. E. T. p. 96).
Here POL = 900° - 840° = 60° ;
.-. the Trigonometrical Ratios of -840° = those of 60°; and in the third
quadrant, the sine and cosine are negative and the tangent positive;
/. sin-^840°=~jV3, cos -^840°= -4, tan-840°= +^/3.
SIGNS OF RATIOS. XXVI. 57
12. 1020° is an angle in the fourth quadrant.
Let the angle i20P= 1020° (fig. iv. E. T. p. 96).
Here FOR = 3 x 360° - 1020° = 60° ;
.'. the Trigonometrical Ratios of 1020° = those of 60°; and in the fourth
quadrant the sine and tangent are each negative, and the cosine is positive;
.-. sin 1020°= -J V^, cos 1020°=+ J, tan 1020°= -^3.
13. 2n7r + Jtt is an angle in the first quadrant.
Let the angle R0P = 2mr + iw = n . 360° + 45° (fig. i. E. T. p. 96).
Then angle POi? = 45°;
.-. the Trigonometrical Ratios of 27?7r + j7r = those of 45° numerically;
and in the first quadrant the sine, cosine and tangent are each positive ;
.'. sin(2?i7r + j7r) = J^2; cos (2n7r + J7r) = J;>^/2; tan (2;i7r + J7r)= +1.
14. (2n + 1) TT - Jtt is an angle in the second quadrant.
Let the angle POP=(27i + l) tt- Jtt, [i.e. w . 360° + 120°] (fig. ii. E. T.
p. 96).
Then angle POL = 180° - 120° = 60° ;
.-. the Trigonometrical Ratios of n . 360° + 120° = those of 60°; and in the
second quadrant the sine is positive and the cosine and tangent are each
negative ;
.\ sin[(2n+l)7r-j7r]=+JV3; cos[(2n + l) tt- i7r]= -J ;
tan[(2n+)7r-j7r]=-^3.
"15. (2w - 1) TT + Jtt is an angle in the third quadrant.
Let the angle JR0P=(2n- 1) tt + ^tt, i.e. w. 360°- 150° (fig. iii. E. T.
p. 96).
Then angle POL = 180° - 150° = 30° ;
.-. the Trigonometrical Ratios of (2n - 1) tt + I^tt = those of 30° ; and in the
third quadrant the sine and cosine are negative and the tangent positive ;
.-. sin{(2/i-l)7r + i7r}=-4; cos {(2n- 1) 7r + i7r} = - J^^ ;
tan {(2n-l) 7r + j7r} = + jV3.
EXAMPLES. XXVII. Page 100.
N.B. The references here are to the figures and construction of
p. 96, E. T.
L cos^ = ~.
As A increases from 0° to 90°, OM decreases from OP to zero and is
positive ; .-. cos A decreases from 1 to zero and is positive.
As A increases from 90° to 180°, OM increases from zero to OP and is
negative ; .-. cos A increases from 2ero to 1 and is negative,
68 TRACING VALUES. XXVII.
As A increases from 180° to 270°, OM decreases from OP to zero and is
negative ; .*. cos A decreases from 1 to zero and is negative.
As A increases from 270° to 360°, OM increases from zero to OP and is
positive ; .'. cos A increases from zero to 1 and is positive.
2. '^''^^OM'
As A increases from 0° to 90°, MP increases from zero to OP and is
positive ; OM decreases from OP to zero and is positive ;
.'. tan A increases from zero to * infinity ' and is positive.
As A increases from 90° to 180°, MP decreases from OP to zero and is
positive, OM increases from zero to OP and is negative ;
.*. tan A decreases from * infinity ' to zero and is negative.
As A increases from 180° to 270°, 3IP increases from zero to OP and is
negative, OM decreases from OP to zero, and is negative ;
.-. tan A increases from zero to infinity and is positive.
As A increases from 270° to 360°, MP decreases from OP to zero and is
negative ; OM increases from zero to OP and is positive ;
/. tan A decreases from infinity to zero and is negative.
3. We may proceed by the figure to trace the changes in the cotangent
in the same way as we have done in (1) and (2) or by means of the formula
cot A = J , we may infer the changes of the cotangent from the known
changes of the tangent.
As A increases from 0° to 90°, tan A increases from zero to infinity and is
positive ; .*. cot A decreases from infinity to zero and is positive.
As A increases from 90° to 180°, tan A decreases from infinity to zero
and is negative ;
.*. cot A increases from zero to infinity and is negative.
As A increases from 180° to 270°, tan A increases from zero to infinity
and is positive ;
.*. cot A decreases from infinity to zero and is positive.
As A increases from 270° to 360°, tan A decreases from infinity to zero
and is negative ;
/. cot A increases from zero to infinity and is negative.
4. We may proceed by the figure or by means of the formula
sec A = , we may infer the changes of sec A from the known changes
of cos A.
As A increases from 0° to 90°, cos A decreases from 1 to and is positive ;
.'. sec A increases from 1 to oo and is positive.
As A increases from 90° to 180°, cos A increases from to 1 and is
negative ; .-. sec A decreases from oo to 1 and is negative.
TRACING VALUES. XXVII. 59
As A increases from 180° to 270°, cos^ decreases from 1 to and is
negative ; /. sec A increases from 1 to oo and is negative.
As A increases from 270° to 360°, cos^ increases from to 1 and is
positive ; .*. sec A decreases from oo to 1 and is positive.
5. cosec^ = -^.
As A increases from 0° to 90°, MP increases from zero to OP and is
positive ; .*. cosec A decreases from infinity to 1 and is positive.
As A increases from 90° to 180°, MP decreases from OP to zero and is
positive ; .•. cosec A increases from 1 to infinity and is positive.
As A increases from 180° to 270°, MP increases from zero to OP and is
negative ; /. cosec A decreases from infinity to 1 and is negative.
As A increases from 270° to 360°, MP decreases from OP to zero and is
negative ; .-. cosec A increases from 1 to infinity and is negative.
6. Since sin^ is never numerically greater than 1, 1 - sin ^ is never
negative. Its least value is when sin^=0, i.e. when ^==0; its greatest
value is when sin^ is negative and numerically greatest, i.e. when
sin ^ = - 1, when A = 270°.
As A changes from 0° to 90°, sin A changes from to 1 and is positive ;
.*. 1 - sin A changes from 1 to and is positive.
As A changes from 90° to 180°, sin A changes from 1 to and is positive ;
.*. 1 - sin A changes from to 1 and is positive.
As A changes from 180° to 270°, sin A changes from to 1 and is negative;
.-. 1 - sin ^ changes from 1 to 2 and is positive.
As A changes from 270° to 360°, sin A changes from 1 to and is negative ;
.*. 1 - sin ^ changes from 2 to 1 and is positive.
7. sin^ A is never negative.
As A changes from 0° to 90°, sin A changes from to 1 and is positive ;
.-. sin^ A changes from to 1 and is positive.
As A changes from 90° to 180°, sin^ changes from 1 to and is positive;
.*. sin^ A changes from 1 to and is positive.
As A changes from 180° to 270°, sin A changes from to 1 and is negative ;
.-. sin2 A changes from to 1 and is positive.
As A changes from 270° to 360°, sin A changes from 1 to and is negative;
.-. sin^ A changes from 1 to and is positive.
8. As A changes from 0° to 90°, sin^ changes from to 1 and is
positive, cos A changes from 1 to and is positive ;
.-. sin A cos A changes from to and is positive,
sin A cos A has its greatest numerical value when sin A = cos A *, i. e. when
^=45° and sin ^ cos ^ = + J.
* For it will be seen in the next Chapter that sin^ cos^ = isin2^, and
.*. sm.4cos.4 has its greatest value when 2^ = 90°.
60 TRACING VALUES. XXVII.
As A changes from 90° to 180°, sin A changes from 1 to and is positive,
cos A changes from to 1 and is negative ;
.-. sin A cos A changes from to and is negative ;
its greatest numerical value is when sin ^ = cos ^4, i.e. when ^4 = 135° and
sin A cos A— -\.
As A changes from 180° to 270°, sin^ changes from to 1 and is
negative, cos A changes from 1 to and is negative ;
.*. sin A cos A changes from to and is positive,
its greatest numerical value is when sin^ = cos^, i.e. when ^ = 225° and
sin^cosyl= + J.
As A changes from 270° to 360°, sin^ changes from 1 to and is positive;
cos A changes from to 1 and is negative ;
.'. sin A cos A changes from to and is negative ;
its greatest numerical value is when sin^ = cosyi, i.e. when ^=315° and
sin A cos A= -^.
9. As ^ changes from 0° to 90°, sin A changes from to 1 and is posi-
tive ; cos A changes from 1 to and is positive ;
.*. sin A + cos A changes from 1 to 1 and is positive ;
its greatest numerical value is when A = 45°, for
(sin^ + cos^)2 + (sin^ -cos^)2=2 (sin^ j: + cos^ ^) = 2 ;
.*. (sin^ + cos^)2 is greatest when (sin^ -cos^)2 = 0, i.e. when sin J = cos ^,
when A = 45° ;
.*. sin ^ + cos ^ has its greatest numerical value when A = 45°,
i. e. when sin A + cos A= + J2.
As A changes from 90° to 180°, sin A changes from 1 to and is positive ;
cos A changes from to 1 and is negative ;
.*. sin A + cos A changes from 1 to - 1 vanishing when sin A=^ - cos A,
i.e. when ^ = 135°.
/ As ^ changes from 180° to 270° sin A changes from to 1 and is nega-
tive, cos A changes from 1 to and is negative ;
.*. sin A -f- cos A changes from - 1 to - 1 ;
it does not change sign ; .*. its greatest numerical value may be found above
to be fJ2 when ^ =225°.
As A changes from 270° to 360° sin A changes from 1 to and is nega-
tive ; cos A changes from to 1 and is positive ;
.'. sin A + cos A changes from - 1 to 1.
Its numerically smallest value being when sin ^ -f-cos^=0, i. e. when
cos A— - sin ^ , i. e. when A = 315°.
10. tan^ + cot^ = tan ^-f- ^.
tan A
As A changes from 0° to 90°, tan A changes from to oo and is positive ;
1
changes from oo to and is positive ;
tan A
TRACING VALUES. XXVII. 61
.*. tan A H , changes from oo to oo and is positive ;
tan A
the least vakie of tan A + , , is when A = 45° ; for
tan A
.-. the least value is when tan A - = 0, i. e. when tan^ A = l. Therefore
tan^
tan ^ + cot ^ changes from oo to 2 as ^ changes from 0° to 45° ; and then
changes from 2 to oo as ^ changes from 45° to 90°.
As A changes from 90° to 180°, tan A changes from oo to and is nega-
tive ; cot A changes from to oo and is negative ;
,'. tan A + -. 7 changes from oo to oo and is negative ;
tan A
its least numerical value being - 2 when A is 135°.
As A changes from 180° to 270°, tan A changes from to oo and is posi-
tive, 7 changes from oo to and is positive ;
tan A
.'. tan A + ^ 7 changes from oo to oo and is positive :
tan A
its least numerical value being 2 when A is 225°.
As A changes from 270° to 360°, tan A changes from oo to and is nega-
tive ; T changes from to oo and is negative ;
tan A
.'. tan A + 7 [i. e. tan A + cot A] changes from oo to oo and is negative ;
its least numerical value being - 2 when A = 315°.
11. As A changes from 0° to 90°, sin^ changes from to 1 and is
positive ; cos A changes from 1 to and is positive ;
.*. sin A - cos A changes from - 1 to 1 vanishing when A = 45°.
As A changes from 90° to 180°, sin A changes from 1 to and is positive ;
cos A changes from to 1 and is negative ;
.'. sin A - cos A changes from 1 to 1, remaining positive,
and having its greatest numerical value when A = 135°.
For (sin A - cos A)^ + (sin ^ + cos A)^ = 2 (sin^ A + cos^ A) = 2;
.'. (sin A - cos A)^ has its greatest value when sin A + cos ^ = 0,
i. e. when A = 135° and sin A - cos A= + ^2.
As A changes from 180° to 270°, sin A changes from to 1 and is nega-
tive, cos A changes from 1 to and is negative ;
/. sin ^ - cos ^ changes from 1 to - 1 vanishing when sin A - cos ^ =0 ;
i.e. when ^ = 225°.
62
TRACING VALUES. XXVII.
As A changes from 270° to 360°, sin A changes from 1 to and is nega-
tive ; cos A changes from to 1 and is positive ;
.•. sin A - cos A changes from - 1 to - 1 remaining negative and having
its greatest numerical value namely - ,^2 when A = 315°.
EXAMPLES. XXVIII. Page 103.
1. With the figure and construction of E. T. p. 102, let J^OP = the angle
which OP describes from the position Oi? = 30° and ROP' the angle described
by OP' = 90° -30°=: 60°;
MP OM'
:. 8mR0P=^-= ^^ = cos ROP'; /. sin 30°= cos 60°.
2. With the figure and construction referred to above let ROP the angle
which OP describes from the position 012= 65°, and ROP' the angle described
by OP' = 90° -65° = 25°;
MP OM'
.-. sinJ20P=^ = ^,=cosJROP'; /. sin 65° = cos 25°.
3. With the construction of p. 102 E. T., let OP and OP' be the two
revolving lines, OP starting from the position OR and turning about in
the positive direction describe the angle EOP=195°; OP' starting from the
position OR and turning about the point O in the negative direction describe
the angle i20P'= -105°.
M
M'
■f"
^^
-^1
\
\
/
\
\
^
0— "^^
the Trigonometrical ratios of
Now angle POL = 195°- 180° = 15°; /
195° are numerically equal to those of 15°.
Angle P'OD = 105°-90°=15°; .'. the Trigonometrical ratios of
= those of 15° numerically.
Since angle POL = angle PO'D, /. MP=N'P' = OM';
MP N'P' OM'
105°
OP OP' OP'
sinPOP = cos-J^OP', i.e. sin 195° = cos (- 105°).
4. Let OPj OP' be two revolving straight lines, OP starting from the
position OR describe the angle POP =+275° and OP' starting from the
same position describe the angle = - 185°. Draw PM, P'M' perpendiculars
on OP, OL respectively, and P'N' perpendicular on OU. Then angle
EQUALITY OF RATIOS. XXVIII.
63
POD = angle POM= (275° -270°) = 5° and angle P'OM' = (185°- 180°) = 5°;
.-. angle POM^=angle FOM' and OP=OP'; .*. the right-angled triangles
0PM and OP'M' are equal in every respect;
D P
OM
OP ''
M'P'
'' OP' ''
P'N\
'' OP' '
cos 275°= sin (-185°).
F' U
5. Let OPy OP' he two revolving straight
lines, OP starting from the position OR de-
scribe the angle JROP=-27°; OP' starting
from the same position describe the angle
i20P' = 117°. Draw PM, P'M' perpendiculars
on OE, Angle
OP'M' = angle P'OU= (117° - 90°) = 27° ;
/. angle OP'il/' = angle POM, and OP=OP';
.-. the right-angled triaiigles 0PM and OP'M'
are equal in every respect ;
OM M'P' , ^_, . ,,_
-op" OF' •••COs(-27°) = sinll7°.
G. Let OP, OP' be two revolving straight lines, OP starting from the
position OR describe the angle POP =300° and OP' starting from the same
position describe the angle ROP'= -210°. Draw PM, P'M' perpendiculars
on OP, OL. Angle POilf =(360°-300°) = 60°; /. angle OPiH=30° angle
P'OM'= (210° -180°) = 30°; /. angle OPi¥= angle P'OM' and OP=OP';
.'. the right-angled triangles 0PM and OP'M' are equal in every respect ;
OM M'P'
.-. cos300°=^in(-210°).
" OP OP'
cos J^=sin(90°-J^)
= 8in{J(^4-P-t-C)-J^}
= sinJ(P-t-C7).
cos4P = sin(90°- JP)
= 8in{^(^+P-fC)-iP}
= sini(^-|-0).
[Art. 140, Ex. 1.]
[^+P+O=180°]
[Art. 140, Ex. 1.]
[^ + P + C= 180°]
64 EQUALITY OF RATIOS. XXVIII.
9. sin i (7 = cos (90° -i (7) [Art. 140, Ex. 1.]
= cos {4(^+5 + C)-iC} [^+JB + a = 180°]
= cos^(A-\-B).
10. sin Jj; = cos (90° -i^) [Art. 140, Ex. 1.]
= cos {i {A + B-i-G)-i A} [^+5 + = 180°]
= G08 ^(B + C).
EXAMPLES. XXIX. Page 105.
1, In fig. E. T. p. 104, let i^OP=60° and i^OP' = 120°;
.-. P'OL = 180° - 120° = 60° and MP = M 'P' ;
MP M'P'
•*• ^ = if' •••sin 60°= sin 120°.
2, In fig. 11 E. T. p. 110, let OP^ starting from the position OR describe
the angle ROP^= +340° and OPo starting from the same position describe
the angle ROP^= - 160°. Draw P-^M^, P^M^ perpendiculars on LOP. Then
angle PiOilfi = (360°-340°)= 20° and angle PgOilfg = (180° - 160°)= 20°;
.-. angle PiOifi = angle P^OM^ and M^P^=M^P^\
MP MP
/. i^i = "^-2. ... sin POPi = sin POP2; .'. sin 340°= sin (- 160°).
3, In fig. 11 E. T. p. 110, let OP^ starting from the position OR describe
the angle ROP^= - 40° and OPo starting from the same position describe
the angle POP2 = +220°. Angle PgOikf 2 = (220° - 180°) = 40° ; /. angle
Pi OiIfi = angle P2OM2 and M^P^ = M^P^\
... ^£l^^A. .., sinP0Pi = sinP0P2; /. sin (-40°) = sin (220°).
OP J OP2
4, In fig. 11 E. T. p. 110, let OP^^ starting from the position OR describe
the angle POPi = 320° and OP^ starting from the same position describe the
angle P0P2= -140°. Angle P^Oil/i^ (360° -320°) = 40°; and angle P^OM^
= (180° -140°) = 40°; .-. angle PjOiV/i = angle P^OM^; .: OM^ = OM^ but of
opposite signs ;
(IM OM
:. r^i= - ^2 . . ^j^g EOP^= - cos POP2 ; /. cos 320°= - cos ( - 140°).
OPi OP2
5, In fig. 1, p. 110 E. T., let OP^ starting from the position OR describe
the angle ROP^= -380°, and OP^ starting from the same position describe
the angle P0P2= 560°.
Angle PjOilfi = (380° - 360°) = 20°. Angle POPg =560° =360° +200°;
.'. the geometrical position of angle 560° is the same as that of 200°; and
angle P20ilf2 = (200° -180°) = 20°; .-. angle PiOilfi=P20i¥2 and OM^=OM^.
but of opposite signs ;
/. 2^= ^^2. . cos POPi=- cos POP2; .-. cos (-380°)= -cos 560°.
EQUALITY OF RATIOS. XXIX.
65
6. In E. T. p. 110, fig. 11, let OP^ starting from the position OR describe
the angle B,OP^= +195°; and OFi starting from the same position describe
the angle llOP^ = - 15°. Angle P^OM,^ = (195° - 180°) = 15° = angle P^OM^ ;
.*. OM^ = OM^, but of opposite signs ;
.-. -^=-;p^; .-. cosl95°=-cos(-lo°).
OP,
OP,
1. Since A^ B, G are the angles of a triangle therefore A + B + C = 1S0°
and {B + G) is the supplement of A ;
.: sin A = sin (B + G) [Art. 141, Ex. 2.]
2. For the same reason (A + B) is the supplement of G ;
.-. sin C=sin (A+B) [Art. 141, Ex. 2.]
3. For the same reason (A + G) is the supplement of B ;
/. cosJ? = cos {180°-(^ + C)}= -cos(^ + C).
4. (C + i?) is supplement oi A-,
:. cos ^ = cos {180° -(0 + 2^)}= -cos ((74-i>').
MISCELLANEOUS EXAMPLES. XXX. Pages 108—9.
1. In figure in E. T. p. 106, let the angle jROP= +60°; and the angle
ROP'= +240°. Angle P'Oilf'= (240°- 180°) ==60° wangle POM;
.*. MP=M^P' but they are of opposite sign ;
MP M'P'
QP = -^'y .-. sin i^OP=- sin JROP'; /.sin 60°=
- sin 240°.
2. In the figure let the angle BOP' = + 170° and the angle ROP= + 350°.
Angle P'OJ/'= (180° -170°) = 10°, angle P01I= (360° - 350°) = 10° ; .-. angle
POiT/= angle P'OM' and MP = M'P' but they are of opposite sign ;
MP M'P'
. Qp=^~-Qp-'^ •*• smjROP= -sin ii OP'; .'. sui 170°= - sin 350".
L.T. K. ,5
EQUALITY OF RATIOS. XXX.
3. Draw a figure similar to that of (2). Let the angle ROP'= - 20° and
the angle jR OP =+160". Then angle POilf = (180° - 160°) = 20° ; /. angle
P'OiJi ' = angle POM and M'P' = MP, but they are of opposite sign;
M'P' MP
; .-. sin ROP'= - sin POP ; /. sin ( - 20°) = - sin 160°.
OP'
OP
4. In fig. E. T. p. 106, let the angle POP= 380° and let the angle
i20P'= +560°.
Now angle POM= (380° - 360°) = 20° and angle
P'OM' = { 560° - (360° + 180°) } = 20° ;
.-. angle POM=P'OM' and OM=OM' but they are of opposite sign;
^=-^';.-.cosW=-oosfiOP';
COS 380°= -cos 560°.
5. Draw a figure similar to that of (2). Let the angle ROP= -225°,
and the angle POP'= -45°. Then angle POiU= (225°- 180°) =45° = angle
P'OM' and OM— OM^ but they are of opposite sign ;
OM OM'
OP'
OP'
cos ROP= - cos POP' ; .'. cos ( - 225°) = - cos ( - 45°).
6. Since 1005° = 2 x 360° + 285°and 1185° = 3 x 360°+ 105° the geometrical
positions of OP and OP' are the same as for 285° and 105° respectively.
PjJl
Let the angles POP, POP' be equal to +285° and +105° respectively.
Draw PN, P'N' perpendiculars on UOD.
Then angle POD = (285° - 270°) ^ 15° and angle P'O U= (105° - 90°) = 15° ;
.-. angle POD = angle P'0C7 and PN=P'N\ but they are of opposite sign;
PNJPN^ 0M_ OM'
•*• 0P~ OP' *' **• 0P~ OP'
:, cos POP= -cos POP'; /. cos 285°= -cos 105°; .'. cos 1005°= -cos 1185°.
7. In fig. E. T. p. 107, Example 4, let the angles POP, POP' be equal
to 60° and 150° respectively. Then angle UOP' = 60° and NT', i. e. OM' = MP
but of opposite sign ;
/. -Qp---Qp> - sin i^OP=- cos JXOP'; .\ sin 60°= -cos 150°.
EQUALITY OF RATIOS. XXX. 67
8. With the figure and construction of (7),
nP^lip' ' •• cosjROP = sinJROP'; .\ cos60° = sin 150^
9. Let the angles ROP, ROP' be equal to + 225° and + 315° respectively.
Angle POM= (225° -180°) = 45° and angle P'OA^' = (315° -270°) = 45°;
.-. angle POil/=angle P'ON' and PM=P'N\ i.e. OM' but of opposite sign ;
... ^lE:^-^^-^ .-. sin POP= -cos POP'; .-. sin 225°= -cos 315°.
10. Let the angles ROP, ROP' be equal to - 60° and + 30° respectively.
Angle PO C7= (90° -30°) = 60° = POil/; .-. MP=-PN' and OM=ON\
i.e. il/'P';
.-. ^=^^; .-. cos POP --r sin POP'; .'. cos (- 60°) = sin 30°.
11. Since A,B,C, are the angles of a triangle, therefore A+B + G = 180°,
sin^= -sin (180° + ^) [E. T. p. 106, Ex. 2]
= -sin(^+P + + ^)= -sin(2^+P + C).
12. sin^=-cos(90° + ^) [E. T. p. 107, Ex. 4]
= -Gos{i{A-hB + C) + A]= -co8i(3A+B + C).
13. cosP = sin(90° + P) = sin{4(^ + P + O)+P} = sini(^ + 3P+O).
14. cos 0= - cos (180° + ) = - cos (^ + P + O + 0) = - cos (^ + P + 20).
15. cos J (P - 0) = sin {90° + i(P-0)} [E. T. p. 107, Ex. 4]
= sin{i(^ + P + 0) + J(P-0)} = sini(^ + 2P).
16. sini(0-.-t)= -cos{90° + i(O-^)} [E. T. p. 107, Ex. 4]
= -cos{4(^+P + 0) + J(0-^)}= -cos4(P+20).
17. In fig. E. T. p. 112, let OP^ and OP^ be two revolving lines, and let
OP starting from OR describe the angle A, and let OP2 starting from OR
describe the angle (- A), then if P1P2 be joined P1P2 will^always be perpen-
dicular to OR; let P^P^ cut OR in M, Then by definition Arts. 81, 132
sm^=^^^,^sm(-^) = -^-^
Now M^Pi is numerically equal to MyP^, but of opposite sign ;
,*. sin A= -sin (- A),
18. With the figure and construction of (7),
^^«^=op;=op;=^^^(-^)-
19. From 18 cos (^ - 90°) = cos (90°-^) = sin ^. [Ex. 1, Art. 110.]
"20. From 17 -sin (^ -90°) = sin (90°-^) = cos A. [Ex. 1, Art. 140.]
5—2
68 EQUALITY OF RATIOS. XXX.
21. cos(f 7r + a)=:C08(7r+i7r + a)= -cos(i7r + a) [E. T. p. 106, Ex. 3]
= sin a [E. T. p. 107, Ex. 4].
22. -sin(f7r + a)= - sin (7r + J7r + a) = sin (^Tr + a) [E. T. p. 106, Ex. 3]
= 008 a [E. T. p. 107, Ex. 4].
23. -cos(f7r-a)= -cos(7r + i7r-a) = cos(47r-a) [E. T. p. 106, Ex. 3]
= sma [Ex. 1, p. 102].
24. sin(|7r-a)= -sin (7r + i7r- a) = sin (jTr- a) [E. T. p. 106, Ex. 3]
= cosa [Ex. 1, p. 102].
25. sin(i7r-a) = cosa [Ex. 1, p. 102]
= sin (iTT + a) [E. T. p. 107, Ex. 4].
26. cos (tt + a) == - cos a [E. T. p. 106, Ex. 3]
= cos (tt - a) [E. T. p. 104, Ex. 2].
27. tan(90°-.4) = «^[^|| = ^4^-^4 [Ex.1, p. 102]
^'' ^ ' cos (90° -J) sin^l L r J
= cot-4.
. sin i4 - sin ( - ^ )
28. *-^=c^=^^^ [17]
29. tan(90o..)=£i|0;ij) = ^^ [E. T. p. 107. E. .]
COS A , ,
= — . — 7 = - cot A.
sin^
30. tana = ^i"-^ = ^^i^ [E. T. p. 104, Ex. 2]
cos a ~ cos (tt - a) ••
= _ sinjTT-a) ^ _ ^^^ _
cos (tt - a) ^ ' "- ' x- J
i sin^ -sin^ sin (180° + ^) r^^ m -.^.^ t^
31. tan A = = = \—- — -( [E. T. p. 106, Ex. 3]
cos^ -cos^ cos (180° + ^) L 1- > J
= tan(180° + ^).,
r%« i. /t \ cos(i7r-a) sin a r-r» ■* ir^.c^i
32. cot ( J TT - a) = -t-tI ( = [Ex. 1, p. 102]
^^ ' sin{j7r-a) cos a l » i- j
= tana.
EXAMPLES. XXXI. Page 112.
1. (i) Since sin ^ = J ; .-. sin A = sin 30°,
and A = nx 180° + ( - 1)** 30°.
Put n=0, 1,-1,-2 successively; and we have 30°, 150°, - 210°, - 330°.
GENERAL VALUE. XXXI. 69
(ii) Since sin ^ = J J2 ; .'. sin A = sin 45^ ;
.-. ^ = wxl80°+(-l)''45°.
Put ri=0, 1,-1,-2 successively; and we have 45°, 135°, -225°, -315°.
(iii) Since sin A = \ ,^/S ; .'. sin A — sin 60°,
/. ^=nxlB0°+(-l)^60^
Putn = 0, 1, -1, - 2 successively, and we have 60°, 120°, -240°, -300°.
(iv) Since ainA=:-h; .'. sin ^ = sin ( - 30°),
and ''^:=nx 180°- (-1)^30°.
Put w = 0,- 1, 1, 2 successively; and we have -30°, - 150°, 210°, 330°.
2. (i) sin ^ = sin 20°; /. ^ = n x 180°+ (- 1)^*20°.
Put w = 0, 1, 2, 3 successively ; and we have 20°, 160°, 380°, 520°.
(ii) sinl9=-i^2; /. sin ^--sin ( - Jtt), ^ = 7t7r- ( - 1)^ Jtt.
Put 71=1, 2, 3, 4 successively, and we have f tt, -Jtt, V t> Y"''^'
(iii) sin ^ = - sin f TT =: sin ( - f tt) ; [vid. Ex. XXX. 17]
.-. e=:mr-{-iy'}'jr.
Put w=l, 2, 3, 4 successively, and we have f tt, ^^tt, \^7r, y-Tr.
3. (i) sin 6/=r -J, and sin (-30°)= -J;
/. sin6> = sin(-30°); /. ^-=n7r + ( - 1)*» (-^tt).
(ii) 2sin2^ + 3sin^ = 2; /. sin^ ^+ ^ sin ^ = 1 ;
.-. sin2^ + |sin(?+i9^=:ff, sin^ + |=±f, sin^ = 4or-2.
The value - 2 is inadmissible ; for there is no angle whose sine is
numerically greater than 1 ;
/. sin (9 = J, but sin30° = i;
.-. sin(9 = sin30°; /. ^ = 7i7r+ (- J)'*i7r.
(iii) sin2^ = cos2^; .-. 2sin2^=l; /. sin^=: ±^ ^2,
^=:ri7r + (-l)»*j7r, and n7r + (- 1)*» ( - Jtt).
In the first case 6 is wir ± J tt as n is even or odd ; in the second case
is nTTijTT as n is odd or even ; .-. 6 = mr:h^Tr.
4. If all the angles have the same sine as 30° they are included in the
general expression n x 180° + ( - 1)** 30°.
Put w = 0, 1, -2, 2, -1 successively, and we have the given angles.
Therefore they have all the same sine.
EXAMPLES. XXXII. Page 115.
1. (i) cos^ = J; .-. cos^ = cosj7r; .*. = 2mr±^Tr.
(ii) tan^ = l; /. tan^ = tanjir; .* d = mr + iir.
(iii) tan (? = - 1 ; /. ' tan 6^ = tan ( - J tt) ; [See Ex. XXX. 28]
.'. = mr- ^w.
70 EQUATIONS. XXXII.
(iv) tan <? = - ^^3 ; /. tan ^ = tan ( - J tt) ; [See Ex. XXX. 28]
(v) cos^ = cosf7r; .*. 6 = 2mr^^w.
(vi) tan^ = tan|7r; .'. d = mr + ^Tr.
2. If all the angles have the same cosine as - 120° they are included in
the general expression for the cosine of - 120° and therefore in the general ex-
pression for the cosine of 120° for cos ( - 120°) = cos 120° [vid. Ex. XXX. 18].
In the general expression 2 x n x 180°±120° put 7i = l, and -1; and we
have the angles.
3. The angle of 60° is in the first quadrant, and the angle of - 120° in
the third quadrant.
The sine and cosine, and cosecant and secant are each positive in the
first quadrant and negative in the third quadrant; therefore the angles of
60° and - 120° cannot have the same sine or cosine, or the same cosecant or
secant. The tangent is positive in the first quadrant and ^\so positive in the
third ; therefore, if the angles of 60° and - 120° have the same tangent they
are included in the p^eneral expression riTr-f 60°. Put n = 0, and -1 succes-
sively and we have 60° and - 120°.
4. If they have the same sine, they are all included in the general
expression n x 180°+ ( - 1)** - 23°, and it will be found that + 157° = 180° - 23°
is not included, therefore it has not the same sine as - 23°.
If they have the same cosine, they are all included in the general
expression 2n x 1S0° ± 23°, - 180° + 23° and 180 - 23° are not included in this,
hence we see they have not the same cosine.
If they have the same tangent they are all included in the general
expression nxl80°-23°; and -180° + 23° is not included in this formula,
therefore they have not the same tangent.
5. (i) From 1. (i) (9 = 2wV ± J tt, that is n x 360° ± 60°,
put w = 0, 1, and - 1 successively, and the four smallest angles are
60° -60°, 300° -300°.
(ii) From 1 . (ii) 6> = titt + J tt, that is n x 180° + 45°,
put « = 0, 1, 2, - 1 successively, and we have 45°, 225°, 405°, - 135°.
(iii) From 1. (iii) ^ = titt - J tt, that is n x 180° - 45°,
put 71 = 0, -1, 1, 2 successively, and we have -45°, -225°, 135°, 315°.
(iv) From 1. (iv) e = mr - Jtt, that is w x 180° - 60°,
put w = 0, -1, 1, 2 successively, and we have -60°, -240, 120°, 300°.
(v) From 1 . (v) (9 = 2w7r =fc f ^, that is w x 360° ± 144°,
put n = 0,l, -1 successively, and we have 144°, -144°, 216°, -216°.
(vi) From 1. (vi) 6> = wtt + f tt, that is n x 180° + 135°,
put n = 0, -1, -^2, 1 successively, and we have 135°, -45°, -225°, 315°.
GENERAL THEOREMS. . 71
EXAMPLES. XXXIII. Page 116.
1. ^ is 07ie of the angles represented by one or the other of the positions
OP^, OP^ of the revolving line in fig. E. T. p. 112. We have two different
M P M P
values for the sine of these angles ; viz. : yrW^ and -7^^ ; these two fractions
are equal in magnitude and opposite in sign.
We also have two different values for the tangent of these angles :
MP MP
riu^ ^^^ nwT^ ' *^®^^ *^^^ fractions are equal in magnitude and opposite
in sign.
2. Since tan ^ — a, ^ is one of the angles represented by one or the other
of the positions OP^, OP^ of the revolving line in fig. E. T. p. 114. We have
MP MP
two different values for the sine of these angles, namely, l^ ^ and 7^5-^ ;
these two fractions are equal in magnitude and opposite in sign.
We also have two different values for the cosine of these angles ; namely
— — ^ and jyp^\ these two fractions are equal in magnitude and opposite
in sign.
3. Since a is the least angle whose sine = <x ; then in figs. i. and ii. E. T.
p. 110, BOP^ in each figure is a as a is positive or negative. In both
OM
figures cos A = ---^ which is positive ; and therefore in the general formula
cos ^ = ±^(1 - sin*-^^), only the positive value is admissible.
4. Since A is the least positive angle with the given cosine, therefore A
is in the first Quadrant; and sin^ is positive. Therefore only tbe positive
value is admissible in any general formula for sin A in terms of cos A ;
.-. sin^=+v/(l-cosM).
EXAMPLES. XXXIV. Page 120.
cos 75° = cos (45° + 30°) = cos 45° cos 30° - sin 45° sin 30°
sJ2' 2 ^2*2 2^2 •
sin 15° = sin (60° - 45°) = sin 60° cos 45° - cos 60° sin 45°
2 V2 2'J2 2^2 '
sin 15° = cos (90° - 15°) = cos 75° (Art. 118)
72 {^+^) FORMULA. XXXIV.
3. cos 15° = cos (45° - 30°) = cos 45° cos 30° + sin 45° sin 30°
X r,Ko sin 75° cos 15° ,. . ^.^x
4. tan 75°= — -^6= ■ -.ro Art. 118),
cos 75° sml5° ^ ^'
= \/^+^ ^ Vir 1 ^ n/^+1 ^ ( v/3 + l)(^3 + l) ^ 4 + 2^3
2 V2 -2 V2 V3 - 1 (^3 - 1) (^3 + 1) ^
=2+^y3.
5. sin (^+£) = sin^ . cos-B + cos^ . sin jB
= sin ^ . J(l - sin2 B) -+ sin B . ^(l - sin^A)
— 4 4a_3 3 — 9il6— 1
cos {A -B) = cos ^ . cos B + sin A .sinB
_3 4 _L 4 3_12 I 12 — 24
6. sin(^ -5) = sin^ .COS J5-cos^ . sin^
= sin ^ . ^(1 - sin2 B) - sin J5 . ^^(1 - sin^ A)
— 3 12_5 4 — 16
— '5"'TTf T^'-g^ — ¥3^*
cos (^ + B) = COS A . cos 5 - sin ^ . sin B
= sj{l - sin2 A) . ^(1 - sin2£) - sin ^ . sin B
— 4 12_3 5—3 3
— Z'TS Z 'TIS— S^'
7. sin {A+B) = sin ^ cos £ + cos ^ sin B
= sin A ^{1 - sin2 J5) + sin5 ^/(l - sin2 A)
~x/5 • ^10 ^ ^10 V5 "* ^50" V 50 " V2 '
/. sin (^ + ^) = ,^ = sin 45° ; /. one value of (^ + 5) is 45°.
8. sin 75° = cos 15° =^^^4^ [Art. 118, and (3) supra],
-V6W2 _ 2^4^94897 + 1.414^136^ J ^3.^^^^^
9. From(2)sinl5<_ ^^^ - ^
4 4
^ ^3-1 ^ ^6-^2
2^2 4
2-4494897 - 1 -4142136 1-035276
4 4
= •2588...
10 tanl5°-'H^-^^^-^ - V3 + 1_V3-1_ 4-2V3
10. ta^l^ -cosl5°- 2^/2 "" 2^72^-^3 + 1 2"^~ ^
= 2 -1-73205 = -2679...
{A -f B) FORMULAE. 73
EXAMPLES. XXXV. Page 121.
1. sin {A-\-B) = sin A cos B + cos A sin B,
sin (A-B) = sin A cos B - cos ^ sin B ;
.-. sin {A + B) + sin (A-B) = 2 sin ^ cos B,
2. sin (^+i^)- sin (.^-5)
= sin A cos £ + cos AsinB - sin ^ cos B + cos ^ sin B = 2 cos A sin 5.
3. cos (A + B)-i- cos (^ - i?)
= cos A cos i? - sin ^ sin B + cos ^ cos B + sin ^ sin 5 = 2 cos A cos B.
4. cos (^ - 5) - cos (A + ^)
= cos A cos 5 + sin i4 sin ^ - cos A cos 5 + sin ^ sin 5 = 2 sin A sin B.
sin (^ + i?) + sin (.4 - P) _ 2 sin ^ cos J5 _
cos(^+B) + cos(^-5) ~2cos^cos.B'~
_ , , ^ sin a sin 8 sin a cos 8 4- cos a sin 8 .
6. tana + tan^ = + — ^ = ^^^ -
^ cos a cos j8 cos a cos p
^8in(a + ^)_ ^^^_j53_
cos a cos /3
_ . ^ sin a sin i3 sin a cos fl - cos a sin 5
7. tan a - tan/3 = ^ = ^^ ^
'^ cos a cos^ cos a cos /3
^sinJa-^_ Art. 153.
cos a cos /3
. ^ ^ cos a siniS cos a cos iS + sin a sin /?
8. cota + tanS^-^ — + — -'^= r-^- ~
^ sin a cos /3 sin a cos j3
cos (a-/3)
sin a cos /3
cos a sin 5 cos a cos 8 - sin a sin 8
9. cot a - tan 8= — ^ = . ^ — -^ ^
'^ sm a cos j3 sin a cos /3
_cos(a + i3)
~ sin a cos /3 *
, ^ sin a cos 5 sin a sin j8 + cos a cos 8
10. tan a + cot /3 = + ^— ^ = ^^— •— ^5 ^
"^ cos a sin/3 cosasin/3
- tan 6 + tan
(sin 6 sin 0\ , /sin sin 0\
COS^ COS0/ * \cos^ cos</>/
tan ^ - tan
_sin(^ + 0) . sin(^-0) _sin(^ + 0)
~cos^cos0 ' cos ^cos<^~~ sin(^-0) *
Art. 153.
Art. 153.
^^08(a-^^ Art. 153.
cos a . sin ^
74 (A + B) FORMULA. XXXV.
_^ tan ^ . tan + 1 _ /sin ^ sin ^\ . fi sin ^ sin 0\
1 - tan 6 . tan <p ~ \cos 6 ' cos <f) ) \ cos 6 ' cos <p/
__ cos{d ~(p) ^ cos{d+<f>)
cos 6 . cos <^ * COS 6 . cos <p
^<^o^ilzA) oo^d-<p). sec (e + <!>),
cos(<9 + 0) \ v'/ V v;
tan^ + cot0 _ /sin ^ cos0\ /cos0 sin ^\
cot 0- tan ^ \cos^ sin 0/ * \sin0 cos 6 J
14.
COS(^-0) cos(^ + 0)
sin cos ^
cos (^ - 0) . sec (0 + 0).
~ cos 6 sin * sin cos ^
cos (^ - 0)
cos(^ + 0)
cot ^ + COt0_ /cos^ COS0\ /cos^ COS0\
cot d - cot ~ \sin sin 0/ " \sin ^ ^ sin 0/
_ sin ^ cos + cos ^ sin sin cos ^ - cos sin ^
"" sin . sin ' sin sin
sin (^ + 0) ^ - sin {0 -</))__ sin (^ + 0)
sin ^ . sin * sin ^ sin ~ sin (^ - 0) '
_ tan ^ . cot + 1 _ /sin ^ cos0 \ /sin ^ cos0 \
tan . cot 0-1 \cos ' sin y ' \cos ^ ' sin J
_ sin (g + 0) sin (^ - 0) _ sin (0 + 0)
cos sin ' cos sin ~ sin (^ - 0) *
l + cot7 . tan 5_ / cosy sin 5\ /cos 7 sin 5\
cot 7 -tan 5 ~\ sin 7* cos 5/ * \sin7 cos 5/
^ sin (7 + a) ^ cos (7 + 3) ^ s in (7 + 3) ^^^^
sin 7 . cos B ' sin 7 . cos 5 cos (7 + 5)
1 - cot 7 . tan S _ / cos 7 sin 5\ /cos 7 sin 5\
cot 7 + tan 5 ~ \ sin 7 * cos dj ' \sin 7 cos 5/
sin7Cos5 sin7Cos5 cos (7 -5) ^' '
tan7 . cot 5- 1 /sin7 cos 5 \ /sin 7 cos5\
tan 7 + cot 5 ~\cos7*sin5 J ' \c0s7 sin 5/
^ s in (7 -3) _^ c os (7 -3) ^ sin (7- a) ^^^ _
cos 7 sin 5 * cos 7 sin 5 cos (7 - 5) '*
tan 7 . cot 5 + 1 _ /sin 7 cos 5 \ /cos 5 sin 7\
cot 5 -tan 7 \cos7'sin5 / * \sin 5 cos 7/
= sm (tH- 3) ^ cos (7 + 8) ^ Bm(v + 5) ^ ^^^ .
COS 7 sin 5 cos 7 sin 5 cos (7+5) ^ '
(A+B) FORMULA. XXXV. 75
cot 5 -cot 7 /cos 5 cos
20. f -I ^
0S7\ /cos 7 cos 5 \ 1/
in 7/ * \sin 7 * sin 5 /
cot 7 . cot d + 1 \sin 5 sin
_ sin (7-5) cos (7-5) _ sin (7 - 5)
= tan (7 -5)
sin 8 . sin 7 * sin 7 . sin 5 cos (7 - 5)
, _ .0^ /sin^a sin^ 8\ sin^ a cos^ j8 - cos^ a sin^ /3
21. tan2a - tan2^=: — . ^ ) = ■■ f — ^
^. \cos^a coa^pj cos^acos^jS
_ (sin a cos ^ + cos a sin /3) (sin a cos j3 - cos a sin /3)
~ cos^ a cos^ j8
_ sin (a + iS) . sin (a - /3)
~ cos^acos^/S
cos2 a cos2 ^ - sin2 a sin^ /3
xo .0^ /cos^a sin^flN
22. cot2 a - tan2 ^ =, . ^ \ ^
'^ \sin2a cos2/3/
sin2 a cos2 ^3
_-(cos a cos /3 + sin a sin /3) (cos a cos /3 - sin a sin p)
sin'^ a cos- ^
_ cos (o - ^) . cos (a + j3)
sin^ a cos*^ /3
tan2a-tan2/3
^^' l-tan2a.tan2^
_ /sin2 a _ sin2 /3 \ ^ / sin^g _ sm2^\
Vcos^a cos^^y ■ \ cos^a cos^^y
_ sin2 a . cos2 ^ - cos^ a . sin^ ^ cos^ a cos^ ^ - sin^ a sin^ /3
cos"^acos2j3 ' cos'^acos^/S
_ sin2 a . cos2 p — cos^ /3 sin^ ^3
~ cos2 a cos2 p - sin2 a sin^ ^
_ (sin a . cos /3 + cos a sin )3) (sin a . cos ^ - cos a sin (B)
(cos a cos j3 - sin a sin /3) (cos a cos /3 + sin a sin /3)
sin (a + 5) sin(a-j3) ^ , r^, j. , ^
= 7 S- 7 S = tan(a + ^ .tan(a-/3).
cos(a + ^) cos(a-^) \ t-f \ t-/
24. sin(a+i3) .sin(a-/3) = sin2acos2/3-cos2asin2^ [Art. 153.]
= sin2 a _ gin2 ^ gin2 ^ _ sin2 p 4. ginS « gin2 ^
= sin2 a - sin2 p
= l-cos2 a- l + cos2/3=cos2^-cos2a.
25. cos (a + ^) . cos (a - /3) = cos2 a cos^ /3 - sin2 a sin2 p [Art. 153.]
= cos2 a - cos2 a sin^ /3 - sin2 /3 f cos- a sin2 p^^ T
= cos2a-sin^'/3 ~"" ' "" [ ^^
= 1 - sin2 a - 1 + cos2 p = cos2 p - sin2 a.
I
26. sin (A - 45°) = sin A cos 45° - cos A sin 45° = -j- (sin ^ - cos A),
76 i^+B) FORMULA. XXXV.
27. sJ2. sin (A + 45°) = ^2 (sin A cos 45° + cos A sin 45°)
= ^ (siny4 + cos^) = sin.4+cos^.
. • ^ /« /cos A 8mA\
28. cos^-sm^^V2.(^^;2---^-j
= y^2 . (cos A . cos 45° - sin A sin 45°)
= J2.cos(^ + 45°).
29. cos {A + 45°) + sin (A - 45°)
= -7^ (cos ^ - sin ^) + -y^ . (sin A - cos ^4) = 0.
30. cos (A - 45°) = sin (90° + A- 45°) = sin (A + 45°) [E. T. p. 107, Ex. 4.]
or, cos (A - 45°) = -7^ (cos A + sin A) = (sin 45° cos A + cos 45° sin A)
= sin (45° + ^).
31. sin(^ + 0) .cos^-cos(^ + 0) . sin ^ = sin (^ + 0-^) [Art. 153.]
= sin 0.
32. sin(^-0) . cos</) + cos(^-0)sin0 = sin(^-0 + 0) [Art. 153.]
= sin 0.
33. cos (e + <f)) . cos ^ + sin (^ + 0) . sin ^ = cos (^ + - 6) [Art. 153.]
= COS0.
34.
35.
tan(^-0) + tan0
1 - tan (0 -(fi) . tan <f>
_/sin(^-0) sin 0\ , /^ sin(^-0) sin 0\
~ \cos(^-0) cos 0/ * \ COS (^ - 0) * cos 0/
_ sin (^ - 0) cos + c os (^ - 0) sin
~ cos (^ - 0) . cos - sin (^ - 0) . sin
_ sin {0-<p + <p) ^ sin ^^^^^ ^
cos (^-0 + 0) cos^
tan(^ + 0)-tan^
1 + tan (^ + 0) . tan
_/ sin (0 -hip) 8m0 \ , /^ sin(<94-0) sin ^ \
~~VCOS(^H-0) COS^y ' V CO8{0 + (f>)' COS0)
_ sin(^ + 0) . cos ^- cos (^ + 0). sing
'~cos(g + 0) .cos^ + sin (^ + 0) . sin
sin (g + 0-g) _sin0_
cos {0 + (f>-0) cos
= tan 0.
{A+B) FORMULA. XXXV. 77
36. 2 sin (a -h i tt) . cos (i^ - J tt)
= (sin a + cos a) (cos /3 + sin /3)
= (cos a cos /3 + sin a sin /3) + (sin a cos /3 + cos a sin /3)
= cos (a - j3) + sin (a + /3).
37. 2sin(i7r-a) .cos(j7r + /3)
= 2 (J-2 . cos a - j-2 sin a) (^ . cos^ - ^-^ • sin/s)
— (cos a - sin a) (cos j8 - sin p)
= (cos a cos ^ + sin a sin ^) - (sin a cos /3 + cos a sin /3)
= cos (a - j8) — sin (a + j8).
38. cos (a + /3) + sin (a - /3)
~ cos a cos ^ - sin a sin ^ + sin a cos )3 - cos a sin /3
= (cos a + sin a) (cos /3 - sin j3)
= 2sin(j7r + a) . cos (i7r + /3).
39. cos (a + p)- sin (a - j3)
= cosa . cos/3-sina . sin^~ sina . cos^ + cosa. sinjS
= (cos a - sin a) (cos /3 + sin /3)
=' (;72 • "°' " " 72 • ''" ") («72 • ''°' ^+7ii • ""^)
= 2 sin ( J TT - a) . cos (i tt - )3).
40. sin nA , cos A + cos n^ . sin A = sin (n^ + A) [Art. 153.]
= sin(M + l)^.
41. cos(n-l)^ .cos^-sin(yi- l)^.sin^=cos {(/i- 1)^+^} [Art. 153.]
= cosw^.
42. sin w^ . cos(n- 1)^ -cosw^ . sin(n-l)^ [Art. 153.]
= sin {nA -{n-l)A\ =sin^.
43. cos {n - 1) ^ . cos (71 + 1) ^ - sin (w - 1) ^ . sin (n + 1) ^ [Art. 153.]
=:cos {(n-l)^4-(n+l)^}=cos2n^.
EXAMPLES. XXXVI. Page 124.
1 tan LI I B)- *^^-^ + ^^"^ - i + i _f_«
^ ' 1 + tan^.tanjB 1 + i.J | ^
78
TAN (A+B).
2. tan{A + B) =
tan A + tan B
1 - tan A . tan A
1 +
1-
XXXVI.
J. "s/3-1
V3
tan 45° - tan 30°
tan 15° = tan (45°- 30°)-
V3
,. , -D^ tan ^ + tan B
tan (A+B)=- — -—
^ ' 1 - tan A . tan B
1-
1 + -
1 + tan 45°. tan 30°
_(V3-1)(V3-1)
3-1
I + tV
= 2-^3.
tan45° = l; /. tan (.4 + i5) = tan 45° ; /. ^ +B = w . 180° + 45°.
1 1
5. tan(^+jB) =
tanyl +tan5
1 - tan A . tan B '
?7l + -
6. cot(^+jB) =
1 - m . -
m
tan 90° = 00 ; /. tan (^ +i^) = tan90°; .-. ^+jB = 7i . 180° + 90°.
cos {A + ^) cos A cos 1? - sin A sin J5
sin (^ + i^) ~ sin A cos jB + cos A sin i>'
cos A cos i^
sin ^ sin B cot ^ . cot ^ - 1
cos A cos B
sin ^ sin i>
7. cot(^-jB) =
COtyl + COt^
COS (^ - B) _ COS ^ COS B + sinA sin jB
sin (4 - ^) ~ sin A cos i? - cos 4 sin 5
COS A cos jB
sin 4 sin JB cot A . cot 5 + 1
cosJB
sin^
cos 4
sin A
8. cot
H)-':
cot B - cot A
COS^
cos ^ + sin _ sin d
"sim^-cos^"
+ 1
cos^
sin^
cot (9 + 1
' f^cot^ *
TAN (A 4- B), XXXVL 79
Go&e ^ ^1.^1
. ^ < -' — ;. - 1 « . ^ cos ^ . — 7^ - sm ^ . ~,^
cot ^ - 1 _ sin^ cos ^ - sm ^ _ J2 J2
cot ^ + 1 ~ COS ^ ~ COS ^ + sin 6 ~ 1 ' ^1 . ^
-. — 3 + 1 -7^ . COS ^ + -^ . sm d
sin ^ ^/2 ,^2
cos ^ COS J TT - sin (? sin i tt _ cos (^ + J tt) _
~sin Jttcos ^ + cos Jtt sin^~ sin (^ + Jtt)"" ^
10. cot((9 + j7r) = tan{47r-((9 + j7r)}=tan(j7r-^) [Art. 140.]
= tan {-(^-J7r)}=: -tan(6>-j7r); [Ex. XXX. 28.]
.-. tan (6 - J tt) 4- cot (<9 + Jtt) =tan (^ - J tt) - tan {d - iir) = 0.
11. tan((9 + j7r) = cot{i7r-(6> + j7r)}=cot(j7r-^) [Art. 140.]
= cot { - (^ - ^tt) }= - cot ((9 - iTT) ; [Ex. XXX. 28.]
.-. cot(<9-j7r)+t^n(l? + i7r) = cot(^-j7r)-cot(<9-47r) = 0.
m 1
U. ^^(^ + P^-i_tana.tan^~ ^ m 1 ~2m-^ + 2m4-i~
m + 1 ' 2wi + 1
13. j5^i^^zi5?J^.=tan{(n + l)0-«0} [Art. 156.]
l + tan(n+l)0.tan?i0 u / v- t-i l j
= tau 0.
tan (/I + 1) + tan (1 - n) </>
14. 1 / , . ix^ X n T^ ^tan { n+ 1 <;6+ 1 - n)<p} [Art. 156.]
1 - tan {n + 1) (f>. tan {l-n)(f> ^^ /-r \ /vj l j
= tan 20.
15. cos a= — = [E. T. p. 73, Ex. 2.]
v/(l + tan^a) Jl + ni^ if. j
[E. T. p. 73, Ex. 2.]
tan a
^(l + tan^a) Jl + m^
similarly cos /3 = . , sin /3 =
Jl + n^ Jl + n'
cos (a + /3) == cos a cos /3 - sin a sin /3
11m
Vl + m2* Jl+n^ Jl + m^' ^l + ri^
_ 1 - mn
Ifi tanra-«\- *55L^_*a:?/3 _ a + l-a+l _ 2
*"• ^ P^ l + tanatan/3~ 1 + («2„X) ~ a^'
I
2 cot (a - /3) = -— A_— = 4 = «'^
'^^ tan(a-/3) 2
80 tan(J^+5). XXXVI.
17. tan7=cot(900-7) = cot(a + ^)
_ 1 1 1 - tan a . tan ^
~ tan (a + /3 ~ tan a 4- tan /3 ~ tan a + tan /3
1-tana. tan^S
EXAMPLES. XXXVn. Page 130.
1. sin 60° + sin 30°= 2 sin i (60° + 30°) cos J (60° - 30°) = 2 sin 45^ cos 15°.
2. sin 60° + sin 20°= 2 sin ^ (60° + 20°) cos i (60° - 20°) = 2 sin 40° cos 20°.
3. sin 40° - sin 10° = 2 cos J (40° + 10°) sin i (40° - 10°) = 2 cos 25° sin 15°.
4. cos Jtt + cos J7r = 2 cos^(j7r + Jtt) cos J(j7r- J7r) = 2 coSy\7rcos jV".
5. cos Jtt-cos J7r = 2 sin J(j7r + Jtt) . sin J (4^^- Jtt)
= 2 sin y\ tt . sin j\ tt.
6. sin 3^ + sin5^ = 2sin4(5yl + 3^)cos J(5yi -3^1) = 2 sin 4tA .cos A.
7. sin7^-sin5^=2cos J(7^ + 5^)sin4(7^-5y() = 2cos6^ . sin^.
8. cos 5A + cos 9^ = 2 cos J (9^ + 5A) cos J (9^ - 5^) = 2 cos 7^ cos 2A.
9. cos 5^ - cos 4^= - 2 sin 4(4^+5^) . sin 4(5^ - 4.1) [E. T. p. 126 iii]
= -2sinf(^)sin4^.
10. cos A - cos 2^ = 2 sin 4(2^ +^) . sin 4(2^ -A) = 2 sin 4(3^) . sin ^A.
11.
d
s in26> + sin 6' _ 2 sin 4(2(9 + ^) cos 4(2i9 - 6) _ ' m i^t7 uua^
cos d + cos~2^ ~ 2 cos * (26 + ^) cos i (2^ - ^) ~ ,, B
^^ ^^ ' cos 1^ cos-
cosf^ ^
sin 2(9 - sin ^ _ 2 cos 4 (26 + ^) sin 4 (2^ - ^) _ cos j ^ _
^^" cos^-cos26'~2sin4(2(9 + ^)sin4(2^-^)"~sinf^~^^ ^ '
sin 3^4- sin 2^ _ 2 sin 4(3^ + 2^) . cos 4(3^ - 26) _ cos \6 _
^^' cos 26 - cos 36> ~ 2 sin 4(3(9 + 2(9) . sin 4(3^ - 2(9) ~ sin \6~^^ * *
14.
sin 6 + sm <
_0 ^ 2 sin 4( ^ + 0) cos 4(^-0) ^ cos4(^-^) ^ ^^^ i /^ _ ^x
cos^-cos0 2sin4(^ + 0)sin4(0-^) sin4(0-^) ^^^ ''
cos ^ + cos _ 2 cos 4(0 + ^) cos 4(0-^) _ cos 4(0-^)
sin 0- sin ^~ 2 cos 4(0 + ^) sin 4(0-^) ""sin 4(0-^)
sin ^ + sin <f) __ cos 6 + cos
cos 6 - cos "~ sin <^ - sin 6 '
= cot 4(0-^);
KATIOS OF TWO ANGLES. XXXVIII. 81
15. cos (60° + ^) + cos (60° - A)
= 2cosi{(60° + ^) + (60°-^)}cosi{(60° + ^)-(60°-^)}
= 2 cos 60° cos A = cos A .
16. cos (45° + ^) + cos (45°-^)
= 2 cos i { (45° + ^) + (45° - A)\ cos J {(45° + ^) - (45° - A)\
2
= 2 cos 45° cos ^ = -^ cos A= ^2, cos A.
17. sin (45° + ^) - sin (45° -A) = 2 cos i (90°) sin i (2.4) = 2 cos 45° sin A
2
= -~ sin ^ = ,^2 . sin ^.
18. cos (30° -A)- cos (30° + ^) = 2 sin ^(60°) sin i(2^) = 2 sin 30° sin A
= sin A .
sing-sin<? ^^ 2cosi(^ + 0).sin^(^-</>) ^ cos^(^ + 0) ^
^^' cos<p-cose 2sini(l? + 0).sini((9-0) sini(^ + 0) ^"''5l^^'V>;-
20.
sin ^ - sin _ 2 cos J (^ + <f>) sin i(d- <t>)
sin ^ + sin "~ 2 sin J (^ + </>) cos ^ (^ - 0)
cosi(^ + 0) sin4(^-0) ^ , ,^
= ■— ? ;^ ^, . — rr^-^, = cot l(d + d>), tan A (^ - 0).
sinJ(^ + 0) cosJ(^-0) ^^ ^^ ^^ ^^
EXAMPLES. XXXVIII. Page 131.
1. 2sin^. cos0 = sin(^ + 0) + sin (^-0).
2. 2cosa.coSj3 = cos (a + /3)+cos (a-j8).
3. 2 sin 2a . cos 3/3 = sin (2a + 3/3) + sin (2a - 3^).
4. 2cos(a + /3). C0B(a-^)=:C0S {(a + /3) + (a -/3)} 4-cos {(a + j8)- (a-|9)}
= cos 2a + cos 2/3.
5. 2 sin 3^ . cos od = sin (3^ + 5^) + sin (3^ - 5^)
= sin 8^ + sin ( - 26) = sin 8^ - sin 20.
6. 2cos43^.cos4^=:cos(i3(9 + J^) + cos(i3(9-i^) = cos2^ + cos^,
7. sin 4:6 . sin 6 = \{2 sin 4^ . sin 6) = l {cos (4(9 - ^) - cos (4.6 + 6) }
= 4 (cos 3^ -cos 5^).
fcos|(9.sin^(?=:J(2cosf ^.sin|(9) = J{ain(f(?+|^)-sin(f^-f^)}
= J (sin 4^- sin ^).
2 cos 10° sin 50° = sin (50° + 10°) + sin (50° - 10°) = sin 60° + sin 40°.
L. T. K, 6
82 RATIOS OF TWO ANGLES. XXXVIII.
10. COS 45° sin 15^ = J (2 cos 45° sin 15°) = i { sin (45° + 15°) - sin (45° - 15°) j
= i (sin 60° -sin 30°).
11. 2 cos 26 cos e = cos (28 + 0) + cos [26 -6) = cos SO + cos 0,
2 sin 40 sin = cos [40 - 0) - cos (4^ + 0) = cos 30 - cos 50 ;
.-. 2cos2^cos^-2sin4^sin^=cos3^ + cos^-cos3^ + cos5^
= cos^ + cos5^ = 2cos J(5^ + ^) . cos 1(50-0)
= 2 cos 3^. cos 2^.
12. sinf ^.cosJ^ = J{sin(|^ + i^) + sin(|(?-i^)}=4(sin3^ + sin2^),
sin|(9.cos|(9 = i{sin(|<? + f^) + sin(|(9-f^)}=i(sin6^ + sin3^)
/. sin I ^ . cos J ^ - sin I ^ . cos f ^ = J (sin 3^ + sin 20 - sin 6^ - sin 3^)
= J (sin 20 - sin 6(9) = J {2 cos J (20-\-Q0) sin J (20 - 60) }
= cos 4:0 sin ( - 2^) = - cos 40 . sin 2^.
13. sin S0 + sin 2(9 = 2 sin J (3(9 + 20) cos J (3(9 - 20) = 2 sin f ^ cos ^ (9 ;
/. sin S0 + sin 2^ + 2 sin f ^ . cos J ^ = 2 cos J (sin f ^ + sin | ^)
= 4cosJ^.sini(|(9 + f^) . cosj (|(9- |^)
= 4cos J^sin2^cosi^ = 4cos2 J^sin2^.
14. sinJ^^.sinJ(9 = J{cos(V-^-J^)-cos(V-<9 + i<?)}=i(cos^(9-cos3^
sini^.sin|(9 = J {cos (J(9-|(9) -cos(J^ + |(9)} =i (cos ^-cos |^) ;
/ sin-V-^ . sin J^ + sin J^ . sin|^= J (cos|^-cos 3^ + cos^-cosf ^)
= i (cos - cos 3^) = 2 sin J {S0 + 0) sin J (3(9 - ^) = sin 20 . sin (9.
MISCELLANEOUS EXAMPLES. XXXIX. Page 132.
1. tan(a + ^) = /"f- + ^^^"^ =^41-^1-^1.
^' 1 -tan a. tan /3 l-J.J f
But tanj7r = l; .*. tan (a + j8)=:tan Jtt ; .-. a + /3=?i7r + Jtt.
Wlien ?i = 0, a + /3 = j7r.
m-1 1
^ , , ^, tan a 4- tan B m 2m - 1
3. tan a + /3 =1 — ; — ^, = —
^' 1 -tan a. tan ^ m-1 1
m * 2m - i
(m-1) (2m - 1) + m _ m (2m - 1) - (m - 1) _
~ m (2m - 1) - (m - 1) ~ m (2m - 1) - (m - l) ~ *
But tanjTT — 1; .'. tan (a + i9) = tan Jtt;
.-. a + j3 = n7r + j7r, when w = 0a + /3 = j7r.
RATIOS OF TWO ANGLES. XXXIX. 83
cos a - cos 5a _ 2 sin J (a + 6a) . sin I (5a- a) _ sin 2a _
* sin a + sin 5a ~" 2 sin J (a + 5a) cos J (5a - a) cos 2a
sin 5x - sin Sx _2 cos J {6x + 3x) . sin J (ox - Sx) _ sin x _
cos ox + COS 3x ~ 2 cos J (5a; + Sx) cos J (5a; - 3a;) ~~ cos x ~~
6.
cos^+co8 3^ _ 2cosi(A-\-dA) ,GOsi(SA-A) _ cos 2A cos J.
cos 3^ + cos 5A ~" 2 cos J (SA + 5 A ) cos ^ (5^ - 3^) "~ cos 4A cos A
_C08 2^
"~ cos 4^ *
sin 3a; - sin a; _ 2 cos J (3a; + x) . sin J (3a; - ^) _ s in a; _
cos 3;r + cos a; " 2 cos J (3a; + a;) . cos ^ (3a; - a;) " cosa;~ '
sin 3a; + sin a; 2 sin J (3a; + a;) cos § (3a; -a;) _ cos a; _
cos 3a; - cos a; ~" - 2 sin ^ (3a; + x) sin ^ (3a; - a;) ~ - sin a; ~ '
sin 3a; - sin x sin 3a; - sin a ;
cos 3a; + cos x cos 3a; - cos x
1 tan-a;-l
= tan X - cot X = tan x -
tan X tan x
2(l-tan2a;) 2 ^ ^^
= 777 =^ - * ^- = - 2cot 2a;.
2 tan X tan 2a;
^ (sin 4^ - sin 2 A) (cos ^ - cos SA )
• (cos 4A + cos 2^) (sin A + sin 3^)
_2 cos ^ (4^ + 2 A) . sin ^ (4^ - 2^) . 2 sin |(3^ + ^) . sin ^ (3^ - A)
~ 2 cos i (4^1 + 2A) . cos J (4.4 - 2^) . 2 sin i (3^ + ^) . cos J (3^ - A.)
4 cos 3^ . sin ^ . sin 2A . sin A sin^ A
4 cos 3^ . cos A . sin 2 A . cos ^ cos^ A
= tanM.
9. 2sin2a . cos a = sin (2a + a) + sin(2a-a) = sin3a+sin a,
2 cos 4a . sin a = sin (4a + a) - sin (4a - a) = sin 5a - sin 3a ;
.*. 2 sin 2a . cos a + 2 cos 4a . sin a=: sin 3a + sin a + sin 5a - sin 3a
= sin 5a + sin a.
10. 2 cos 2a . cos a = cos (2a + a) + cos (2a - a) = cos 3a + cos a,
2 sin 4a . sin a = cos (4a - a) - cos (4a + a) = cos 3a - cos 5a ;
2 cos 2a . cos a - 2 sin 4a . sin a = (cos 3a + cos a) - (cos 3a - cos 5a)
= cos a + cos 5a = 2 cos 3a cos 2a
= 2 cos 3a cos 2a.
11. tan 5A - tan SA - tan 2A
= tan (3^ + 2A) - (tan SA + tan 2A)
ta.nSA+iSi,n2A ,^ ^ , ^ ^ ,^
-^ — I — 5-7—1 — iTT - (*an SA + tan 2A)
1 - tan SA . tan 2A ^ '
6—2
13
84 RATIOS OF TWO ANGLES. XXXIX.
_ tan 3^ + tan 2yii - (1 - tan 3^ . tan 2A) (tan SA + tan 2A)
1 - tan 'dA . tan 2 A
_ (tan SA + tan 2 A ) . tan 3^ . tan 2 A
~ 1 - tan SA . tan 2 A
tan 3^ + tan 2^ , ^ . , _ .
= ,— 7 — 5-.— -r — ^r-. . tan 3^ . tan 2A
1 - tan 3^ . tan 2 A
= tan 5 A . tan SA . tan 2 A [Art. 156.]
12. 4sin(^ + 0).cos((?-0) = 3, 4cos (<9 + 0) . sin (i?- 0) = 1.
By addition 4 { sin (d-\-<p) . cos (^ - 0) + cos (^ + 0) . sin (^ - 0) } = 4;
.-. sin {((9 + 0) + (^-0)}=l; .-. sin2^ = l; .-. 2<9 = 90°; /. 19 = 45°.
By subtraction 4 { sin {d + <f>) . cos (^ - 0) - cos (^ + ^) . sin (^ - ^)} = 2;
.-. sin {((9 + 0)- ((9-0)} =J; .-. sin 20 = J; .-.20 = 30°; .*. = 15°.
sin A . sin 2 A + sin 2 A . sin 5yi + sin SA . sin 10^
cos ^ . sin 2^ + sin 2yi . cos 5^ - cos 3^ . sin 10^
_ sin 2A (sin A + sin 5^) + sin SA . sin 10^
~" sin 2A (cos A + cos 5^) - cos 3^1 . sin 10^
_ 2 sin SA . cos 2A . sin 2^ + sin SA . sin 10^
~ 2 cos 3^ . cos 2A . sin 2A - cos 3^4 . sin 10^
_ sin 3^ (2 cos 2 A . sin 2 A + sin 10^)
~ cos SA (2 cos 2A . sin 2A - sin 10^)
_ sin SA (sin 4^ + sin 10^) _ sin 3^ (sin 4^ + sin 10.4)
~ cos SA (sin 4^ - sin 10^) ~ cos 3^ (sin 10^ - sin \A)
_ sin 3^ . 2 sin 7^ . cos SA _ sin 7^ _
~~ cos 3^ . 2 sin SA . cos lA ~ cos 1A~
,. . ^+J5 . A-B
14. tan — 2 tan — 2—
sin^(^ + ^) sin^(J-^ )
~cos4(^ + ii) cosJ(^-£)
_ sin i(^ +^) . cos \[A-B)- cos |(^ +^) . sinj^ (^ - B)
cosi(J+jB).cosi(^-i?)
_ sin \\^^-^^)-\{^-^)\ _ sini?
~ cos 4 (^ + 5) . cos \[A-B)~ cos 4 (^ + ^) . cos J (^ - 5)
2 sin J5 2 sin J5
*2cos J(^+i?). cos4(^ -5) cos^+cosi?*
[Art. 161.]
2 cosec 2A = —
EXAMPLES. XL. Page 136.
2 2 1
sin 2 A 2 sin A . cos ^ sin A . cos -4
= sec^, cosec^. [Art. 164, (1).]
MULTIPLE ANGLES. XL. 85
cos ecM _ 1 ^ / _1 \ _ 1
= ci [Art. 164. (4).]
= sec2A,
2 - sec' A f 1 \ 1
4. oos2 ^ (1 - tan" ^) = 0082 ^ f 1 - ®"'^^^ = cos^ A - sinM
\ COS •" /
= cos2^. [Art. 164, (2).]
.^A COS 2^ cos^^-sin^^ ..,..,,.
5- '""2^ = sTrr2-2=2¥iir2T^o-r^ [Art. 164. (2). (D.J
cos2 A
sinM~ cot2^-l
" 2 sin ^ cos ^ 2 cot A
cos^^
2 tan J5 2 tan B 2 sin i^
cos2 j5 = 2 sin 5 . cos B
' l + tan^^ sec^^ cos 5
= sin2B. [Art. 164, (1).]
7, tanJ5 + cot J5 = tani^+- - =
tsinB t&YiB
sec^B coaB 2
sin B sin 5 . cos^ B 2 sin B . cos B
= 2cosec25. [Art. 164, (1).]
cos£
2
' sin 2B
sin^ B cos^B-ain^B
^ l~tan'-7i cos^B cos2^~"~ . .
o. rrr — T~r3 = j „ = j =cos^ jK - sm^ B
1 + tan^ B sec^ ^ 1
cos21^
= cos2B. [Art. 164, (2).]
n i. T» ^ X. cos 5 sin B cos2 B - sin2 B
9. cot B - tan B = - — =: ^ =
sin B cos B sin ii cos B
cos 2B 2 cos 2B 2 cos 22?
sin B . cosB 2 sin Z? .cos B sin 2i>*
= 2cot2Z?.
86 MULTIPLE ANGLES. XL.
1
cot2 B + 1 _ cosec^ B __ sin^jB __ 1
^^' cof^B-i ~ cos'^ ^ _ 1 ~ cos^B-sin^-B "" cos^ B - sin^ B
sin^B " sin2 B
= J— = sec2B. [Art. 164, (2).]
cos 2B
11. (sinJ(? + cosi^)2 = sin2J(9 + cos2Jl9 + 2sin J^cosi^
= l-f 2sinJ^cosi<9 = l + sin^. [Art. 164, (1).]
12. (sinJ^-cosJ^)2 = sin2J(9 + cos2i^-2sin4^.cosJ<?
= l-2sinJ^.cosi^=l-sin^. [Art. 164, (1).]
13. cosH^(l + tani^)2 = ^cosi^ + cosi(?.?^|^Y = (^^^
= cos2i^ + sin2J^ + 2sinJ^cosi^
= l + 2sinJ^cosi(9 = l + sin(?. [Art. 164, (1).]
14. sin2J<?(coti^-l)2=^sini(?.^^i^-sini(9^ = (cos J ^ - sin i <?)2
= cos2 J^ + sin2 J^ - 2 sin i^ . cos ^6
= l-2sini<9.cosJI? = l-sin^. [Art. 164, (1).]
/sinij y
/ tan^ + i y^/ cos^g^ \ / fiinp + cos^^ Y
■*'^- VtanJ^-1/ ~l sini^_^ I Vsini^- cosi(?y
\ cos J ^ /
_ sin2 ^(^ + cos2 ^^ + 2 sin ^g . cos 1^
~sin2^^ + cos2j^-2sini^.cos i^'
_ 1 + 2 sin i ^ . cos ^e _ 1 + sin ^
~ 1 - 2 sin 4 ^ . cos fe ~ 1 - siiT^ *
16 _jmg_ ^ 2 sin ^^ cos j^ ^ sjn||^ [Art. 162, (2).]
^^- l + cosi8 2cos2J.i3 cosj^ ^^ ^ ' ' ^ •*
17 _^1^^ = 2sinJ^^^ ^ ??^^ = cot i/3. [Art. 162, (2).]
•*•'• 1-cos^ 2sin2i^ smi^S ^^ •• " ' ^
18 l-cos^^2sin2^/3^ [Art. 162, (2.)]
■*-°- 1 + COS^ 2C0S2J^ ^^ L > \ /J
19- ^5^=(s-i-^+0^^"^^ + ^ = ^^"^'i^- [Art. 162, (2).]
„ ^„ 1 cosiS l-cosfi 2sin2i«
20. cosec ^ - cot ^= ^-j^ - ^-j^ = -^5^ = 2 sin i^ . cosl^
=^i^=tanjj8. [Art, 162, (2).]
21.
22.
23.
24.
25.
26.
MULTIPLE ANGLES. XL. 87
cos 2x _ cos^ X - sin^ x _ (cos x + sin x) (cos x - sin a;)
1 + sin 2x ~ sin^a; + cos^x + 2 sin x . cos ar ~" (cos x + sin a;)^
sin X
cos a; - sin x cos a; 1 - tan x
cos a: + sin a; ^ sin a: 1 + tan x '
cos a;
cos X _ cos^ J a? - sin2 J a;
1 - sin a: ~ sin^ J a; + cos^ \x-2 sin J a; cos J a;
_(co8 Ja^ + sinja;) (cos Ja;- sin Ja;) _ cos ^x-ein^x
~ (oosjaj + sin Ja;)2 ~ cosja^ + sin Ja;
sin \x
cos \x 1 - tan J x
sin Ja; ~ l + tanja;*
1-
1 +
cos J a;
cos^Jaj-sin^Ja;
1 + sin X ~ sin^ ^x + cos^ \x+2 sin J x cos J a;
_(cos4a; + sin \x) (cos 4a;-sin Ja;) _cos Ja;-sin Ja;
~ (cos^x + sin Ja:)2 "" cos J a; + sin J a;
cos \x
sin J a: _cotJa;-l
"" cos ^x ^~ cot \x-\-\'
-^-T" +1 ^
sm \x
_ cos2^a;-sin2 Ja;
1 - sin a; ~ cos^ ^a; 4- sin^ Ja; - 2 sin ^ x . cos J a;
_ (cos 4 a; + sin J a;) (cos Ja; - sin Ja;) __ cosjar + sin Ja;
~" (cos J a; - sin J a;)2 "" cos ^ a; - sin J a;
cosia; ^
^ — I- 1
_ sin j^a; cot J a; + 1
"cosja; "cotja;-!'
sin^a;
1 + sin a; + cos a; _ (l + cosa;)4-sina; __ 2cos2 Ja; + 2 sinja^cos Ja;
1 + sin X - cos x~ (1- cos x) + sin a; "" 2 sin^ Ja; + 2 sin Ja; cos ^x
cos i a; (cos i a; + sin Ax) cosia;
sm Ja: (cos^x + sin Ja;) sin^a;
cos^ a 4- sin^ a _ (cos o + sin a) (cos^ a + sin^ a - sin a cos a)
cos a + sin a ~ cos a + sin a
= cos^ a + sin^ a - sin a cos a = 1 - sin a cos a
=:i (2 - 2 sin a cos a) = i (2 - sin 2a). [Art. 164, (1.)]
88 MULTIPLE ANGLES. XL.
cos' a - sin^ a __ (cos a ~ sin a)(cos2 a + sin^ a + sin a cos a)
cos a - sin a ~~ cos a - sin a
= cos^ a + sin^ a + sin a cos a = 1 + sin a cos a
= J (2 + 2 sin a cos a) = i (2 + sin 2a).
28. cos* a - sin* a = (cos^ a + sin^ a) (cos^ a - sin^ a)
= cos2 a - sin2 a = cos 2a. [Art. 164, (2).]
29. cos^ a + sin* a = (cos^ a + sin^ a) (cos* a - cos^ a sin^ a + sin* a)
= (cos* a - cos^ a sin^ a + sin* a)
= {(cos^ a — sin^ a)2 + sin^ a cos2 a}
= cos2 2a + sin2 a cos'^ a [Art. 162, (2). ]
= J {4 cos^ 2a + (2 sin a cos a)-} = J (4 cos^ 2a + sin^ 2a)
.= J(4cos2 2a + l-cos22a) = i(l + 3cos2 2a).
30. cos^ a - sin^ a = (cos^ a - sin^ a) (cos* a + cos^ a sin* a + sin* a)
= cos 2a {(cos2 a + sin2 a)'^ - sin* a cos* a} [Art. 164, (2).]
= cos 2a ( 1 - sin* a cos* a) = cos 2a - cos 2a sin* a cos* a
= J {4 cos 2a - cos 2a (2 sin a . cos a)*}
= J (4 cos 2a - cos 2a . sin* 2a)
= J {4 cos 2a - cos 2a (1 - cos* 2a)} = J (3 cos 2a + cos' 2a)
= J {(3 + cos* 2a) cos 2a}.
sin 3/3 cos 3j8 _ sin 3/3 cos /3 - cos 3/3 sin /3 _ sin 2/3
sin /3 cos/3 ^ sin /3 cos /3 ~~ sin /3 . cos j8
_ 2 sin 2/3 _ 2 sin 2/3
2 sin /3 . cos /8 sin 2/3
= 2.
cos 3j8 sin 3/3 _ cos 3/8 cos /3 + sin 3j8 sin /3 _ cos 2/3
■ sin /3 cos ^ ~ sin /3 cos /8 ~ sin /3 cos /3
2 sin /8 cos /3 sin 2/3
sin 4)3 2 sin 2j8 cos 2)3 ^ ^^ , » ^ ^.>i .-.^ -.
33. -^-7i^ = ^--oo — ^ = 2 cos 2^. [Art. 164, 1.]
sm 2/3 sm 2/3 ^ ' ^ ^ J
. Bin5j8 cos 5/3 _ sin 5/3 cos /3 - cos 5/3 sin /3 _ sin 4^8
sin /3 cos /3 ~~ sin /3 . cos /3 ~~ sin /3 cos j3
_ 4 sin 2/3 cos 2/3 _ 4 sin 2/3 cos 2j8
2 sin /3 cos /3 sin 2^3
= 4 cos 2/3.
sin -/y ^ cos y\ TT _ sin ^^^ tt . cos yV ^ - cos ^^ tt . sin ^V 71
sin ^ IT cos y^ TT sin jV ^ • cos y^^ tt
MULTIPLE ANGLES. XL. 89
_ sin (y^tt - 1^5 tt) _ 2 sin ^ TT
~ sin j^^ TT . cos yV 't 2 sin ^^ir . cos iV ^
2 sin J TT _ 4 sin ^ TT . cos ^ TT
~ sin^TT ~ sin Jtt
= 4cosJ^7r = 4xi ^3 = 2^3.
36. tan(45° + ^)-tan(45°-^) = |-t^ - ^"^''l [Art. 156.]
^ ' ^ ' 1 - tan ^ 1 + tan A
_ (l + tanyl)2-(l-tan.4)2 _ 4tan^
"" 1 - tan*^ A ~~ 1- tan^ A
= 2,:r^^i^=2tan2A. [Art. 163, (5).]
l-tan^^
37. tan (45° -.4) + cot (45°-^)
38.
= tan(45°-
^^"^tan(45°-
1-tan^
^)~l + tan^"^
1
1 - tan A
1 + tan^
1-tan^
1 + tan A
~ 1 + tan A 1 - tan A
(1-tan^)
2 + (l + tan^)'^
^_2(l + tan2^)
1-
tan2^
l-tan2^
26ec2^
2 1
• C0S2^
~ sin2 A ~
cos^A
2
~l--tan2^
cos2^-sin2^
_ 2 _,
sec 2A,
cos 2^
tan2(45° + ^)-l
tan2(45° + J) + l
8in2(45° + ^) . sin^
!(45° + ^)-cos2(45° + ^)
cos2(45° + ^)
cos2(45° + ^)
sec2(45° + ^)
1
cos2(45° + il)
= sin2 (45° -{-A)- cos2 (45° + A)
= { sin (45° + ^ ) + cos (45° + A)\ { sin (45° + ^) - cos (45° + A) \
/cosA+sinA coSi4 - sin^\ /cos-4 +sin^ cos /I- sin ^>
- V ;/2 + J2 ; V V2 J2 ;
4 sin A . cos A
V2 . \/2~
= 2 sin A cos A = sin 2 A ,
90 MULTIPLE ANGLES. XL,
sec A + tan A _ cos A cos A _ 1 + sin ^ _ 1 + 2 sin J ^ . cos J ^
sec^-tanyl~ 1 sin^i ~ 1 -sin^ "" 1 -2 sin J^ . cosj^
cos A cos A
1 2 sin J^ cos J^
cosH^^ cos^^ secH^ + 2tan^^
2 sm ^ ^ . cos iA~ sec'^ \A — 2 tan J A
cos2 J ^ cos2 J ^
_ 1 + tan^^+^tanj ^ _ (l + tani^)^
~ iTtan^ 4^-2 tan 1^ ~ (1 - tan \Af
1 + tanJ^
_ l-tanj^ _ tan (45°+ U) rxrt 156 1
- l-tan|^ - tan (45° -J^) ^^^** ^^^'J
1 + tanJ^
= tan (45° + i^) . cot (45° - i^).
cos A - sin A
cos (v4 -+- 45°) J2 _ cos ^ - sin ^
cos (^ - 45°) "~ cos ^ + sm -4 ~ cos ^ + sin ^
V2
(cos ^ - sin AY _ cos2 ^ + sin^ ^ - 2 sin ^ . cos A
cos^ A - sin2 ^ ~ cos 2 A
1 - sin 2 A 1 sin 2^
cos 2A cos2 ^ QQg 2^
= sec 2 A - tan 2 A,
41.
sinB + sin2jB sini^ + 2 sin5 .cosB .
1 + C0SE + C0S2B - l + cos5 + 2cos2£-l ^^'*- ^^*' <^^' ^^^•■'
sin B (1 + 2 cos jB) sinB ^ „
= _L — ' = ^ = tan B.
cosB(l + 2cosB) cosZ?
sin2B-sinJ5 2sin B .cosB-sinB
42- l-cosi? + cos2i? = l-cosB + 2cos^B-l [Art. 164, (1U3).]
sin B (2 cos ^ - 1) sin JB ^ ^
_ V ^^ ( = ^ = tan JB.
cos B (2 cos B-1) cos jB
EXAMPLES. XLL Page 139.
sin 3^ 3 sin ^ - 4 sin* A ^ . , „ .
__ — ^ = . — = 3-4 sin2 A
sin A sin A
= 3-2(l-cos2^) = 2cos2^ + l. [Art. 164, (4).]
MULTIPLE ANGLES. XLL 91
2 cos 3^ 4 cos^ A -3 cos A . « ^ *>
, = = 4cos2^-3
cos A cos A
= 2(1 + cos 2^) -3 = 2 cos 2.4-1. [Art. 164, (4).]
3 sin ^ - sin 3^ _ 3 sin A - (3 sin A -4: sin* A) _4: sin* A _ 3
cos 2 A + 3 cos A~ A cos* A -S cos ^ + 3 cos ^ ~ 4 cos* ^ ~
cos SA 4 cos* ^ - 3 cos A
4. cot 3^ = -^ — -. = -^—. — - — . . „ ■ .
sin SA 3 sin A - 4: sin** A
Divide both numerator and denominator by sin*^.
4 cot A ——
,^ . siYi^A 4cot*-4 -3cot^ .cosec^^
cot 6 A = o = o ^-^ — —A
3 , 3cosec2u4-4
— 4
sin^ A
4 cot* ^ - 3 cot ^ (1 4- cot2 A)
3(l + cot2^)-
_ cot* ^ - 3 cot ^
■" ~3 cot2"I -T" *
sin 3^ - sin ^ _ 3 sin ^ - 4 sin* ^ - sin ^
cos3^ + cos J[ ~ 4cos*^ -3cos^ + cos^
_ sin ^ (2 - 4 sinM) _ sin ^ (2 - 4 sin'-* A)
~ co8A(4:Coa^A-'2) ~" cos^ (2-4sin2^)
sin^ ,
= T = tan^.
cos^
^ sin SA - sin ^ 2 cos 2 A sin ^ sin ^ ,
Or ^-— 7 = -. = = tan A .
cos SA + cos A 2 cos 2 A cos A cos A
sin 3^ - cos SA __ 3 (sin A + cos ^) - 4 (sin* A + cos* A)
sin ^+ cos ^ "" sin ^ + cos ^
= 3-4 (sin^ A + cos^ ^ - sin ^ cos ^)
= 4 sin ^ . cos ^ - 1 = 2 sin 2^ - 1.
sin SA + cos 3^ _ 4 (cos* A - sin* ^) - 3 (cos .1 - sin ^)
cos A - sin A ~ cos A - sin A
= 4 (cos^^ + sin2 ^ + sin ^ cos ^) - 3
= 1 + 4 sin ^ . cos ^ = 2 sin 2^ + 1.
^1 1
Q I
' tan 3^ - tan A cot A - cot 3^
1 1
"~ sin SA sin A cosJ^ cos 3^
cos 3^ cos A sin A sin SA
cos A cos SA sin A sin SA
- +
[Art. 105, iii.]
sin 2^ sin 2^
cos A cos 3^ + sin A sin 3^ _ cos 2 A
sin 2 A ~ sin 2^
= cot2^.
92 MULTIPLE ANGLES. XLI.
f'S sin A - sin SA
SAy _ | 3 sin ^ - (B sin^ - 4 sin^ A) }^ _ sin«_^
SaJ ~ ( 3 cos^+4cos3^ -3cos^ ) ~ cos^ A
= (^£Y = ( J-^^IT- [Art. 164, (4). (3).]
Vcos2^y \l + cos2.4/ •■ » V / V /J
3 cos A + cos
3
Divide both numerator and denominator by cos 2^.
1 - cos 3^ _ 1 + 3 cos A -A cos^ A
1 - cos A ~ 1 - cos A
Divide numerator by denominator.
l-cos3^
1 - cos A
= 1 + 4 cos ^ + 4 cos2 A ==(1 + 2 cos A)^,
MISCELLANEOUS EXAMPLES. XLIL Page 140.
sin A + cos A _ sin^ A + cos- A +2 sin A .cos A
cos A - sin A ~ cos- A - sm'-^ A
1 -f- sin 2^ 1 sin 2 A
cos 2^ cos 2 A cos 2^
= 8ec2i4 + tan2^.
tan 4^ + 1 _ sin J^ + cos J^ _ sin^^^ H-cos^ J^ + 2 sin Ji4cos J^
" 1-tan J^ ~ cos Jil -sin^^ ~ cos^J^ - sin^^^
1 + sin^ , . .
— — = tan A + sec A .
cos J^
3. 2 sin (71 + 1) a . cos (n - 1) a - 2 sin 2a
= sin 27ia + sin 2a - 2 sin 2a = sin 2?2a - sin 2a
= 2 cos (n + 1) a . sin (n - 1) a = 2 sin (n-1) a cos (n + 1) a.
sina + sin^ ^ 2 sin i(a + ^) . cos |(a-^ ) ^ sin^(a + ^) , ^
^' cosa + cos^ 2cosi(a + /3).cosi(a-^) cos i(a + /3) ^^ '^^^
5.
cos 2a + cos 12a 2 cos 7a. cos 5a cos 5a
cos 6a + cos 8a 2 cos 7a . cos a cos a
cos 7a - cos 3a - 2 sin 5a . sin 2a sin 5a
cos a - cos 3a 2 sin 2a . sin a sin a '
cos 2a + cos 12a cos 7a -cos 3a cos 5a sin 5a
cos 6a + cos 8a cos a - cos 3a ~ cos a sin a
cos 5a sin a - sin 5a cos a sin 4a 2 sin 4a
cos a sin a sin a cos a sin 2a
cos 2a + cos 12a cos 7a - cos 3a ^ sin 4a ^ sin 4a ^ sin 4a _
. — __ : 1_ — . — . 1- 2 . = — 2 -. h 2 ~. = 0.
cos 6a + cos 8a cos a - cos 3a sin 2a sin 2a sin 2a
MULTIPLE ANGLES. XLIL 93
6. Since A = 18° therefore 2A = 36° and 3^ = 54°.
2A + SA= 90° therefore 2A is the complement of SA ;
and .-. sin 2A = cos 3^. [Art. 118. ]
Because sin 2^ = cos 3^,
therefore 2 sin A cos A = 4: cos^ A -3 cos A ;
divide both sides by cos 4, •'• 2 sin A = 4: cos^ ^ - 3 ;
.-. 4sin2^ + 2siny(-l = 0.
From this quadratic we have &mA=^(- 1 ±^5),
i.e. sinl8° = i(-l±^/5).
But as the angle of 18° is in the first quadrant, its sine is positive,
.-. sinl8°=:J(V5"-l).
sin a + sin ^ + sin (a + )3)
' sin a + sin ^ - sin (a + /3)
_ 2 sin ^ (g + i3) . cos ^ (a - ^) + 2 sin |(a + ^) . cos ^ (a + /3)
~2sin J(a + /3) . cos J (a-/3) -2sin J (a-f /3) .cosi(a + /3j
^co s^(a-^) + cosi(tt + /3)^2 cos^acos^/3 ^^^^ ^^^ ^
cosJ(a-^)-cos4(a + /3) 2sin4asinJ/3 ^ * ^^*
8. sin 2^ .sin 25 = 4 {cos 2 (^ - J5) -cos2 (^ + J5)}
= i[l-2sin2(^-jB)-{l-2sin2(^+5)}] [Art. 164, (4).]
= 8m^AA-B) - sin2 (A-B),
9. cos 4^ = 2 cos2 2^ - 1 = 2 (2 cos^ ^ - 1)2 - 1 [Art. 164, (4).]
= 2(4cos'*^-4cos2^ + l)-l = 8cos4^-8cos2^ + l.
sin 50° cos 50° sin^ 50° + cos2 50° 2 2
J-**" rrvo '
COS 50° sin 50° sin 50° cos 50° 2 sin 50° cos 50° sin 100°
2 2
. = 2secl0°
sin (90°+ 10°) cos 10°"
[E. T. p. 107, Ex. 4.]
11. 4sin^ .sin(60° + ^)sin(60°-^)
= 4 sin A (sin2 60° - sin2 ^) = 4 sin ^ (| - sin2 A) Examples XXXV. 24.
= 3 sin ^ - 4 sin3 A =sin 3 A, [Art. 167.]
12. (coiiA~t^niAr=('^\4^'^^!^X=('^^
^ ^ ^ ' \amiA cosily V sin^^ . cosj^ /
_/ CQS^ y / 2co8A \
~ \sin J^ .cos J^y ~ \ sin^ /
s^sin J^ .cos J^y ~ \ sin^ /
cot i4 - 2 cot 2A = -
sin2^
cos A 2 cos 2A
' sin A sin 2^
_2 cos2^4 2(co82^-sin2^)
2sin A cos ^ 2 sin ^ . cos A
94 MULTIPLE ANGLES. XLII
~ sin A cos A ~ cos A '
/. (coti^-tani^)2(cot^-2cot^) = — 7-— J- .- -= 4-^— =4 cot ^.
^ ^ ^ / V gin^ ^ cos A sin ^
13. cos 3a - sin ^ . sin 5a - cos 7a
= cos 3a - cos 7a - sin /3 . sin 5a = 2 sin 5a . sin 2a - sin j8 sin 5a
= sin 5a (2 sin 2a - sin p) ,
sin 3a + sin ^ . cos 5a - sin 7a
= sin 3a - sin 7a + sin /3 . cos 5a = - 2 cos 5a sin 2a + sin j3 cos 5a
= - cos 5a (2 sin 2a - sin j3),
cos 3a - sin /3 . sin 5a - cos 7a
sin 3a - sin j3 . cos 5a - sin 7a
sin 5a (2 sin 2a - sin j8) ^ ^ , . , j i. i- «
= g )c, -~.^ r— ^; = - tan 5a a value independent of p,
cos 6a (2 sm 2a - sm /3)
14. (cosa: + cosi/)2 + (sina; + sin2/)2
= (cos^ X + sin^ x) + (cos^ y + sin'^ 2/) + 2 (cos x . cos y + sinx, sin y)
= 2 + 2cos(a:-i/) = 2 {l + cos(x-2/)}
= 4 cos2 ii^x- y). [Art. 164, (3).]
15. 2cos2^cos2 5 + 2sinM .sin2^
= J(2cosM . 2cos2J5 + 2sin'M .2sin2i^)
=:i{(l + cos2.4)(l + cos2^) + (l-cos2^)(l-cos2B)} [Art. 164, (3), (4).]
=|(2 + 2co82^ .cos2.B) = l + cos2^ .cos2J5.
, , cos 4 TT sin 4 ir cos^ 4 tt - sin*-^ 4 t
16. coti7r-tan|7r=^-^^ 1— =— r-^ r^-
smjTT cosJtt sm ^ tt . cos I^ tt
= — £2M£_= 2cosi5^2coti. = 2.
Sm^TT . COS^^TT SinjTT
17. tan4. = ji^^ [Art. 164. (5).]
4 tan 6 4 tan &
1 - tan^ d 1-tan-^^ _ 4 tan S (1 - tan2 6)
/ 2^an e y " 1-
Vl-tan2(9y
-6tan2^ + tan^^ 1-6 tan^ d + tan^'di *
(l-tan'^f?)-
18. COS^ 1^ TT - Sin^ I TT = cos J TT = -7^ , COS^ ^ TT + sin^ J TT = 1 .
By addition 2 cos2 1 7r = l + -4 = "^^^^ ;
.-. 4cos2i-7r=:?-y4^^ = 2 + s/2; /. 2 cos i tt = V^W^-
MULTIPLE ANGLES. XLII. 95
19. cosm°15' -sinnr 15' = cos22°30' = cos^7r = i V2T\/2 (18),
cosm°15' + sin211°15' = l.
By addition 2 cos^ 11° 15' = 1 + i J2 + J2 ;
.-. 4cos2 11°15' = 2+>/2Hhv/2;
20.
21.
.-. 2cosll"15'=\/2 + x/2 + ^2.
sin A . sin 2A + sin ^ . sin 4^ + sin 2 A . sin lA
sin A . cos 2A + sin 2 A . cos 5 A + sin ^ . cos 8^
_ sin A (sin 2A + sin 4:A ) + sin 2 A . sin 7 A
~ sin A (cos 2 A + cos 8^ ) + sin 2 A . cos 5 A
_ 2 sin ^ . cos A . sin 3^1 + sin 2 A . sin 7 A
" 2 sin ^ . cos 5 A . cos SA + sin 2 A . cos 5^
_ sin2^ (sin3^+sin7^) _ 2 sin 2^ , cos 2^ . sin 5^
2 sin A . cos 5^ (cos SA + cos ^) 2 sin A cos ^ . 2 cos 2^1 . cos 5A
_ sin 4^ . sin oA _ sin 4^ . sin 5^ _
~ 2 sin 2^ cos 2A cos 5^ ~ sin 4^ . cos 5^ ~
sin ^ + sin (^ + 0) + sin (^ + 20)
cos e + cos (e + 0) + cos (^ + 20)
_ sin ^ + sin (8 + 20) + sin {0 + 0) _ 2 sin {0 + 0) cos + sin {0 + 0)
~cos ^ + cos (^ + 20) + cos (^ + 0) ~ 2 cos {0 + 0) cos + cos {0 + 0)
_ sin(^ + 0){2cos0 + l} _ sin(^ + 0)_ .
cos(6> + 0) {2 cos + 1} cos(d + 0)~ ^ "^*^^*
22. 2 cosM - 2 sins^ = 2 (cosM + sin^^) (cos^^ - sin^A) (cos^A-\- sin^ ^)
= 2 (cos* ^ + sin* A) (cos^ ^4 - sin^ ^)
= 2 { (cos2 A + sin2 ^)2 - 2 sin^ ^ cos2 ^ } cos 2 A
= 2 (1 - 2 sin2 A cos2 A) cos 2^
=:(2-2sin2^ .2cos2^)cos2^
= { 2 - (1 - cos 2^) (1 + cos 2 A) } cos 2A [Art. 164.]
= (l + cos2 2^)cos2^.
23. (3 sin A-i sin3^)2+ (4 cos^^ - 3 cos^)2 = sin2 3^ +cos2 SA = 1.
OA sin 2a . cos a _ 2 sin a cos^ a _ sin a r A + i ra. i
' (l + cos2a)(l + cosaj~~2cos2a(l+cosa) ""i + cosa ^ *■'
2 sin A a . cos i a sin A a ^
= -^- ^ = ^ =tan i a.
2cos2Ja cosja
96 MULTIPLE ANGLES. XLII.
cos {n-2)a . cos na
^ cot (n - 2) g . cot na + 1 ^ sin {n-2 ) a . sin na
cot {n-2) a- cot na ~~ cos {n-2) a cos na
sin (n - 2) a sin na
_ cos (n - 2) g cos na + sin (n - 2) a . sin na
cos (?i - 2) a sin na - sin (n - 2) a cos na
_ cos { {n - 2) a - na } _ cos 2a
sin { na - (n - 2) a } ~ sin 2a
_ 2 (cos2 a - sin^ a) _ cos a sin a
~ 2 sin a cos a ~ sin a cos a
= cot a - tan a.
^^ ^ ^ 2tana f I ^
tanr2a + 5^- tan 2a + tan^ __ A+A _ilf_..n__.
tan(2a + ^)-^_-^2a.tan)3~l-/j.A~IM"^^''~'''
27. We have to shew that {x - tan J^) (a; - cot ^A) is identical with
x^ - 2x cosec ^ + 1.
The first expression is x^ - (tan J^ + cot J ^) a; + 1.
Hence we have to shew that tan ^A +cot J^ = 2cosec^,
XI ^ ^ 1 . 1 tan^i^ + l
tani^ + coti^ = tanA^ + - —-= , \ , —
^ ^ ^ tanj^ tanj^
sinH^ + cosH^ 2
= — -. — i— i ~ — =— — 7 = 2cosec^. Q. E. D.
smj^Acos^A Bin A
Or J proceed to solve the given equation
x'^-2x . cosec ^ + 1 = 0;
.-. ic2 - 2a; . cosec ^ + cosec2^=cosec2^ -l = -^—-—-l= -^— — .
sin'' A sm^ A
cos^
.*. x - cosec ^ = ± ;
sin^
1 cos A 1 + cos A 1 - cos A
,\ x = - — j^. — ,= — -. — J- or — ; — -—
sin A sm A sm A sin A
2cos2ij: 2sin2i^
^ —7r~- — T~T r~i o^ n~ — 1 J — 7— :=coti^ or tan i^.
2smJ^cosJ^ 2smJ-4cosJ^ ^ ^
28. IftanJ5=-, /. 6 = atani?;
/a + b /a-&_ /1 + tan^ /I -
tan 5
tan 5
2 cos B 2 cos B
" ^(1 - tan2\B) // sin^ x/(cos2 B - sin2 pj ^cos 2B *
V V COS2J5;
MULTIPLE ANGLES. XLIII. 97
EXAMPLES. XLIII. Page 142.
1. cos(a + ^ + 7) = cos(a+/3) cos 7 - sin (a + j3) sin 7
= cosa . cos/3 . cos 7 -cos a . sin j9 . sin 7 - cos j8 sin 7 sin a - cos 7. sin a. sin j3.
2. sin (a + /S - 7) = sin (a + /3) cos 7 - cos (a + /3) sin 7
= sin a . cos /3 . cos 7 + sin /3 . cos 7 . cos a - sin 7 . cos a cos /3 + sin a sin p sin 7.
3. cos (a - /3 + 7) = cos (a - j8) cos 7 - sin (a - j3) sin 7
= cos a . cos j8 . cos 7 + cosa . sin /3 . sin7 - cos /3 . sina . sin7 + C0S7 . sinjS . sin a.
4. sina-sin (a + /3-7)= -2cosi(2a + j3-7) sin^ (j8-7).
And sin j3 - sin 7 = 2 cos J (j3 + 7) sin | (jS - 7) ; [Art. 158.
'. sin a + sin /3 - sin 7 - sin (a + j8 - 7)
=:2 cos J (/3 + 7) . sin |- (/3 - 7) - 2 cos J (2a + ^3 - 7) . sin i (^ - 7)
-2sini(/3-7){cosi(^ + 7)-cosi(2a + ^-7)}
= 2sini(/3-7) .2sini(a + /3) .sinl(a-7) [Art. 158.
= 4 sin ^ (tt - 7) . sin J (j8 - 7) . sin ^ (a + /3).
5. sin(a-/3-7)-sina= -2cos^ (2a -/3- 7) sin ^(p + y),
sin ^ + sin 7 = 2 sin J^ (^ + 7) cos |- (jS - 7) ; [Art. 158.
'. sin (a - j3 - 7) - sin a + sin /3 + sin 7
--=2 sin U^ + 7) • cos i (/3 - 7) - 2 cos ^ (2a - /S - 7) . sin ^(^3 + 7)
= 2sin^(^ + 7){cosi(^-7)-cosn2a-^-7)t
= 2sinHi3 + 7) .2sin J(a-7) .sini(a-/3)
= 4 sin ^ (a - /3) . sin J (a - 7) . sin J (i8 + 7).
6. sin2a-sin2(a + j3 + 7)= - 2 cos (2a + j3 + 7) sin (j3 + 7)
sin 2/3 + sin 27 = 2 sin (/S + 7) . cos (/3 - 7) ; [Art. 158.
•. sin 2a + sin 2/3 + sin 27 - sin 2 (a + /3 + 7)
= 2sin(/3 + 7) . cos(/3-7) -2 cos(2a + i3 + 7) . sin (/3 + 7)
= 2 sin (/3 + 7) {cos (i8-7) - cos (2a + /3 + 7)}
= 2sin(/3 + 7) .2sin(a + /3). sin(7 + a)
= 2 sin (a + /3) . sin (/3 + 7) . sin (7 + a).
7. sin(/3-7) + sin(7-a)= -28in J(a-/3) . cos J(a + j8-27)
sin (a - /3) = 2 sin ^ (a - /3) . cos i (a - j8) ;
*. sin (/3 - 7) + sin (7 - a) + sin (a - /3)
= 2 sin J (a - p) {cos J (a - j3) - cos J (a 4- /3 - 27)}
= 2 sin i (a - /3) . 2 sin 4 (a - 7) . sin i (j3 ~ 7)
= 4 sin 4 (a - /3) . sin J (/3 - 7) . sin J (a - 7)
= - 4 sin 4 (a - /3) . sin J (i^ " 7) • sin J (7 - aj ;
.-. sin (jS - 7) + sin (7 - a) + sin (a - /3)
+ 4sini(/3-7) .sini(7-a) . sinj (a-^)=0.
L. T. K. 7
98 MULTIPLE ANGLES. XLIII.
8. sin (^ + 7 - a) + sin (7 + a - ^) = 2 sin 7 cos (a - p)
sin (a + i3 - 7) - sin (a + /3 + 7) = - 2 sin 7 cos (a + j8) ;
.-. sin (a + jS + 7) + sin (7 + a - j8) + sin (a + j3 - 7) - sin (a + ^ - 7)
= 2 sin 7 {cos (a - /3) - cos (a + /3)}
= 2 sin 7. 2 sin a . sin /3 = 4 sin a . sin /3. sin 7.
9. sin (a + i3 + 7) + sin (j3 + 7-a) = 2 sin (j3 + 7) cosa
sin (7 + a - ^) - sin (a + jS - 7) = - 2 cos a sin (j3 - 7) ;
/. sin (a + j8 + 7) + sin (/3 + 7 - a) + sin (7 -f a - /3) - sin (a + jS - 7)
= 2 cos a {sin (p + y)- sin (/3 - 7)}
= 2 cos a . 2 cos /3 . sin 7 = 4 cos a . cos p . sin 7.
10. cos ic + cos y = 2 cos ^ (a; + r/) cos i (x-y)
cos 2 + cos (a; + 1/ + 2) = 2 cos J (2^; + a; + 2/) . cos i{x-{-y);
.'. cos a; + cos 2/ + cos ^; + cos (a; + 2/+ ^^)
= 2 cos i {x + y) {cos i (a; - y) + cos J (2-2 + x + 1/)}
= 2 cos J (a; + y) 2 cos J (2 + x) cos i (2/ + 2)
= 4 cos i{x-\-y) . cos J (2/ + ^) • cos i(z + x).
11. cos 2a; + cos 2y = 2 cos (a? + 2/) . cos (x - y)
cos 2z + cos 2 (a; + y + z) = 2 cos (2z-{-x + y) . cos (« + 2/) ;
/. cos 2a; + cos 2y + cos 2^ + cos 2 (a; + 2/ + -2)
= 2 cos (a; + y) {cos (a; - ?/) + cos {2z-\-x + y)}
= 2 cos (a; + y) .2 cos (x-\-z) . cos (y + z)
= 4 cos (x + y) . Cos (2^ + 2) . cos (z-\-x).
12. cos (y + 2 - a;) + cos {z + x~y) = 2 cos 2 . cos (x - y)
cos (a; + 2/ ~ -2^) + cos (a; + 2^ + 2) = 2 cos z . cos (x + y);
:. cos (2/ + 2 - «) + cos {z + x-y) + cos (a;+ 2/ - -z) + cos (x + y + z)
— 2 cos -2 {cos {x-y) + cos (x + ?/)}
= 2 cos 2 . 2 cos a; . cos y := 4 cos a; . cosy . cos 2.
13. cos- X + cos'-^ y + cos^ z + cos^ [x + y + z)
= J {1 + cos 2x + 1 + cos 22/ + 1 + cos 22; + 1 + cos 2 (a; + 2/ + 2)}
= J {4 + cos 2x + cos 2?/ + cos 2z + cos 2 (a; + 1/ + 2;)}
= J {4 + 4 cos (2/ + -z) . cos (z + a;) . cos (a; + 1/)} [Example 11.
= 2 {1 + cos (y + z) . cos (z + x) . cos (a; + y)}.
14. sin^ X + sin2 y + sin^ 2 + sin^ (x + y + z)
= J {1 - cos 2a: + 1 - cos 2y + l- cos 2z + l - cos 2 (a; + 2/ + -s)}
= J [4 - {cos 2a; + cos 2y + cos 2z + cos 2 (a; + 2/ + -s)}]
= J {4 - 4 cos {y + z) . cos (2; + x) . cos {x + y)} [Example 11.
= 2{l-cos (y + z) . cos(;2; + a;) . cos (x + y)}.
MULTIPLE ANGLES. XLIII. 99
15. cos2 X + cos^ y + cos^ z + cos^ (x + y-z)
= 4 {1 + cos 2x + 1 + cos 21/ + 1 + cos 22 + 1 + cos 2 (x -{-y -z)}
= J {4 + cos 2x + cos 2y + cos 2z + cos 2 (x + y - z)\.
Now cos 2x + cos 2i/ = 2 cos {x + y) . cos (x - y),
and cos2<2; + cos2(a; + 2/-z) = 2cos (x + y) .cos (a5 + y-22);
.-. cos 2x + cos 2?/ 4- cos 22 + cos 2(x-{-y-z)
= 2 cos (x + y) {cos (« - y) + cos [x + y- 2z)}
= 2 cos (ac + y) . 2 cos (x-z) . cos (y - z)
= 4 cos (x-z) . cos (2/ - ^) . cos (x + y) ;
.*. cos^ X + cos^ y + cos'^ z + cos^ (x + y - z)
= J {4 + 4 cos (a; -i;) .cos(y-2;) . cos (x + y)}
= 2{l + cos(ic-2) . cos (y-z) . cos(a: + y)}.
16. cos a . sin (p-y) + cos j8 . sin (7 - a) + cos 7 . sin (a - p)
= 4 {2 cos a . sin (/3 - 7) + 2 cos j9 . sin (7 - a) + 2 cos 7 . sin (a - /3)}
= i {sin (a + /3 - 7) - sin (a - /3 + 7) + sin (/3 + 7 - a) - sin (/3 - 7 + a)
+ sin (7 + a -^) - sin (7 - a + ^)} = 0.
17. sin a . sin (^3 - 7) + sin p . sin (y - a) + sin 7 . sin (a - p)
= i {2 sin a . sin (^ - 7) + 2 sin /3 . sin (7 - a) + 2 sin 7 . sin (a-p)]
= i{cos (a-/3 + 7) -cos (a + j3-7) + cos (j3-7 + a) - cos (^3 + 7- a)
+ COS (7-a + i3)-cos(7 + a-|3)} = 0.
18. cos(a + j3) . cos (a - j3) + sin (j8 + 7) . sin (^ - 7) - cos (a + 7) .cos (a -7)
= cos2 a - sin^ p + sin" p - sin- 7 - cos^ a + sin- 7 [Examples XXXV. 24, 25.
= 0.
19. cos (5 - a) . sin (jS - 7) + cos (5-p) , sin (7 - a) - cos (5 - 7) . sin (p - a)
= J {2 cos (5 - a) . sin (^ - 7) + 2 cos (5 - ,3) . sin (7 - a) - 2 cos (5 - 7) . sin (p - a)}
= A {sin (5 - a + /3 - 7) - sin (5 - a - j8 + 7) + sin (5 - /3 + 7 - a)
- sin (5 - j3 - 7 + a) = sin (5 - 7 + /3 - a) 4- sin (5 - 7 - /3 + a)} = 0.
20. 8cosi(l9 + + x) .cosi(0 + x-^) .cosi(x + ^-0) .cos^ (d + <p-x)
= 2{2cosi((? + + x)-cosi(0 + x-<?)}x{2cosi(x + 6?-0).cos4((? + (/>-x)}.
Now 2 cos 4 (^ + + x) • cos J (0 + X - ^) = cos (0 + x) + cos ^,
and 2 cos i (x + ^ - 0) • cos i (^ + - x) = cos (0 - x) + cos ^ ;
2{2cosi(^ + + x)-cosi(0 + X-<^)}x{2cosJ(x + (9-0).cosi((? + 0-x)}
>= 2 {cos (0 + x) + cos 6} X {cos (0 - x) + cos 6]
= 2 [cos (0 + x) cos (0 - x) + cos {cos (0 + x) + cos (0 - x)} + cos^ 0]
= 2 cos (0 + x) cos (0 - x) + 4 cos cos cos x + 2 cos^
= cos 20 + cos 2x + 4 cos . cos cos x + 1 + cos 20.
Therefore
8cosi(^+0 + x)-cosi(0 + x + ^) .cosJ(x4-^-0).cosi(^ + 0-x)
= cos 20 + cos 20 4- cos 2x + 4 cos . cos . cos x + 1-
7—2
I
100 ON ANGLES UNLIMITED IN MAGNITUDE.
EXAMPLES. XLIV. Page 145.
1. 2.
5. cos (90° + A) = cos 90° cos A - sin 90° sin A
= X cos ^ - 1 X sin A— - sin A,
6. sin (90° -A) = sin 90° cos A + cos 90° sin A
= 1 xcos^ -i-Ox sin ^ = cos ^.
7. cos (90° -A)=^ cos 90° cos A + sin 90° sin A
= X cos ^ + 1 x sin ^ = sin ^.
8. sin (180° - ^) = sin 180° cos A - cos 180° sin A
= X cos ^ - ( - 1 X sin A) = sin A .
9. cos (180° -A) = cos 180° cos A + sin 180° sin A
= - 1 X cos ^ + X sin A= - cos A .
10. sin (180° + A)= sin 180° cos A ;+ cos 180° sin A
= 0xcos^ + (-l xsin^)= -sin^.
11. cos {A+B) = Qm (90° + ^ +1?)
= sin (90° + A) cos B + cos (90° + A)smB
= cos A cos B - sin A sin B,
sin {A-B) = sin J cos ( - jB) + cos ^ sin ( - B)
= sin A cos B - cos A sin B,
cos (A - B) = sin (90° + A-B)
= sin (90° + A) cos (-B) + cos (90° + ^) sin ( - 5)
= cos A cos i)* + sin A sin £".
ox ANGLES UNLIMITED IN MAGNITUDE. 101
EXAMPLES. XLV. Page 147.
1. sin 180" = sin (90^ + 00^) = sin 90° cos 90° + cos 90° sin 90°
= 1x0 + 1x0 = 0.
cos 180° = cos (90° + 90°) = cos 90° cos 90° - sin 90° sin 90°
= 0x0-1x1- -1.
2. In fig. E. T. p. 72, let OiV/=2, PM=1; /. OP = ^5;
.'. tanPOJ/=i, sinP0M=-4, cos POM =:-^.
Since A is between 180° and 270°, tan A is positive and sin A and cos A
1 2
are negative, .*. sin ^ = — r-, cos A=z ,
1 2 4
sin 2^ = 2 sin A cos A = 2x r-x r- = - ,
\/o vo 5
. o I o • . ^ •. . 3 4 11 11 ,^
sm3^=3 sin^ -4sin'^^ = 7^+ f— 7^= - i^—r^— -irJ^'
3. Since is the fourth quadrant sin is negative,
sin^=-V(l-cos2(9)=-^,
sin2^ = 2sin^cos^ = 2x --^l_ x Jz^ - ^^15,
sin3^ = 3sin^-4sin3^==-?-^ + l^^ = T3^^15^
cos3(9 = 4cos3(9-3cos6>=tV-|^ - H-
To determine in what quadrant 3^ lies, it should be noticed that cos 3^ is
negative and therefore 3^ may lie in the second or the third quadrant, but
sin 3^ is positive therefore 3^ viust lie in the second quadrant.
4. cosp0 + coBq0 = O; .'. 2 cos ^ (p + q) , co8^ (2)'^q) = 0;
.-. cosj(^ + g)^=0 or cos J (^'-g) ^ = 0.
If coR4(^ + fy)^ = 0, then (^ + g)6^ = (2n + l)7r; /. = ^^!^Tr.
If cos^(p'^q) = Of then (p <^ q) = (2n -\- 1) w \ .'. 0= r by giving
integral values to n in order tt and ir are evidently two series in
p. with common differences and .
1
102 ON SUBMULTlPLE ANGLES.
EXAMPLES. XLVL Page 149.
1. When A lies between - 180° and 180°, ^A lies between - 90° and 90°,
, , . ... , ■ /1 + cosJ^
1. e. cos J^ IS positive ; .'. cos iA=: -\- /
2
2. When A lies between 180° and 540°, ^A lies between 90° and 270°,
T , . ,. 1 , /1 + cos^
1. e. cos J^ IS negative ; .*. cos J J = - / — .
3. When A lies between 180° and 360°, ^A lies between 90° and 180°,
. , . . ... . 1 ^ /I -cos^
I.e. sin J ^ IS positive; .*. sinj^ = + / — .
4. When A lies between (4n + 1) ir and (4w + 3) tt its trigonometrical
ratios have the same signs as when it lies between tt and Sir ; /. the trigono-
metrical ratios of ^A have the same signs as when it is between ^w and Itt,
i.e. cos J^ is negative;
, . /1 + cos^
.-. cosi^=-^ ^— •
5. When A lies between Anir and (4n + 2)7r its trigonometrical ratios
have the same signs as when it lies between and 2ir; therefore the trigono-
metrical ratios of ^A have the same signs as when it lies between and tt,
i. e. sin J^ is positive ;
. , , /1-cos^
EXAMPLES. XL VII. Page 154.
1. (i) When i^ is 22° sin J J is positive and less than cos J^, cos J^
is also positive ; .*. sin iA + cos J^ is positive,
sin ^A- cos J^ is negative.
(ii) sin 191° = sin (180°- 191°)= -sin 11°; .-. sin 191° is negative and
numeriqally less than cos 191°,
cos 191°= - cos (180° - 191°) = - cos 11° ; /. cos 191° is also negative ;
sin 191° -h cos 191° is negative,
sin 191° - cos 191° is positive.
(iii) sin 290° = sin (360° - 70°) = - sin 70° ; .'. sin 290° is negative and
numerically greater than cos 290°,
cos 290°= cos (360° -70°) = cos 70°; .-. cos 270° is positive;
.-. sin 290° + cos 290° is negative,
sin 290° - cos 290° is negative.
(iv) sin 345° = sin (360° - 15°) = - sin 15° ; /. sin 345° is negative and
numerically less than cos 345°,
cos 345°= (360° - 15°) = cos 15° is positive ;
.*. sin 345° -f cos 345° is positive,
sin 345° - cos 345° is negative.
ON SUBMULTIPLE ANGLES. XL VII. 103
(v) sin -22°= -sin 22° is negative and numerically less than
cos - 22°; cos - 22° = cos 22° is positive ;
.-. sin - 22° + cos - 22° is positive,
sin - 22° - cos - 22° is negative.
(vi) sin -275° = sin (360° -275°) = sin 85°; /. sin -275° is positive
and numerically greater than cos - 275° ;
cos - 275° = cos (360° - 275°) = cos 85° is positive;
.-. sin - 275° + cos - 275° is positive,
sin - 275° - cos - 275° is positive.
(vii) sin - 470° = sin (360° - 470°) = sin ~ 110°= - sin (180° - 110°)
= - sin 70°; cos - 470° = cos (360° - 470°) = cos - 110°= - cos (180° - 110°)
= - cos 70° ; .*. sin - 470° is negative and numerically greater than cos - 470° ;
cos - 470° is also negative ;
.*. sin - 470° + cos - 470° is negative,
sin - 470° - cos - 470° is negative.
(viii) sin 1000° = sin (3 x 360° - 80°) = - sin 80° ;
cos 1000° = cos (3 X 360° - 80°) = cos 80° ;
.-. sin 1000° is negative and numerically greater than cos 1000° ; cos 1000° is
positive, sin 1000° + cos 1000° is negative,
sin 1000° - cos 1000° is negative.
2. Consider the values of (sin J ^4 + cos J A) and (sin ^A- cos J A) when
(i) A = 92°, 268°, 900°, 4mr + f tt, (4w + 2) tt - j tt.
We see that when J .4 = 46° sin J^ is > cos J^ and positive,
also when J ^ = 134°, when ^ 4 = (360° + 90°) , when i ^ = 2w7r + 1 tt,
and when J ^ = (2w + 1) tt - f tt the same is tr^e ;
hence in all these cases sin J ^ + cos ^A is positive,
and sin ^A- cos ^ ^ is positive,
and .'.the formulcB for sin J^ and for cosj^ in terms of sin J^ are unaltered.
When (ii) ^ = 88°, -88°, 770°, -770°, or 4n±i7r,
it may be shewn that when J^ = 44°, cos J^ is greater than sin^^i and
is positive ; the same statement is also true when J i = - 44°,
when iA = (360° + 25°) , when iA= - (360° + 25°) and when iA = 2mr .-t ^\ w.
Hence in all these cases sin J ^ + cos J ^ is positive,
sin \A- cos J ^ is negative,
and the formulae for sin^^ and cosj^ in terms of sin^i have the same
form in each case.
3. sin 9° is positive and numerically less than cos 9° ; cos 9° is positive ;
.-. sin9° + cos9°=+V(H-sinl8°) = V{l + i(v/5-l)}=W(3 + v/5)
sin 9° - cos 9° = - ^(1 - sin 18°) = - V { 1 " i (V^ - 1) } = i \/(^ - V^) i
/. (i) sin9° = lW(3 + V5)-V(5-^5)}.
(ii) cos 9° = i { V(3 + v/5) + V(5 • V5) } .
104 ON SUBMULTIPLE ANGLES. XLVIL
(iii) sin 81° r= cos 9°.
(iv) cos 189° = cos (180° + 9°) = - cos 9°.
(v) tan 2024° = tan (180° + 22 J°) = tan 22i°.
224° is in the first quadrant, therefore its tangent is positive.
From E. T. Art. 181, tan 22 J := ^-^i^^^^^^^^^i^L ^2 - 1.
(vi) tan 974°= -cot 74° =
tan 45°
1
tan (74°) '
^/3-l
/. tan 974° =
tan 7i°- ^^iLli°- - V^ - V^-1
^ ~ 1 + COS 15° V3 + 1 ~ 2 ^2 + ^3 + i
"^ 2^2
(V3-1)(2V2 + 1-V3) ^ 2^6-2^2-4 + 2^3
"(2^2 + ^3 + 1) (2V2 + 1--V3) 6 + 4^2
^ n/6-J2-2 + V3 ^ ( V6- ^ 2 -2 + ^3) ( 3-2^2)
3 + 2^2 (3 + 2^2) (3-2^2)
= x/6-^/3 + ^2-2;
1 1
v/6-v/3 + V2-2 (V3-V2)(^2-l)
._ (v/i+V2Hx/2 + l) ./3. /2W/2+n
(V3-^/2) (V2-1) (V3 + V2) (x/2 + 1)- W3 + x/^) (^2+1).
4. (i) If ^ = 200°, 4 ^ = 100°. Now sin 100° = sin (180° - 80°) = sin 80°,
sin 80° is positive and numerically greater than cos ^A ;
.-. sin4^+4cos^= +;^(l + sin^), sin4^ - cos 4^== +>/(l-sin^);
.-. 2sin4^=: +^(l + sin^) + ^(l-sin^).
(ii) The tangent of the angle 100° is negative ; therefore we have to
take the negative value in the formula tan ^A= ^^ — ^^ ; the
negative value is tan ^A =
tan^
. 1-^^(1 + tan2^) _-{l + V(l + tan^^)}
tanJ^ tan -4 * ^
tan ^ =tan 200° is positive.)
5. (i) If A Ues between 270° and 360°, ^A lies between 135° and 180°;
cos 4-4 is negative and numerically greater than sin ^A ;
.-. sin4^+cos4^= -\/(l + sin^), sin4^ -cos4-4= +\/(l - sin J);
.*. 2sin44=;^(l-sinJ[)-;^(l + sin^).
(ii) Now tan A is negative when A is between 270° and 360° ; also
tan 4^ is negative when ^A is between 135° and 180°; we have therefore to
take the negative value for tan ^A in the formula
ON SUBMULTIPLE ANGLES. XL VII. 105
tan ^A= V~~~j ' negative value is
^(tan2.:t + l)-l ,. ^ ., ... 1 secM , . .
^^-5^ — —^ (for tan A is negative) = - , + , — -— = - cot ^ + cosec ^.
tan A o / ^^^ ^ ^^^ ^
6. If A lies between 450° and 630°, \A lies between 225° and 315°.
When J^ is between 225° and 270° sin^^ is greater than cos J^; and is
negative ;
.'. sin \A-Y cos J ^ is negative, sin \A- cos J ^ is negative.
When \A is between 270° and 315°, sin \A\% negative and greater than
cos \A ; cos J^ is positive and less than sin J^ ;
/. sin J^ + cos J^ is negative, sin J^ - cos \A'\^ negative;
.-. 2sin Ji4= -^(l + sin^)-;^(l-sin^).
7. 2sini^=^(l + sin^)-^(l-sin^).
When sin J^+cos J^= +v/(l + sin ^), sin J^ - cos J^= -^,^(1 - sin ^).
These two statements are each satisfied when cos J^ is >sin J^ and is
positive ; that is, when the revolving line OF turning in the positive direction
is between - Jir and + Jtt.
8. See (6) above, by subtraction,
2cosiu4= - /^1+sin^ -^1-sin^.
9. When A lies between wx360°-90° and nx 360° tan J is negative
and tanj^ is negative. When A lies between nx360° and nx360° + 90°
tan^ is positive and so also is tanj^. So that when A lies between
nx360°-90° and wx360° + 90° tan ^ and tanj^ have the same sign, so
that tan \A x tan A is positive and .". =^(l-htB>n^ A) - 1. Similarly it may
be shewn that when A lies between wx360° + 90° and n x 360° + 270° tan ^
and tanj^ have opposite signs, so that tanj^xtan^ is negative and
.•. = -V(l + tan2^)-l.
10. When the sine of an angle is given by Art. 144 if A is the least
positive angle which has the given sine, then the angle may be any one of
sB A. p
^4
the angles n x 360° -{-A orn x 360° f 180° -A; /. sin J ^ is the sine of any of
the angles n x 120° + iA or nxl20° + m° -^A; that is the sine of any of the
angles given by the dotted lines OP^, OP^, OP^, OP^, OP^, OPq in the figure.
106
ON StlBMULTiPLE ANGLllS. XLVII.
The lines OR, OA, OB, OL, OG, OB divide the four right angles at into
equal angles each of 60° ; the angle ROP = A, and the angles ROP^, ^OP^,
BOPgy LOP^, GOP 5, DOPq are each equal to ^A. From the symmetry of
the figure it will be seen that sin J^ may have any one of three different
values and no more, for
sin ROP^ = sin ROP^, sin ROP^ = sin ROP^ and sin ROP^ = sin ROP^ ;
but cos ROPi= - cos ROP^ etc., /. there are six different values for cos ^A.
11. When the cosine of an angle is given, by Art. 146 if A is the least
positive angle which has the given cosine, then the angle may be any one of
the angles nx360° + ^ or nx360°-^.
Therefore J A may be any one of the angles n x 120° 4- J ^ or « x 120° -^A,
that is any one of the angles given by the dotted lines in the figure
OP^, OP^, OP^, OP^, OP^y OPq. From the symmetry of the figure it will be
seen that cos ^A may have any one of three different values and no more, for
cos ROPj = cos ROPq, cos ROP^=:=i cos ROPfi, cos ROP^ = cos ROP^.
But sin ROPj^= - sin BOP ^ and so on, .-. there are six different values
12. When tan A is given the angle may be (by Art. 148) any one of the
angles wxl80° + ^, and .*. ^A may be any one of the angles wx60° + J^;
OK SUBSIDIARY ANGLES. XLVII. 107
hence tan J ^ is the tangent of any one of the angles given by the dotted
lines OPt^, OP^, OP^, OP^, OP^, OP^; hence it will be seen that tan ^A has
three different values and no more, for
tani?0Pi = tanjR0P4, tan EOPg = tan IJOPg, tan E OP3 = t an POPg.
13. In the figure of Art. 182 the sines of the angles POP3, POP4,
POP5, ROPq (which are the possible values of sin^A when tan^ is given)
are all different and the result follows. Also the values of the cosines of
these angles are all different.
14. In the figure of Art. 179,
tanPOP3 = tanPOP5 and tan P0P4=: tan POP^;
hence the result follows.
EXAMPLES. XLVIII. Page 157.
1. 2sin^ + 2cos^ = ;iy2.
Divide both sides by 2^2, then — j— -\ =|;
.-. sin^cosjTT + cos^sin Jtt^J, sin (^4- Jw) = J = sin^7r;
2. sin ^ + ^3. cos = 1.
Divide both sides by 2, then —^ \- ^ . cos d = ^;
.: sin^cos jTT + sin J7rcos0 = j, sin (0 + Jir) = J = sin|7r;
.-. + j7r = W7r+(-l)'»|7r; .♦. 6= - iw + mr + {-!)'' ^w,
3. >/2 sin + ^/2 cos 0=>/3.
1 1 /3
Divide both sides by 2, then — - . sin + -t^ . cos $ = '^ ;
.-. sin(j7r + (9)='^=sinj7r, j7r + (? = W7r + (- 1)« Jtt;
.'. 0=_J^ + n7r+(-l)^j7r.
4. sin 0- cos 0=1.
Divide both sides by ^2, then -^ sin - -^ cos = — - ;
V^ v^ v2
.-. sin(0^j7r) = -^ = sinj7r; .-. l9- j7r = ?i7r + (- l)'47r;
.-. = j7r + W7r + (-l)"J'r.
5. sin + cos = 1.
Divide both sides by ^2, then -^ . sin + --- . cos 0= — - ,
cos (0 - Jtt) = -- = cos J TT ;
.-. 0- J7r = 2/t7r±i7r; .'. = J7r + 2w7r±47r.
108 ON SUBSIDIARY ANGLES. XLIX.
'6. ^/3sin^-cos^-^2 = 0.
v/3 1
Divide both sides by 2, then —■ sin 6 -^ . cos ^= — - ;
2 f^2
.'. sin(^-|7r)=:-^ = sinj7r; /. ^-^7r = W7r+ (- l)** Jtt;
7. 2 sin a; + 5 cos a; =: 2, sin a; + 2*5 cos a; =1, sin a: + tan 68^12' cos x = l,
sin X cos 68° 12' + sin 68° 12' cos x = cos 68° 12',
sin (x + 68° 12') = cos 68° 12' = sin 21° 48' ;
.-. X + 68° 12' = nx 180° + ( - 1)^ (21° 48') ;
/. x=- 68° 12' + w X 180° + ( - 1)" (21° 48').
8. 3 cos a; - 8 sin a; = 3, cos a; -2*6 sin a; = 1,
cos X - tan 69° 26' 30" sin a: = 1,
cos X cos 69° 26' 30" - sin 69° 26' 30" sin x = cos 69° 26' 30",
cos (a; + 69° 26' 30") = cos (69° 26' 30"),
X + 69° 26' 30"= 2w x 180° ± 69° 26' 30" ;
,\ x=- 69° 26' 30" + 2w X 180° ± 69° 26' 30".
9. 4 sin a: - 15 cos a; = 4, sin a;- 3*75 cos a; = 1,
sin X - tan 75° 4' cos a; = 1,
sin X cos 75° 4' - sin 75° 4' cos x = cos 75° 4',
sin (x ~ 75° 4') = cos 75° 4' = sin 14° 56' ;
/. a; - 75° 4' = n X 180° + ( - 1)** (14° 56') ;
.-. a: = 75° 4' + ?i X 180° + ( - 1)~ (14° 56').
10. cos (a + x)- sin {a. + x)=^ ^2 cos p.
Divide both sides by ^2, then -t^cos (a + a:) - - sin (a + a;) = cos j3 ;
/. cos (a -f a; + J tt) = cos ^; .'. a + x + lir = 2mr db p ;
.-. x= —a-l'jr-\- 2?i7r ± p.
EXAMPLES. XLIX. Page 158.
1. Leta = sin-U; .-. sina = f, cos a= ^yi - (|)^ = |, i.e. a = cos~^ |,
tana = ^^— =f--f = J, i.e. a = tan-i|;
cosa "^ ^ *' ^*
.'. sin-if = cos-i| = tan~i|.
2. Leta = sin-iJ; .-. sin a = J and cos a= />yi - J = '^- , i.e.a = cos-^^;
cot a = 5£^ = ^^_^J^ /3, i.e. a = cot-V3;
sm a 2 ^ ^ ^
.*. sin-i \ = cos-i '^^- = cot-i JS.
ON SUBSIDIARY ANGLES. XLIX. 109
3. Let sin-^a = (?; /. sin(? = a and co8 = ^(l-Bin^ 0) = ^(l-a%
i.e. ^=.cos-V(l-a^); *^^^ = c^ = ;/Xi3^)' i-^- ^=*^^"'^/(CT) '
/. sin-i a = cos-i v/(l - «^) = tan"! ^ ^
4. a = sin-if, i.e. sina = f. j3 = C03~H, i.e. cos)3 = f ;
.-. sin a = cos /3 = sin (90°-^); .'. a = 47r-i8, i.e. a + /3 = i7r.
5. ^=sin-^a, i.e. sin ^— a. B = cos~^ a, i.e. coaB — a;
.'. sin^ = cosB = sin(90°-B); .'. ^ = 90°-B, i.e. ^ + jB = 90^
6. Let a=tan-if, i.e. tana = f, ^ = tan-i^, i.e. tan j8-^.
XT X / . o\ tana + tan/3 f + J ^ , .
Now tan {a + p) = , — 7 — - . ^ = .——^ = 1 = tan J tt ;
^ ^' 1- tan a . tan /3 1 - f . J^
.*. a + i3 = j7r, i.e. tan"! f + tan-^ J = Jtt.
7. Let a = tan-i ^2^.^ i,e, tan a= j\, and j3 = tan-if, i.e. tan/3 = f ;
i.e. 2/3=tan-V¥ = 2tan-if
xr . / , o^\ tana + tan2|8 t\ + A i
Now tan (a + 2^) = - — — -^ = .. ^ or = i »
^ ^^ 1 -- tan a . tan 2/J 1 - tt • ttV
.;. a + 2^ = tan-i4, i.e. tan-i^i +2 tan-i f^tan-^^.
8. Let a=:tan~^ wii, i.e. tana = mi, j3=tan~^m2, i.e. tan /3= wig'
^, tan a + tan 8 m. + m^
tan (a+B) =^ — —^^ = .— ^^ ^ ;
^ ^' 1- tan a . tan ji 1 - m^mg
.-. a + i3 = tan-^ — -^ , i.e. tan ^ m, + tan~-^m2= tan~i ~ -.
'^ 1 - m-^pi^ ^ 1 - mjWijj
9. Let a; = sin~^a, i.e. sinx — a, Goax = ^(l-a^),
sin 2x = 2 sin a; cos x — 2a^J(l - a?), i. e. 2x = sin-^ 2(i J(l - a^) ;
.-. 2 sin-i a = sin"! 2a ^(1 - a^),
sin (2 sin-i a) = sin (sin-i 2a^(l - a^)) = 2a;^(l - a^).
10. Let cos~^a = a;, i.e. cosa; = a; cos2a; = 2 cos^^^c- l = 2a2- 1;
.-. 2a;=:cos-i (2a2- 1), i.e. 2 cos-^ a = cos-i (2a2_ 1).
11. From (9) we obtain 2 sin-^ a = sin-i 2a^(l -a^) putting J for a we
have 2 sin"^ .J = sin-^ ^ ;
.-. cos-^ 4 + 2 sin"i i = cos"^ J + sin"* ^ .
Let a = cos-^4, i.e. cosa = 4; .'• a = 60°.
110 ON SUBSIDIARY ANGLES. XLIX.
Let i3=sin-i'^^, i.e. sin^=^-; .-. /3=60°;
:. cos-i i + 2 sin-i J = a + /3 = 60° + 60° = 120°.
12. From (9) 2 sin-i a^sin-^ 2a ^^(1 - a?), write % for a ;
.-. 2 sin-i 4^sin-i s ^ji _ (4)2} ^gin-i |4. .
.*. 2 sin~i 4 - sin-i f | = sin-
13. Let tan-i(cos2a) = a;, i.e. tana: = cos2a.
2 tan a; 2 cos 2a 2 cos 2a
Now tan 2ic =
1 - tan^ x~ \- cos^ 2a ~ sin"^ 2a
2 cos 2a _ (cos^ a - sin^ a) (cos^ a + sin^ a)
4 sin^ a cos'-^ a ~ 2 sin^ a cos^ a
cot^ a - tan^ o
_ cos^ a - sin^ " _ i /cos^ a sin^ ^ \ _ ^
~ 2 sin*^ a cos^ « ~ \ sin^ a cos'^ q.j~~
J /cot2 a - tan^ a\
" V 2 j'
.'. 2a;=tan-
i.e. 2 tan-i (cos 2a) = tan-^ ( J .
14. tan- 1 re + tan-^ v = t an-^ ^
^ \-xy
1 . 1 ^ ^1-x-y-xy
tan--^ a; + tan--^ v + tan--^ .i ^^
l + x-\-y-xy
x + y 1-x-y -xy
= tan-i ^-^y l + j: + y-:rt/
■^ (a; + ?/)(l-a;-i/-a:y7
(l-xy){l + x-i-y~xy)
^tan-^ ( ^+ yHi +^+y - ^y) + (1 - ^y) (1 - ^' - y - ^y )
(1 - xy) (1 + x + y - xy) - (x + y) (l-x-y-xij)
^^^^-i (^-^y){{x + y) + {l-xy)}+{l-xy){(l-xy)-{x-]-y)}
(l-xy) {(l-xy) + (x+y)\-(x + y){(l-xy)-(x + y)}
(x + y)^-{-(l-xy)^
15. 4 tan-i i - tan-i ^ J^ = 2 tan-i i + 2 tan-i i - tan-^ ^^^
2. 2
= tan-1 ^ ° , + tan-
= tan-i T^j + tan-i yV - tan~i tj^ J ^
5
= tan-i —^ -tan-i ^^^^
1 20 _ 1
_+or,-l 120 _ fori -1 1 — t,fln-l--53Z ^^^^
— lan iYjT-ian tj-^^ — idn i , 120 1
= tan-1 fim = tan-i 1 = i tt.
ON SUBSIDIARY ANGLES. XLIX. Ill
16. Let a = sin-i4, i.e. sina=:i; cos a = /^(l -sin'-a)=;y/{l - (l)'-^} =|.
Let /S^sin-iyV i- e. sin/3 = ^V; oos/?=V(l " sin-i3) = Vil " (it)'"} = xf
Let7=sin-iif,i.e. sin7 = Hcos7=x/(l-sin27)=^{l-(U)2}=|i
sin (a + /3) = sinacos/3^ cosasinj3 = | . If + i . YV = lf = cos7 = sin (47r-7),
i.e. a + j3 + 7=j7r or sin~i| + sin-i ^^ + sin-i-|^ = j7r.
17. Let tan-i >J5 (2 - ^^3) = a, i. e. tan a = ^6 {2- JS) ,
and cot-V5 (2 + ^/3)= ft i.e. cot /3=V5 (2 + ^/3);
1 _^5(2-^3)
.-. tan/3=
Now tan (a - /3) =
^5(2 + ^3)'
tan a — tan p
1 + tan a . tan p
x/5(2-x/3)- ^ - 4^5(2-^/3) ^ 1
l + (2-^3)2 20(2-^3) ^5'
.-. cot (a - /3) = ^5, i.e. a-p = cot -^ ^^5,
or tan-i ^5 (2 - ;^3) - cot-i >^o (2 + ^3) = cot-i ^^5.
18. Let a = sin"im, i. e. sin a=:m and cosa = ^(l -m^),
and ^=rsin~i/i, i.e. smp=n and coSj8=;^(l -n^),
sin (a + P) = sin a cos j8 + cos a sin /3 = mij(l - n^) + nfj{l — rnP)^
i.e. a + j3 = sin-i{w^/(l-7i2) + n;^(l-m2)},
or sin~i m + sin~i 71= sin-^ [in ^(1 - 7i2) + n ,>J(1 - m^)}.
If sin~^m + sin-in = i7r or sin~^l;
/. sin-i {m ^(1 - n2) +n ^{1 - m^)} = sin-i 1 ;
.-. m^{l-n-) + n^(l-7n^) = l.
MISCELLANEOUS EXAMPLES. L. Page 159.
1 _1
T . 1 1 , ,1 ^ 'l + a"*'i-a , ,2
1. tan-i — — + tan-i = tan"! ,— = tan-i - -, ;
1 + a 1-a 1 1 a^
1+a '1-a
I
1 12
.-. tan-i ^ + tan-i 4- tan-^ — ,
1 4- a 1 — a a^
= tan-i - -5 + tan~i -^ = tan'i — -— — = tan-i 0.
a^ a^ ^22
a^ a^
The least angle whose tangent is is 0°, therefore (Art. 149, E, T.) all the
angles whose tangent is are included in the expression mr,
.-. tan-i + tan~^ :; + tan-^ —,=mr.
1 + a 1-a a^
112 ON SUBSIDIARY ANGLES. L.
_ ^ , a-1 ^ , 1
2. tan-i ^ tan-i .
a 2a-l
g-l 1
, , ~^"^2a-l , _,a{2a-l)-{a-l) , .^
= tan-i— — -— =tan ^—^ -{ — :p( = tan-il.
a-1 a(2a-l) -(a-1)
The least angle whose tangent is 1 is J tt ; therefore all the angles whose
tangent is 1 are included in the expression titt + J ir ;
/. tan~i + tan--^ = wtt + i tt.
a 2a -1
3. sina = a;, smp = y, .: cosa= ±,^l-a;^ cos j3= ±^1 -y^
cos (a- p) = cos a cos ^ + sin a sin ^
:. a- p = COS- 1 {.rt/ ± ^ (1 - a;2 _ 2^2 -f x^y^)}y
i. e. sin-^ a; - sin-^ y = cos~i {xy ± ^^(1 - a;^ - 1/^ ^ «^y^)}.
x+1 x-1
M X 1^ + 1 X 1^-1 ^ 1 ^ + 2"^^?^ . , 4-2a;2
4. tan-i 3^ + tan"-^ — ^= tan-^ ., ., = tan ^ — ^^ — ,
^' a; + 2 a;-2 i_^ + ^ a;-l 3
~^T2*x-2
Now tan-i — — -_j^_tan-il; /. — ^ =1; :. x^=\,
1
1 ^■*"a
5. tan-i a + cot-^ a = tan"^ a + tan~^ - = tan"^ -— - =tan-i oo .
" a 1-1
Let a; = tan~ico; /. tanic = oo;
.*. x = mr-\-^'ir = (2n+l)lir.
6. tan-ia + tan-^i3 = tan-ii^^ — ^;
"' '^ 1 - a/3
.-. tan-ia + tan-i/3+tan-i7 = tan-i
a + 8 + y-a^y , ,.
= tanz ^-zr—^ ^ =7r=:tan 10;
1-a^-ay-py
:, i'±^tlZL^^^o, i.e. a+i3 + 7-a/37 = 0, or a + ^ + y=^a^y,
1 — ap — ay — py
ON SUBSIDIARY ANGLES. L.
113
tan"^i z - tan~^ -^— ^ = tan-i
x~l x+l
_1 1
x-1 x+l
1 +
1 x^
x-1' x+i
Now
tan~
IT . ^ _i /v/3 - 1 _ 2
.,..-ii^±i):i(^zl).tan-^.
8. tan-i (X + 1) - tan-i (^ - 1) ^ tan-i ^^^^^ ^ -^ _-^
Now
r2
2 1
tan-i -2 = cot-i (a:2 - 1) = tan-i
x^~l'
•• J^^' ''J = ^'-^^ ^'^ = 2, a:=iV2.
9.
2a;
2x
i+a-'-^^'r^^'''
Let tan~^x = a, i.e. tana = ic; and sin 2a =
2 tana
1 + tan^ a '
1. e. sin 2a =
2x
sin-
1+x:-
Therefore the equation may be written
2tan ■^- ;, = 7r; .*. tan^
1-x^
' n^2 = 2a = 2 tan-i x = tan-i ^-^^ •
l-a;'^~2'
2x
1-x^
:oo; .'. ic*-^- 1 = 0, a;=: ±1.
10- *--'^2^K^^+t--'"^-v;*4^
2;^(a + l) 2;^(a + l)
2^(a + l) • 2V(a+l)
, tan- ^^^J^) ^ tan- -^^ + tan- ^^V(^
1 ^^"+^>+7(^> ■
^(a + l)
.fi:i2^=ta„-...
The least angle whose tangent is oo is J tt ; therefore all the angles whose
tangent is oo are included in the expression nir + i tt.
T^ T. K.
114 ON SUBSIDIARY ANGLES. L.
11. All the angles whose sine is a are included in the expression
W7r + (-l)**a; and all the angles whose cosine is a are included in the
expression 2mr =t ( J tt - a) ; for sin a = cos ( J tt - a) ;
.*. sin-^ a + cos-i a = ?i7r + ( - l)"- a + 2mr ± (i tt - a) = Stitt + ( - 1)*^ a ± (Jtt - a).
The expression 37i7r + (- l)**a±(j7r- a) is included in the expression
WTT + ( - 1)** a dt ( Jtt - a) whatever integer odd or even n may be, for when 3?i is
odd n is odd and when %n is even n is even ;
.'. sin~i a + cos-^ a = titt + ( - 1)** a ± (J ir - a).
12. tan ^ = ; therefore its sign depends on the signs of sin A and
cos A
cos A ; being positive when they have like signs ; and negative when they
have unlike signs.
sin 2 A = 2 sin A cos A ; therefore its sign depends on the signs of sin A
and cos A ; being positive when they have like signs, and negative when they
have unlike signs.
13. cos A + cos ^A + cos 5^ = ; .*. cos 3^ + (cos bA + cos -4) = ;
.-. cos 3^ + 2 cos 3^ cos 2^=0; /. cos 3^ (1 + 2 cos 2^) = 0;
:. cos3^=0, or 1 + 2 cos 2^ = 0.
If cos 3^ = 0, then cos 3^ = cos 4 TT ;
.*. 3^ = 2w7r±47r, i. e. W7r + Jtt; .*. A = ^(mr + \Tr) = \ (2?i + l)7r.
If l + 2cos2^ = 0, /. cos2^=: -4 = cos§7r;
.-. 2^==2;i7r±§7r, i.e. ^=: J (3?t±l) tt.
14. sin5^ + sin3(9 + sin^ = 3-4sin2^,
sin %d + (sin 5^ + sin (9) + 4 sin' ^ - 3 -0,
sin 3^ + 2 sin 3(9 cos 26 -i- 4 sin2 (9-3 = 0,
sin 3^ (1+ 2 cos 2(9) + 4 (1 - cos2^) - 3 = 0,
sin 3^ (4 cos2 ^ - 1) - (4 cos^ (9 - 1) =0,
(4 cos2^-l) (sin 3(9-1)^0;
.-. 4cos2(9=:l or sin 3^ = 1.
If 4cos2^ = l; .*. cos2^ = J = cos2j7r; .*. cos^= icos^Tr.
If cos^=cos^7r, d = 2mr^^Tr.
If cos ^= -cos Jtt, then ^^tt-Jtt; :. 6 = 2mr^[Tr-^Tr).
The two expressions are included in the expression ?i7rrfcj7r, when n is
any integer whatever. ,
If sin 3^- 1 = 0, sin 3(9 = sin Itt; .'. 3(9 = 7i7r + ( - l)"47r. 1
15. 2sin2 3^ + sin2 6^ = 2;
.-. sin2 6^ = 2 (1 - sin2 3^ ), 4 sin^ 3^ cos^ 3^ = 2 cos^ 3^ ; ^
therefore either cos2 3^=0, or sin2 3J=J.
If cos2 3^ = 0, 3^=W7r + i7r.
If sin2 3^ = i = sin2i7r, 3^=7^7^ + ( - If jTr.
ON SUBSIDIARY ANGLES. L. 115
16. a(GOs2x-l) + 2b(QOSx + l)=:0;
.-. 2a(cos2a;-l) + 2&(cosa; + l) = 0;
.*. 2{cosx-T-l){acosx-a + b) = ;
therefore either cos x + 1 = 0, or a cos x-a-\-b = 0.
If cos jc f 1 = 0, cos x= -1\ .'. x = 2)nr + tt.
Tfi .i,A ^-^ ,a-b
If acosx-a-\-b = 0. cosa; = , /. a; = cos~^ .
a a
17. sin (m + n) 6 + sin 2nid + sin {in -n)d = 0;
.'. {sin(w + w) ^ + sin(m-w) ^} +sin2m^=0;
.*. 2 sin md cos w^ + 2 sin md cos m^ = ;
/. 2 sin md (cos w^ + cos md) = ;
therefore either sin md = 0, or cos nO + cos md = 0.
If sinm^ = 0, md = mr, .'. d = — .
m
If coS7i^ + cosm^ = 0,
then 2cos J^(m + w)cos J^(w-w) = 0;
therefore either cos ^d(m + n) = Oy or cos ^d {m-n) = 0.
If cosi^(m + w)=0, J^ (m + w) = r7r + j7r;
. . = TT.
m+ n
If cosj^(m-n)=0, J^{m-7i) = ?'7r + j7r;
Both expressions are included in the expression
(f= — , TT.
m±n
18. sin{Trx (x + y)} + sin {try (x + y)} = (i),
i.e. 2sinj7r(a: + y)2cos Jtt (x'^-y^) = (ii),
sin irx^ + sin iry^ = ( iii) ,
i.e. 2sin4'n-(a;2 + 2/2)cos j7r(a;2-y2)_o (iv).
Now (i) and (iii) are simultaneously true if (ii) and (iv) are simultaneously
true; that is, if the same values of x and y satisfy the two equations,
sin 4 TT (x + ?/)2 cos J T (a;2 - y2) — 0,
sin J TT (a:2 + 2/2) cos J TT (:c2 - y 2) = 0.
Both equations together become zero if either
cosi9r(a:2-2/2)=:0 or if sin J tt (x + y)2 = and sin J tt (x^ + ^2) = o.
If cos J9r(a;2- 2/3) = 0, then Jtt (x2^2/^) = wir + Jtt;
.*. x^-y^=2n + 1, i. e. an odd number.
If siniT(ar + 2/)2 = 0, ^ir(x- + 2/)2=:n7r and (a; + y)2 = 2n.
If sin J IT (x2 + y2) - 0, 4 TT (ar2 + y2^ _ ^^ ^nd x- + y^ = 2m.
8—2
116 ON LOGARITHMS. LI.
Combine these two last equations, for in order that the general equations
should be true, these two are to be true together,
{x + yY - [x^ + y2) = 2n - 2m = 2xy,
(x- yY^ = x^ + y^- 2x7/ = 2m-(2n- 2m) = 4:171 -2n.
For 19, 20, 21 see Ans. E. T.
EXAMPLES. LI. Page 162.
1. If m = a/*, n = a^.
(i) m2 X w3 = (a*)2 x {a^f = aP' x a^* = a^^+^K
(ii) m'^-^-n^ = (a*)4-r- [aJ^f = a-^^-v-a^* = a^-s*.
(iii) ^/Im-i X 7i5) = 4/{(a*)* x (a*)^} = ^{a^ x a^*) = ^/(a'^^+s*) ^ (^4^+5*)^ = a » *
(iv) {4/(m5 X w3)}2={4/[(a*)5 X (a*)3]}2 ={4/(^5* ^ rt:iA)j2^{^a5*+aA:}2
5M;:3fc
2 (i) 453 X 650 = lO^'^^^^^^^ X 10^ 8I29134 _ 1026560982+2-8129134 _ 1()6-4690116^
(ii) (453)4— n02'^^^^2\4_]^()2-6560982x4_ 2^010-6243928^
(iii) (650)3 X (453)2 = (102-8129134)3 ^ (102-6560982)2 ^ 108'*387402 ^ IQS 3121964
_ l()8-4387402+5-3121964 _ ;[Q137509366^
(iv) /^4:53 = ^102^^60982 _ n ()2-6560982\4 — IQ J (2 6560982) _ 1Q-8853»)61^
(V) ^453 X 4/650 = ^102-6560982 ^ ^102-8129134
= (102-6560982)4 X (102-8129134)^
— 102 (2-6560982) ^ 10^ (2-8129134) _ 1013280491 ^ 10-4688189
_ ]^01-3-80491+-4688l89 — 101-7968680^
(Vi) 4/453 X (1(550)-* = 4/102 6560982 X (102-8129134)3
_ n 02-65€0982)i ^ (102812134)3
= lOi (2-6560982) ^ 102812134 x 3 _ 10-53121964 ^ 1084387402
_. 1 5312196+8 -4387402 _ 1 08-9699598^
(Vii) ^/(453 X 650)=V'(102"«^«^^«' X 102-8l29134):^^(102-6560982+2-8129134) j
= v/105-4690116_ n05-46901l6)i_ lo4 (5 4690116)= 102-7345068^
3. 8 = 2x2x2 = 23, 32 = 8x4 = 23x22 = 23+2 = 25,
i-?- ?. =2— 22=21-2 = 2-1 —= — = -, = 2h- 25 = 21-5 = 2-4
2~4~22 • ^ '16 32 25
.125— i?^ = - = -- = - =2-- 24=21-4=2-3
^"^^""1000 8 16 24 ^ '
128 = 16 X 8 = 24 x23 = 24-+3 = 27.
ON LOGARITHMS. LII. 117
4. 9 = 3x3 = 32,81 = 9x9 = 32x32 = 32+2 = 3^
1
81
3
~ 81 X 3 ~
3
34 X 3 "
exa:
3
- 34+1
mpl:
3
BS.
-3-^3^:
LII.
= 31
P.
-6 _
163.
1 22 = (10'3<>10300\2 _ ;[Q-60206^ 32 _ n ()-4771213\2 _ J^Q-9542426^
23 _ n O'^^^^^^^M'*^ = 10^^^^^
2x3 = 10'3<^^**3 X 10''*''''1213 _ 1Q-30103+-4771213 _ 1()-7781513
24 _.n03010300y4_101 2041200^ 72=:(10*^"^^^^^*^)2 = 10^'^^^^^^.
2 14 = 7x2 = 10'8^50^<) X lO-*^^^^ = 10^"*^^^^+ 3030300 _ IQI -146128^
16 = 24 = (10-30103)4 ^ lQl-20412^
18 = 9X2 = 32X2 = (10-4771213)2 ^ 10-30103
_.lQ-9542426 ^ 10-30103 _ 10'9542426+-30103— 101-2552726
24 = 3x8 = 3x23 = 10-477i?i3 x (10-30103)3
_ 10-4771213 X 10-90309_ 10-4771213+-90309 _ 101 '3802113
27 — 33 — (10 4771213)3 _ 1014313639^
42 = 7x6 = 7x2x3 = lO'^^sogs x lo soios x 10-4771213
— 10-845098+-30103+-4771213 _ 101-6232493^
3. 10=10', 5 = 10-^2 = 10'-v-10•30103.-^101-30103^10-69897^
15 = 10 X 34-2 = 10' X 10-4771213^ 10-30103 ^101+-4771213--30103^ 1011760913^
25 =^ 100-^-4 = 102 -J- 22 = 102 -^ (10 30103)2"^ IO2- 60206^ 101-39794^
30 = 10 X 3 = 10' X 10-4771213^101-4771213^
35__Y X 10-T-2 = 10'^-'^^^^ X 10'-f-10'30103_. 10 845098+l--30103_ 101-544068^
4. 36 = 9 X 4 = 32 X 22= (10 4771213)2 x (10-30103)2
_ 10-9542426 x 1060206 — 10 9542420+-60206 _ 101*556 :026
40 = 10 X 4 = 10 X 22 = 10 ' X (10-30103)2 ^ 101+ 60-J06 ^ 101-60206^
50 = 100 4- 2 = 102 -^ 10-30103 = 102- 30103 ^ 101 69897^
200 = 2 X 100 = 2 X 102 =. 10 30103 x 102 = 10^ 3oio3^ 1000 = 103.
5, 310 X 710.^220 -(10-4771213)10 X (10-^45098)10^(10 30103)20
_ 104-771213 X 10^ 45098 _^ 1060206
__ 104-771-213+8-45098-60206 _ 107201593
212 X 320_^7]1 _ (10-30103^12 x (104771213)20_^/10-846098
_ 103-61236 X 109'54242«_i_ 109-296078
_ 103-61236+9 542426-9 296078 — 103-858708^ '
118 ON LOGARITHMS. LII.
6. 4/21 X 4^18 = 4/(7 X 3) X 4/(9x2)
_^(10-845098 X 10'*771213) ^ ^| (IQ 4771213)2 x IQ-SOlOS J
— 3/10-845098+-4771213 ^ 4/X0"9542426+-30103
_ HQl -3222193 i J ^ (IQ^ 2552726\i == JQ '*^^^'*^^+'36381815_ 10'7645579
4/(-49 X 45) X 4/(34 X 210)
= 4/(72-f-102 X 2^0) X 4/(3* X 2i«)
= 4/{(10845098)2_^102 x (lOSOlOS^lOj x 4/{(10'477l213)4 x (1030103)10|
_ 2/]^Q-845098-2+3 0l03 x 3/][Ql-9084852+30103
— n 02700496\ J X (10-*'^1*^'852\ J _ ^Ql 350248 x 101'639595
— IQl ■•^50348+1 -639595 __ J Q2 989843^
7. 3iy42 ^ jy(7 X B X 2) =: ]1J/(10S45098 x 10-4771213 x 1030103)
— 10/1Q845098+-4771213+ -30103
_ 10/XOl -6232493 ^=(10^ •6232493\tV = 10*1623249^
But 101623249^1.4532, ... 14/42 = 1-4532.
8. 4^(42)4x4/(42)3
= 4/(7x3x2)4x4/(7x3x2)3
— 3//10-845098 X 104771213 x 10'30103W
X 4/(10 845098 X 10-4771213 x 10-30103)3
— 3//10-845098+-4771213+ 30103)4 x Vn0-845098+-4771213+-30103)3
= 4/(101623249)4 X ^(101-623249)3^(106-492996)4 x (104-869747)J
_ 102164332 X 101-21743675 _ 102-1643324-1-21743675_ 10338177
But 103 38177^2408-6, .-. 4/(42)4x4/(42)3 = 2408-6.
9. (i) 4/6x4/7 + 4/9
= 4/(3 X 2) X 4/7x4/32
= 4^(10'4771213 X 10-30103) x 4/10846098 x 4/(10-47n213)2
_ 3/IQ-4771213+-30103 x 4^10845098 x 5/10-9542426
= (10-7781613)J X (10-845098) J x (10»542426)i
_ 10-25938377 x 10-2112745 x 10 19084832
_ 10-25938377+-21 12745+ -19084832 _ 106615067.
But 10-66i5«67^ 4.5868; /. 4/6x4/7x4/9 = 4-5868.
(ii) i4/2x3-*'x7A
_ (10-30103) tV X (10-4771213) -i x (10"846098)TT
_ 10030103 X 10--5964016 x 10-5377896
_ 10030103— -5964016+-5377896 _ 10-028509
But 10-028509^.93(546^ . w/2 x 3-^x 7Tr = .93646.
ON LOGARITHMS. LII. 119
10. (67-21)^ X (49-62)^ x (3-971)-"^
_ QQl 8274339\f ^ (10' '^^^^^^S) 5 ^ /lQ-5988!)!»tn " i
_ 1010»646034 X lO'^^'^^^^SS X 10--83845986
_ 101'^9646034+-33913136--83845986 — X0'59713184^
But 10-5»7i3i0:= 3.9549^ ... (67-21)*x (49-62)* X (3-971)- » = 3-9549 nearly.
11. Area of field = 640-12 x 640-12 = (640-12)2 =(10*^-80626i4)2^ 105-6125228.
But 105 6125228^40975.3;
.-. area of fields 40975-3 square feet.
12. Edge of cube = ^42601 inches
_ 3/ j^Q4-6294l98 — IqA (4-6294198) _ 101 5431399 .
But 101-5^1=^99 = 34-925;
.-. the edge of the cube = 34*925 inches.
13. The edge of the cube = 4/34-701 inches
— 3/lQl-5403420_ ]^q4 (1-5403420) _ ]^Q-5134473.
But 10 5i3'*473^ 3.2617;
.-. edge of the cube = 3*2617 inches.
14. Volume of cube = (47-931)3 cubic yards = (lOiesoeiesjs = 10504i84y5^
But 10-5 0418495^110115;
.-. volume of cube = 110115 cubic yards.
EXAMPLES. LIII. Page 166.
1, The index of the power of a which = a^ is 3, .-.3 = log„ a^,
10
= (i'^^' >. 3", .*. 3-3 = log„aV-,
= 4/«(i-e. ai)is j , .-. ^ = log„4/a,
= 4/a2(i.e. aS)is | , .'. | = log« ^/a^,
1 ,. -I . 5 5 , 1
= -(i.e.a i^ IS--, .-. -- = log^— .
2. 8 = 23, .^ 3 = log2 8, 64 = 26, ... 6 = log2 64,
1=2-1,... -l^log^i,
•^^^ = nlo=l = |3 = 2-3,.-. -3 = log,.125,
•015625 = 1= i = 2A .-. -6 = log2 -015625,
^64 = (64)i=(2«)4 = 2^ .-. 2 = logo 4/64.
120 ON LOGARITHMS. LIII.
3. 9=33, .-. log39 = 2; 81 = 3«, .-. log3 81 = 4,
1 = 3-1, • log3i=-l; ^7 = p=3-^ .•.log3l=-3,
■i=5 = P=^"' ••••°g3-i=-2; 8-i=p=3-, .Mog3^=-4.
4. 8 = V64=(64)i = (4»)i = 4i .-. log,8 = |,
4/16=^42 = (4=^i = 4i .-. log,4/16=|,
4/.015625=^l=(l) =(4-3)4 = 4-1, .-.log, ^^015625= -1.
5. log2 8 = log2 23=:31og2 2 = 3,
log2 -6 = logg i = log2 2-1 = - 1,
log3 243=:log3 35 = 5,
logs (-04) =log5 ^V = log5 5-^= - 2,
logiolOOO=logio 103 = 3,
logio(-100)=logiolO-3=-3.
6. ^ogaai = ^logaa = i; forlogaa=l,
log^4/62=log56§ = §log56 = t,
I0g8 2 = l0g8 8i = 4l0g8 8=:i,
l0g27 3 = l0g27 27* = J l0g27 27 = J,
logioo 10 = logioo 100* = i logioo 100 = J.
7. logio 6 = logjo (3x2) = logio 2 + log^o 3 = -4771213 + -30103 = -7781513,
logio 42 = logio (2x3x7) = log^o 2 + log^o 3 + logj^ 7
= -30103 + -4771213 + -845098 = 1-6232493,
logjo 16 = logip (24) = 4 logj^) 2 = 4 X (-30103) = 1-2041200.
ON LOGARITHMS. LIII. 121
8. logio 49 = logio 72 = 2 logjo 7 = 2 X (-845098) = 1-690196,
logjo 36 = logio (4x9) = logjo (2^ x 3^) = log.^ (22) + log^o (3^) = 2 log^o 2 + 2 log^^ 3
= -6020600 + -9542426 = 1-5563026,
logio 63 = logio (9x7)= logio (3^) + logjo 7 = 2 logio ^ + logio 7
= -954246 4- -8450980 = 1-7993406.
9. logio 200 = logjo (100 X 2) = 2 logio 10 + logi^ 2 = 2 + -30103 = 2-30103,
log 600 = los (100 X 3 X 2) = log 100 + log 3 + log 2
= 2 + -4771213 + -3010300 = 2-7781513,
log 70 = log 10 X 7 = log 10 + log 7 - 1 + -8450980 = 1-8450980.
10. log 5 = log (10 -=- 2) = log 10 - log 2 = 1 - -30103 = -6989700,
log 3-3 = log y= log (104- 3) = log 10 - log 3
= 1- -4771213 = -5228787,
log 50 = log (100--2) = log 100 - log 2
= 2 - -3010300 = 1-6989700.
11. log 35 = log (70--2) = log (10 X 7-^2) = log 10 + log 7 -log 2
= 1 + -845098 - -3010300 = 1-544068,
log 150=log (100 X 3+-2) = log 100 + log 3 - log2
= 2 + -4771213 - -30103 = 2-1760913,
log (-2) = log (2-+ 10) = log 2 - log 10= -30103 ~ 1= - 1 + -3010300.
12. log3-5 = log(7^2)=log7~log2
= -8459800 - -3010300 = -5440680,
log7-29 = log(729--100) = log(3«--102)
= 6 log 3 - 2 log 10 = 2-8627278 - 2 = -8627278,
log -081 = log (81 -+ 1000) = log (3^ -- 103)
= 41og3- 3 log 10 = 1-9084852 -3= -2 + -9084852.
13. Iogio{y6x^7 X V9} = log,o{(3 x 2)4 x 7* x (3-^)^}
= J logio (3x2) + J logio 7 + f logio 3
= J logio 3 + i logio 2 + i logio 7 + 1 logio 3
= J logio 2 + H logjo 3 + i logio 7
-30103 11 X '4771213 -845098
3 ^ 15 ^ 4
= -1003433 + -3498889 + -2112745 = -6615067.
But -6615067 = logio 4-5868; .-. logio {^6 x 4/7 x ^9} = logjo 4-5868;
.-. -V^x ^7x^9 = 4-5868,
logio {JJ/2 X 3-^ X 7^^} =log 2tV + log 3~i + log 7^^"i
= tV logio 2-1 logio 3 + i^T logio 7
•30103 5 X -4771213 7 x -845098
+
10 4 ' 11
= -030103 - -5964016 + -5377896
= - -028509 = logio -93646 ;
/. 5J/2 X ^-i X 7^T = -93646,
122 ON LOGARITHMS. LIV.
14. (i) log{y2x^7-^V9}
=log {2J X 7i-^9^} =log {2* X 7i-^(32)i}
=:log {2^ X 7i-h3« } =:log 2^ + log 7^ - log 3^
= 41og2 + Jlog7-|log3.
(ii) Vid. (13).
15. (i) logioa& = logjoa + logio/>
= 2-6560982 + 2-8129134 = 5-4690116.
(ii) logioa'* = 41ogi^a=4 x 2-6560982
= 10-6243928.
(iii) \ogiQa^b^ = 2\oga-\-Slogb
= 5-3121964 + 8-4387402= 13-7509366.
,. , , :w- , , 2-6560982
(iv) logio V'^ = ilogio«= 3
= -8853661.
(v) Iogio(a3&)^ = i(31ogioa + logio6)
7-9682946 + 2-8129134
~ 6
^10-7812080^^.^^^^^^^
D
(vi) logioai63 = logioa^ + logio?>^
=:ilogioa + 31ogio6
5
= 8-9699598.
16. (i) log JO ( V^2rx4/l8) = logio 4/7^ +logio 4/3^x2
^ log^o7-^log^o3 , 21ogio3 + log,o2
6 ' 4
1-3222193 1-2552726
=— ^-+— T—
= -4407397 + -3138182 = -7545579.
(ii) logio^(-49x 4-^) X ^(3^x210)
= J (2 logjo 7 - logio 102 + 10 logio 2) + i (4 log^o 3 + 10 log^o 2)
= 1-3502480 + 1 -6395951 = 2-989483.
EXAMPLES. LIV. Pages 169, 170.
1. 17601 is between 10-* and 10^; .-. logj^ 17601 = 4 + a decimal.
361-1 is between 10^ and 10»; .'. log^o 361-1 = 2 + a decimal.
4-01 is between 10^ and 10 ; .'. log^Q 4-01 is a decimal.
Integral part = 0.
723000 is between lO-^ and 10^; .'. log^o 723000 is 5 + a decimal.
29 is between 10 and 10^ ; .-. log^) 29 is 1 + a decimal.
ON LOGARITHMS. LIV. 123
2. '04 is between 10-^ and 10~^; .*. logiQ '04= - 2 + a positive decimal;
.*. -2 = integral part.
•0000612 is between lO-^ and 10-^;
.'. logjo '0000612 = - 5 + a positive decimal ; .*. - 5 = iiitegral parts.
•7963 is between 10-^ and 10" (i. e. 1);
.-. log^o '7963=: - 1 + a positive decimal ; /. integral part= - 1.
•001201 is between 10-3 and 10-2; .-. logj^ -001201 = - 3 + a positive decimal;
.*. integral part= -3.
3. 7963 is between 10** and 10-*;
.-. logjo 7963 = 3 + a decimal ; .'. integral part = 3.
•1 is between 10"^ and 10"; .*. logj^^l^ - 1 + a positive decimal;
.-. integral part= - 1.
2-61 is between 10" and 10^; .*. log^o 2-61 = + a decimal;
.-. integral part = 0.
79-6341 is between 10^ and 10^;
.-. logjo 79*6341 = 1 + a decimal ; .*. integral part = 1.
1-0006 is between 10" and 10^; .*. log^o 1-0006 = + a decimal;
integral part = 0.
-00000079 is between 10-^ and 10"^;
.-. logio -00000079= - 7 + a positive decimal; .*. integral part= - 7.
4. 103-461 is between 103 and 10^ ;
.'. there are 4 digits in integral part of the number.
10-30203 is between 10" and 10^ ;
.*. there is 1 digit in integral part of the number.
105-4712301 is between 10" and 10^;
.*. there are 6 digits in integral part of the number,
102-6710100 is between 10^ and 103 ;
.*. there are 3 digits in integral part of the number.
5. The logarithm of the number is between - 2 and - 1 ;
.-. the number is between 10"^ and 10-^,
,, ,, ,, -01 and -1;
.'. the first significant figure is in second decimal place.
The logarithm of the number is between - 1 and ;
.•. the number is between 10~^ and 10",
-1 and 1 ;
.-. the first significant figure is in first decimal place.
The logarithm of the number is between - 6 and - 5 ;
.-. the number is between 10"^ and 10"^,
•000001 and -00001 ;
/. the first significant figure is in the sixth decimal place.
124 ON LOGARITHMS. LIV.
6. The number is logiolO^'^^^'i^^;
/. the number is between 10"* and 10'\
„ 10000 and 100000;
.*. the first significant figure is ten thousands.
The given number is 10 '"^o^^s^;
/. „ „ „ is between lO** and 10^,
1 and 10;
.*. first significant figure is in units place.
The given number is 102-5860244 .
.-, the number is between 10"^ and 10^,
„ 100 and 1000.
1, The given number is 10-3+-1760913.
.-. the number is between 10~^ and 10"^;
„ „ „ -001 and -01.
The given number is 10~^+^
.-. the number is between 10~^ and 10'\
„ -1 and 1.
The given number is 10-^80347.
.-. the number is between 10^ and 10^,
„ 1 and 10.
7. log 8.10 ^ 10 log 8 = 30 log 2 = 9-0309 ; /. S^^ = 109t>309 .
.-. 8^0 is between 10^ and lO^o.
log 212=: 12 log 2 = 3-61236 ; .-. 2^2 is between 10^ and lO^.
log 16-0 = 80 log 2 = 24-0824; .'. 16^0 is between lO^-* and 1026.
log 21^ = 30-103 ; .*. 2^^ is between lO^o and lO^i.
8. log 7^0 = 8*45098 ; .'. 710 is between lO^ and 10«.
log 496 = 12 log 7 = 10-141176 ; .'. 49« is between lOio and 10".
, ;, . log 343-^- = log (73)"^~ = log 7100 = 84-5098;
.-. 343-^^- is between lO^^ and lO^^.
log (V )^^ = 20 log 10 - 20 log 7 = 3-09804 ; .-. ( Y)^ is between 10^ and 10*
log (4-9)12 = log (11)12 = 24 log 7 - 12 log 10 = 8-282352;
.-. (4-9)12 is between 10^ and 10^.
log (3-43)10 = log (f ^f )io = 30 log 7-20 log 10 --= 5-35294 ;
/. (3-43)10 is between 10^ and W,
ON LOGARITHMS. LIV. 125
9. logjo '^2 = -^V logio 2 = -030103 ; /. '^2 = lO-^^oios .
.-. iJ/2 is between W and lO^;
„ 1 and 10.
log (J) 10= - 10 log 2= -4 + -9897;
/. (i)io(=10-4+-9897) is between 10"^ and lO"-.
log ( V-)2o = 20 log 10 - 20 log 7 = 3-09804 ;
. (y.)2o^ 10309804. .^ (.y>)-o is between 103 and 10^,
log (-02)^ = 4 log -2-4 log 100= - 7 + -20412 ;
.-. (-02)^ = 10-7+20412. .^ (-02)4 is between 10"^ and 10-«.
log(-49)6 = log(49-M00)6=121og7-121ogl0= -2 + -141170;
/. (-49)6 is between 10"* and lO-^.
10. log (20)' =7 log 2 + 7 log 10 = 9-10721;
.-. 20^ is between 10^ and lO^o.
(•02)7 = (2 -f- 100)7 = (10'30103_^ 102)7 = (10-2+30103)7 _ 10-14+210721 _- 10-12+10721
.-. (-02)7 is between 10-^2 and 10-^^
log (-007)2 = log (7-^ 1000)2 2 log 7 - 6 log 10= - 5 + -690196;
.-. (-007)2 is between 10-^ and 10-^.
log (3-43)TV=:log (343-t-100)tV^.0535294;
.-. (3-43)iV is between I and 10.
log (-0343)8= log {73-f-10'*}8= - 12 + -282352;
.-. (-0343)8 is between 10-^2 and lO-^i.
log (-0343)A = log(73^10'*)TV== - 1 + -8535294;
.-. (-0343)^1^ is between -1 and 1.
EXAMPLES. LV. Pages 172, 173.
1. 776-43 =^mi- = log 77643 - 2 = 2-8901023 ;
7-7643 = mil = log 77643 - 4, -00077643= r^H§U77^ = log 77643-8,
logio 776430 = 5-8901023.
2. logio 5908200 =6-7714552, log^^ 5-9082 = -7714552,
logio -00059082 = 4-7714552, log^^ 590-82 =2-7714552,
logio 5908-2 =3-7714552.
3. log 4/(-0059082) = i {log -0059082 - log 107}
= i{-3 + -7714552} = i{- 4 + 1-7714552}
= -1 + -4428638= - 1 + log 2-7724 = log -27724.
4. log {-00059082 X -027724} = - 4 + -7714552 - 2 + -4428638
= - 5 + -2143190 = log -00001638.
5. log 5J/(-077643) = tV { - 2 + -8901023 } = ^-^ { - 10 + 8-890123 }
= -1 + -88901023= - 1 + log 7-7448=log -77448.
126 ON LOGARITHMS. LV.
6. log {(-27724)2 X -077643} = - 2 + -8857270 - 2 + -8901023
= - 3 + -7758299 = - 3 + log 5-9680
= log -005968.
EXAMPLES. LVI. Page 175.
1. Let m be any number, and let x be its log to base 8.
The log of m to base 2 is supposed to be known.
Now m = 8^ = (23)^ = 23^, or 3a; = logg"* ;
.-. x = ^ of the log of m to base 2.
2. Let m be any number, and let x be its log to base 3.
The log of m to base 9 is supposed to be known.
Now m=3^=:(94)^=9^ or | = logj,"»;
.-. .r = 2 X the log of m to base 9.
3. Let m be any number, and let x be its log to base 2.
The log of m to base 10 is given.
Now m = 2^=-(103oio3...)x,^10xx-3oio3...^ oj. a;xlogio2 = logio^;
.-. X = log of m to base 10 divided by logj^^.
4. Let m be any number, and let x be its log to base 10.
The log of m to base 3 is known.
Now m=:10=«=2 (3»«g3l«)^:=3^x^o83l^ or aj x logg 10 = log3^;
.*. X — log of m to base 3 divided by logg^^.
5. Let m be any number, and let x be \o^.^.
The log of m to the base 10 is known.
Now, 7rt = 3«rz:(10^ogio3)^=10^'%o3; z. ic logjo 3 = log^o*" ;
.-. the log of m to base 3 = log of m to base 10 divided by log^^^^
6. Let a^rrlogglO; then2^-10; .-. 2 = 10^
But 2 = 103oio3oo. . _:= -30103.
x
1
7. Let X = log7 10 ; then 7^ = 10 ; .'. 7 = 10^.
But 7^10-8450980 . ^:=. -8450980.
X
8. Let x = logs 10; ^^^^ 8^ = 10; /. 23^=10, or 2 = 103*^.
But 2 = 10-30103. ... i-=. 30103.
6x
1
Let y = log32l0; then 32?' = 10 or 2 = 105^^;
.-. -^ = -30103.
ON LOGARITHMS. 127
MISCELLANEOUS EXAMPLES. LVII. Pages 175, 176.
1. Let log28 = a;; .'. 8 = 2% but 8 = 23; . 2^ = 2^;
.-. x = S and logg 8 = 3.
Let loggl^o:; /. 1 = 5=^; but 1 = 5°; /. 5^ = 5°;
.'. .T = 0, and log5l = 0.
Let a; = log82, then 8^ = 2; but 8^ = 2; /. x = ^.
Let log7l = a:; /. 1 = 7^; but 1 = 7°; /. 7^=7°;
/. x = and log7l = 0.
Let a; = log32l28; then 32^=128; or 2^^= 2^; /. x = t.
2. In every system of logarithms the log of 1 is 0; for a°=l,
log 2, and log 3 are given,
4 = 22; .-. log4 = 21og2,
5 = 10-^2; .-. log 5 = log 10 -log 2,
6 = 3x2; .-. log 6 = log 3 + log 2,
log 7 is given,
8 = 23; ., log8 = 31og2,
9 = 32; . log9 = 21og3,
log 10 = 1, log 11 cannot be found from our data,
12 = 22x3; .-. log 12 = 2 log 2 + 3,
log 13 cannot be found from our data,
14 = 7x2; .-. log 14 = log 7 + log 2,
15 = 10x3^2; .-.log 15 = log 10 + log 3 -log 2,
16 = 24; .-. log 16 = 4 log 2,
log 17 cannot be found from our data,
18 = 32x2; .-. log 18 = 2 log 3 + log 2,
log 19 cannot be found from our data,
20 = 10x2; .-. log 20 = log 10 + log 2,
21 = 7x3; .-. log 21 = log 7+ log 3,
log 22 cannot be found from our data,
log 23 cannot be found from our data,
24 = 23x3; .-. log 24 = 3 log 2 + log 3,
25 = 102+- 22; .-. log 25 = 2 log 10 + 2 log 2,
log 26 cannot be found from our data,
27 = 33; .-. log 27 = 3 log 3,
log 29 cannot be found from our data,
30 = 10x3; .-. log 30 = log 10 + log 3.
The eight numbers are 11, 13, 17, 19, 22, 23, 26, 29.
3. log 1 = 0,
log2 = log^8 = ilog8,
128 ON LOGARITHMS. LVII.
log 3 = log ?i^ = log 21 + i log 8 -log 14,
log4 = 21og2=:flog8,
log5 = log-V-=l-log2 = l-ilog8,
log6 = log(3x2) = log3+ log2
= log 21 + 1 log 8 -log 14,
log 7 = log -V = log 14 - log 2 = log 14 - i log 8,
log 9 = 2 log 3 = 2 (log 21 + ^ log 8 - log 14).
4. log 85762 = 4-9332949;
.-. log •0085762 = 3-9332949,
log5y-0085762 = 3i, log -0085762 = ^^5^^
-11 + 8+ -9332949
11
= -1 + -8121177
= 1-8121177,
log (85762)11 = 11 log 85762 = 11 x 4-9332949
= 54-2662439;
.-. (85762)11 = 1054-2662439 .
.-. (85762)11 is between 10^4 and lO^S;
.'. there are 55 figures in the integral part of this power.
5. log {47-609 X 476-09 x -47609 x -000047609}
= log {4-7609^ X 10 X 102 x lO-i x 10-^}
= 41og4-7609 + l + 2-l-5
-14.2-1-5 + 4X-6776891
= -3 + 2-7107564
= - 1 + -7107564 = 1-7107564 = log -51375 ;
.-. 47-609 X 476-09 x -47609 x -000047609 = -51375.
6. By trial 3^ = 3087 and 3^ = 9261;
.-. 3742 is between 3^ and 3^;
.-. logg 3742 = 7 + a decimal.
By trial 6^ = 1296 and 6^ = 7776 ;
.-. 3742 is between 6^ and 6^;
.-. logg 3742 = 4 + a decimal.
By trial 10^ = 1000 and 10^ = 10000 ;
.'. 3742 is between 10* and 10^;
.-. logio ^742 = 3 + a decimal.
By trial 12^ = 1728 and 12^= 20736 ;
.-. 3742 is between 12^ and 12^;
.-. logi2 3742 = 3 + a decimal.
ON LOGARITHMS. LVII. 129
7, (i) 2^x3^* = 72,
.-. log (2^x3'**) = log 72,
.-. log2« + log34a^ = log72,
.-. a;log2 + 4a;log3 = 21og7,
.-. a;(log2 + 41og3) = 21og7,
21og7
log2 + 41og3*
(ii) 32a:=128x7-*-^
.-. log 32^ = log (128 X 74-^) = log (27 X 7^-^) ;
.-. 2a;log3 = 7log2 + (4-a;)log7;
.-. a;(2log3 + log7) = 71og2 + 41og7;
_7lo g2 + 41og7
•'•'^~ 21og3 + log7 •
(iii) 12^ = 49 = T\ :, log 12^ = log 1\ /. x log (4x3) = 2 log 7 ;
/. a:(log4 + log3) = 21og7;
.-. a;(21og2+log3) = 21og7;
2 log 7
•'• '^~21og2 + log3*
(iv) 28^ = 2l4-3a;, .-. log 28«=:log 21'*-3^,
.-. So; log 2 = (4 - 3a;) log 21 = (4 - %x) (log 3 + log 7);
.-. 8a;log2 = 41og3-3a;log7-3xlog3 + 41og7;
/. a;{81og2 + 31og7 + 3log3}=4{log3 + log7};
. ,^ 4 (log 3 + log 7)
81og2 + 31og7 + 31og3'
8. Let X = log^ 490, then T = 490,
.-. 7^ = 7^x10; .-. 7^-2=10.
Equate the logs of each ; /. (x - 2) logj^ 7 = log^^ 10 = 1 ;
1
.-. x = 2 +
logio7 ■
9. Let a;=log9270; then 9^ = 270,
/. 32^ = 3^x10 or 32^-3 = 10.
Equate the logs of each ; .-. (2x - 3) logj^ 3 = log^^ 10 = 1 ;
3 1
•'•''-2 + 2Togio3-
10. Let a; = log5l0, then 5^=10,
/. 5»^ X 2* = 10 x 2^ ; /. 10^= 10 x 2^.
Equate the logs of each, then x logj^ 10 = logio 10 + a: log^, 2;
L. T. K.
130 ON LOGARITHMS. LVII.
11. loggQ^a, i.e. 9 = 8«; log^5 = h, i.e. 5 = 2&.
Since 5=:2&, /. 10 = 2&+i, /. 2 = 10&+r
••• logio2 = ^
1
"6 + 1 *
a 3a
Again 3=:v/9 = /^8« = 82=2 ^ .
1 3a 3a
But 2 = 10&+1, /. 3 = 2 2 = 102&+2 ;
3a
.-. Iogio3 =
Now Iogio4=:21ogio2=:
gio5 = logiol0-logio2r=l
Iogio6 = logio3 + logio2 =
26 + 2*
2
6+r
_ J_- ^
6 + l~6 + l'
3a 2 3a + 2
^<^ 2(6 + 1) ^2(6 + l)~2(6 + l)'
c6
log57 = c, 7 = 5^ .-. Iogio7 = clogio5 = ^-j-^.
12. 25 = 32 and 2^ = 64, .-. all the integers from 32 to 63 have 5 for the
characteristic of their logarithms ; all these integers will be found to be
(63-31), i.e. 32.
13. All the integers having 10 for the characteristic of their logarithms
begin with a^^ and the integer next before a^^; all these integers will be
found to be of the number (a^^ - a^^).
14. logjiy {(39-2)2} =-^\ log -W-= r\{log (7^ x 2^) - 1}
= ^2i-{2 X -845098 + 3 X '30103 - 1}
= XT (3*18572) = -289688 = log 1*9485.
15. The expression
= 7 (log 15 - log 16) + 6 (log 8 - log 3) + 5 (log 2 - log 5) + log 32 - log 35
= 71og3 + 71og5-281og2 + 181og2-61og3
+ 5 log 2 - 5 log 5 + 5 log 2 - 2 log 5
= log3.
1 6 . The expression = 2 { log a + log a2 + log a^ + . . . + log a"}
= 2{loga + 21oga + 31oga...+wloga}
= 21oga{l+2 + 3+...+n}
= i{n(n + l)] 21oga(by A.p.)
= n (n + 1) log a.
ON LOGARITHMS. LVII. 131
I 1
17. 'Letlogab = x, .'. b = a^; .'. b^ = a, :. - =\ogi,a.
Thus logabx\ogi,a = xx- = l.
Let x = logfib and y = logi,c.
Then & = a=^, c = by = a=^, .'. xy = log„ c.
But log^c X logca = l, /. \ogca=— ;
xy
.\logabx\ogj,cxlogca = xxyx— = l.
xy
18. By continuing as in 17, it can be shewn that
log„6 . logftC . log^d loggr . log^ a=:l ;
.-. log„6 . log^c. loggti log3r= J— -^ = log„r, since log„r xlogya = l.
19. Let log 3-456 = x, then 3-456 = lOo;,
.*. (3'456)i<^<^ooo_no*)i<^"^^<^=:10^^^^^^*^,
but (3-456)i<>««oo lies between 10^3856 and lO^^sss^
.'. 100000 xx = 53855 + a proper fraction ;
/. a:=:-53855... ; /. log 345-6 = 2-53855.
20. Let log3-981 = ic, then 3-981 = 10*; .-. (3-981)i«o<>oo^(10*)i«<>ooo^
but (3-981)i»«ooo lies between lO^woo and 10^99»«;
.-. 100000 xx = 50999 + a proper fraction ;
.-. a; =-59999...; .'. log 39810 = 4-59999.
21. Let P be the number of people living at the beginning of any year;
F P
then at the end of the year the number will be P -\--r^ - wj., i.e. Mi P.
48 bO
Similarly at the end of two years the number is iH ^ iii^ = (Hi)^ ^ y ^^^
at the end of x years the number will be (|U)*P.
Let the number be doubled at the end of x years,
.-. (|UfP = 2P; or, (|Hr=2;
••• ^ log 11^ = log 2 ; /. X (log 241 - log 240) = log 2,
log 2 log 2
'• ^ ~ log 241 - log 240 "" log 241 - log 3 - 3 log 2 - 1
'30103
= •0018057 = ^^^^'
.'. the population will be doubled within 167 years.
I
22. log 8 + log (s-a)- log b - log c =log s (s - a) - log be
9—2
132 ON LOGARITHMS. LVII.
23. log(a^-{-x^)-\-log(a-\-x) + log(a-x)
= log (a^ + x^) (a + x)(a-x) = log (a* - x*) .
24, log sin 4tA = log 2 sin 2 A cos 2^
=: log 4 sin ^ cos A cos 2^
= log 4 + log sin ^ + log cos A + log cos 2 A .
EXAMPLES. LVIII. Pages 181, 182.
1. -8839112) The differences in the numbers are '0001 and -00002,
•8839055 { the differences in the logs -0000057 and d ;
-0000057 ) -00002
•*• ^""^M^ *^^^^ -0000057= -0000011...
.-. log 7-6432 == -8839055 + -0000011 = -8839066.
2. Here, d=^ of -0000077 = -0000023 . . . ;
log 5-64123 = -7513715 -f -0000023,
.-. log 504-123 = 2-7513738.
3. Here, d = -^W of -0000050 = -0000008. . . ;
.-. log 8-736416= -9413325 + 0000008.
4. Here, <f=r^A of -0000067 = -0000017... ;
.-. log 6-437125 = -80B6903 + -0000017.
5. Here, d=^ of -0000116 = -000007... ;
.-. log 37245-6 = 4-5710680+ -000007.
6. -5686827 -5686760 Here, d = Ti^ pf -0001 = -00004... ;
•^^^^'^^^ -5686710 .-. -5686760 = log(3-7040 + -00004)
•0000117 -0000050 =log 3-70404.
7. Here,^ = JSgof-0001 = -000026...;
.-. -6602987 = log (4-5740 + -000026);
.-. 4-6602987 = log 45740-26.
8. Here, d = ^Ys of -0001 = -000037... ;
.-. -3966938 = log (2 -4928 + -000037) ;
.-. 6-3966938 = log2492 37.
9. Here, d = U of -0001 = -00008... ;
.-. -6431150 = log (4-3965 + -00008);
.-. 4*6431150 = log -000439658.
10. d = ^ot -0001 = -000058...;
.-. log -7550480=log (5-6891 + -000058).
ON LOGARITHMS. LIX. 133
EXAMPLES. LIX. Pages 184, 185.
-6736577 ^ = ^' ^^ -0002150 = -0001075 ;
"0002150 . sin 42° 21' 30" = -6736577 + -0001075.
2. Here, d = ^ of -0002150= -0001075;
.-. cos 47° 38' 30"= 6738727 - -0001075.
N.B. Corresponding to an increase in the angle there is a diminution in
the cosine.
3. Here, d='^, oi -0001064 = -0000798 ;
/. cos 21° 27' 45" = -9307370 - -0000798.
4. -6667493 -6666666 D=i^li of 60" = 37";
•^^^5325 -6665325 -6666666 = sin (41° 48' + D").
•0002168 -0001341
5. Here, D = ^y^ of 60" = 16-4" ;
-3333333 = cos (70° 32' - 16-4").
N. B. The angle diminishes as the cosine increases.
6. Here, D = ^f f of 60" = 39" ;
.-. -25 = cos(75°32'-39").
SO"
7. Here, d = g— , of -0001251 = -0000625 ;
.-. L sin 45° 16' 30" = 9-8514969 + -0000625.
45"
8. Here, d = r^. of -0003106 = -0002329 ;
.-. L tan 27° 13' 45" = 9-7112148 + -0002329.
20"
9. Here, d = -— of -0002647 = -0000882 ;
.-. L cot 36° 18' 20" = 10-1339650 - -0000882.
N. B. The cotangent diminishes as the angle increases.
10. 9 -8465705 9-8464028 .-. d = f H^ of 60" = 23" ;
9- 8463018 9-8463018 9-8464028 = L tan (35° 4' + 23").
-0002687 -0001010
11. Here, D = if| of 60" = 32-5'';
.-. 9-9448230 = L (cos 28° 17' - 32-5").
N. B. The cosine diminishes as the angle increases.
134 ON LOGARITHMS. LX.
12. Here, D = -^%% of 60" = 19" ;
10-4274623 = L cosec (21° 57' - 19").
N.B. The cosec diminishes as the angle increases.
EXAMPLES. LX. Pages 188, 189.
a 1046-7
sin^ =
c 1856-2 •
log sin ^ = log - =log a -log c
c
= 3-019822 -3-2686248= 17511974.
Here, d = ^At of 60" ; /. d =S1'S" ;
/. 9-7511974 = L sin (34° 19' + 31-8") ;
.-. ^=34° 19' 31-8".
2. - = cosec ^ = cosec 34° 15';
a
log c = log a + L cosec 34° 15' - 10
= 2 -9259306 + 10-2496421 - 10
= 3-1755727 = log 1498-2;
.-. c = 1498-2.
3. *^^^-^=4-7r2'
L tan ^ = 10 + log a - log 6 = 10-0093323.
Here d = f f |f of 60" = 56" ;
.-. 10-0093323 = L tan (45° 36' + 56") ;
.-. ^ = 45°36'56".
4. - = sin A;
c
:. log a = log 8762 + L sin 37° 10' - 10
= 3-9426032 + 9-7811344-10
= 3-7237376 = log 5293-4;
.-. a = 5293-4.
-=cos^;
c
log fe = log c + log cos 37° 10'
= 3-9426032 + 9-9013938-10
= 3-843997 = log 6982-3.
ON LOGARITHMS. LXI. 135
5. - = cot A\
.'. log a =log b - log cot A
= 3-2289647 - 10-4683893 + 10
= 2-7605754 = log 576-2.
6. h^ = (c2 - a2) = {c + a)(c- a) ,
21ogfe = log(c + a) + log (c-a)
= 3-7723951 + 3-5771470 = 7-3495421,-
.-. log 6 = 3-67477 = log 4729;
.-. fe = 4729 chains.
7. a3 = c2-/>2;
.-. 2 log a = log (c + b) + log (c-b)
= 3-6630410 + 3-4655316 = 7 1285726 ;
.-. log a = 3-5642863 = log 3666 -8.
8. log tan A = loga- log 6,
L tan ^ = 10 + log 7694-5 - log 8471
= 10 + 3-8861804 - 39279347
= 9-9582457 = L tan 42° 15'.
- = cosec ^ = cosec 42° 15' ;
a
.-. log c= log a + L cosec 42° 15' - 10
= log 7694-5 + L cosec 42° 15' - 10
= 3-8861804 + 10-1723937 - 10
= 4 -058574 = log 11444.
EXAMPLES. LXI. Page 190.
1. Vide fig. E. T. p. 186. Let AB be the distance; then, since
c = a cosec J,
.-. log c = log a + log cosec A
= log 2500 4- L cosec ^ - 10
= 3-3979400 + 10-1867171 ~ 10
= 3-5846571 =log 3842-9;
.-. c = 3842-9.
2. Vide fig. E. T. p. 186. Let the height he BC; then since a = &tan^;
.-. log a = log 369-5 + L tan 37° 19' 30" - 10
= 2-5676144 + 9-8822317 - 10
= 2-4498461 = log 281-74.
136 ON LOGARITHMS. LXI.
3. Figure as above. Let a be the height, then since
a = 5 tan ^ = 6 tan 32° 12'
.-. loga = logl76-23 + Ltan32°12'-10
= 2-2460798 + 9-8158311 - 10
= 2-061910=. log 115-32.
4. Figure as above. Let & be the required distance.
Then fc = a cot ^ = 163-5 x cot 29° 47' 18'';
log 6 = log 163-5 + L cot 29° 47' 18" - 10
= 2-2135178 4- 10-2422738 - 10
= 2-4557916 = log 285-6;
.-. & = 285-6.
L cot 29° 47' = 10-2423617, L cot 29° 48' = 10-2420687.
18"
Hence, d=:— of -0002930= -0000879...
.-. L cot 29°4r 18^=10-2423617 --0000879 = 10-2422738.
The logarithmic cotangent diminishes as the angle increases.
r T^. 1, « 673-12 , ,
5. Figure as above. - = pj-k.^q = tan A ;
.-. Z tan ^ = 10 + log 673 -12 -log 415-89
= 10 + 2 -8280925 - 2*6189785
= 10-2091140.
From the tables 10-2090013 = L tan 58° 17',
10-2092839 = L tan 58° 18'.
Hence, d = \\l^ of 60" = 24"...;
.-. Z tan ^ = L tan 58° 17' 24", A = 58° 17' 24".
J5 = 90° - ^ = 90° - 58° 17' 24" = 31° 42' 36".
^ ^. , • < « 576-12
6. Figure as above. sm ^ = - = - .
L sin ^ = 10 + log 576-12 - log 873-14
= 10 + 2-7605130 - 2-9410839
= 9-8194291.
From Tables, 9-8194012 = L sin 41° 17', 9*8195450 = Iv sin 41° 18',
^ = T¥Aof 60" = ll-6";
.-. 98194291 = Lsiu41°17'll-6",
h'^=zc^-a'^ = {c + a) (c-a)',
.-. 2 log 6 = log 1449-26 + log 297-02
= 3-1611463 + 2-4727857 = 5-63393 ;
.-. log & = 2-816966 = log 656-1.
ON LOGARITHMS. LXI.
137
7, Vid. fig. E. T. p. 62. Let OM be the lighthouse, Q and P fhe two
ships, Z OQM=27° 18', Z 0P3f= 20° 36',
PQ = PM - QM= OM cot 0PM - OM cot OQM
= 112-5 (cot 20° 36' - cot 27° 18')
= 112-5 (2-6604569 - 1-9374645)
= 112-5 X -72 = 81 feet.
8. Let AB be the cliff, CD the lighthouse, angle ^C£ = 23°17',
CD = 97-25 and Z ADD = 24° 19'.
AC^
CD
sin ADC
Since
sinD^C"
sin^2)O = sin(90°-^DB)=cosJD^.
ADC = 90°-ADB and ACD = 90°-\-ACE]
.-. (7^D = 180°-(^CD + ^X>(7)
= 180° - (90° + 23° 7' + 90° - 24° 19') = 1° 2' ;
^ AC _ cos^4° 19'
•'• CI)~ sini°2' '
,^ 97-25 X cos 24° 19'
^^^ sinl°2'
But -4J5 the height of the cliff above the light-house
. ^ sin 230 ir = ?I:Hi£^^° i9;_f i>^^?!il' ;
sin 1° 2'
.-. log AE = log 97-25 + L cos 24° 19' + L sin 23° 17' - L sin 1° 2' - 10
= 1-9878896 + 9-9596535 + 9-596903 - 8-2560943 - 10
= 3-2883518 = log 1942-4;
.-. ^£ = 1942 ft.
9. Draw a figure similar to figure on p. 79 E. T., and let POM= 51° 25' ;
the diameter of the circle described by St Paul's
= 2NP = 20P cos POM= 7914 cos 51° 25' ;
the circumference of this circle is 3-1416 x 7914 cos 51° 25'. Let x be the no.
of miles travelled by the cathedral in an hour in consequence of the revolu-
tion of the earth ; since the cathedral makes a complete revolution in a day
24a; -3-1416 x 7914 cos 51° 25';
138 ON LOGARITHMS. LXI.
/. Iogic=log3-I416 + Iog7914 + Lcos51°25'-Iog24-I0
= -4971509 + 3-8983960 + 9*7949425 - 1-3802112 - 10
= 2-7102782 = log 646-7 nearly;
.-. a; = 646*7 miles.
10. Let h = ihe height of the balloon, x the horizontal distance of the
first station from the vertical line through the balloon, and y the horizontal
distance of the second station from the vertical line through the balloon.
Then - = tan 47° 18' 30", 1/^ = 0:24- (671-38)2, ^ = tan 41° 14',
X y
tan 47° 18 = 1-0836896 ; tan 47° 19'= 1-084323 ; difference for 60" is -0006327 ;
.-. 60" : 30" :: -0006327 : -0003164;
.*. tan 47° 18' 30"= 1-0836896 + *0003164 = 1*084, tan41° 14' = *876462.
Now ^ = 1-084; .-. a: = ,-^ ; - = -876462, - ^
x~ ' " 1-084' y '^•^ " ' •• ^ -876420'
Since the second station is due west of the first,
.-. a:2 + (671-38)2 = 2/2.
•••^Mc87W^-(r^
.-. h^{ (1-084)2 - (-87642)2 } = (671-38)2 x (-87642)2 x (1-084)2 .
.-. 7i2 I (1-084 + -87642) (1*084 - *87642) } = (671*38)2 x (-87642)2 x (1-084)2;
.*. 7i2 X 1-96042 X -20758 = (671-38)2 x (-87642)3 x (1-084)2;
2 log A = 2 (log 671-38 + log -87642 + log 1*084) - log 1-96042 - log -20758
= 2 (2-8269684 +1*9427123 + -0350293) - -2923447 - 1-3171855 = 6 ;
.-. log /i = 3 = log 1000; .-. ;i = 1000ft.
EXAMPLES. LXI. a. Page 190 (vi).
1. log 4/451 = J log 451 = i (2-65418) = -88472
= log 7-669 [from the Tables] ;
.-.4/451 = 7*669.
2. log 4/8O2 = i log 802 = 4 X (2 -90417) = -58083
= log 3-809 [from the Table].
3. The log of the expression = | x log 273 + J log 234
= |x2*43616 + Jx 2-36922
= 1-08274 + -59230
= 1-67504 = log 47*32.
4. The log of the expression = | x log 451 + 1 log 231
= 1x2-65418 + 4x2-36361
= 1-59250 + 3-15148
= 4-74390 = log 55460.
ON LOGARITHMS. LXla. 139
5. The log of the expression = 3 (log 192*5 - log 84)
= 3x2-28443- 3x1-92428
= 6-85329-5-77284
= 1-08945 = log 12-03 [by the Table].
6. The log of the expression = f x log 34-79 - f x log 41-25
=f X (1-54145) - f X (1-61542)
= 1-02763 - 2-42313 = 2*6045
= log -04023. ;
7. The log of the expression = f x log 24-76 - | x log -0045
= 1 X (1-39375) - 1 X (3*653213)
= •39821-4*47981=3-91840
= log 8287.
8. The log of the expression = log 7*89 - log -0345 + f log 89130
= 1 -89707 - 2*53781 + f x 4-95002
= 3-0664 = log 1165.
9. The log of the expression
= log3-log2 + ilog5-2-log5-ilogll-31-Jlog3 + ilog7
= i (-47712) - -30103 + J x -71600 - -6989/- 4 (1-05346) + J (-84509)
= -23856 - -30103 + -35800 - -69897 - -52673 + -42254
= 1*49237 = log -3107.
10. The log of the expression = 4 {log 2 + i log 34 - log 3 - i log 791 }
i X { -30103 + i (1-53147) - '47712 - 4 (2 -89817) }
= J X { -30103 + -76573 - -47712 - 1-44908}
= 1*8281= log -6731.
11. The log of the expression = J log 3 - ^ log 3
= h (log 3) = •03976=log 1-096 nearly.
12. The log of the expression = 4 X {3 log 21 + 5 log 45 - 7 log 2 -9 log 3}
= 4 X {3 X 1-32221 + 5 X 1-65321 - 7 x -30103 - 9 x -47712}
= 4x {3-96663 + 8-26605-2-10721-4-29408}
= 2-91569 = log 823-6.
13. 10^ = 421, a; = log 421 = 2-624.
14. (H)" = 3.
X (log 21 - log 20) = log 3, XX -02118 = -47712 ;
-47712
a;=_-!_L!r = 22-52.
-02118
140 ON LOGARITHMS. LXI a.
15. (IKP-2;
/. 1x (log 203 - log 200) = log 2 ,
2a; X -00646 = -30103, x^ ~^_^ = 23-29.
16. (Iff = 3,
X (log 26 -log 25) = log 3, xy. -01703 = -47712;
-01703
17. log 37^+3 ^3 -412,
(a: + 3) log 37 = 3-412, (a: + 3) x 1-56820=3-412,
o 3-412
^ + ^=r5682'
a: = 2-1757-3= --8243.
18. 0^ = 10^3^2,
log 10 ;^3r2 = log 10 + J log 31-2
= 1+ J X 1-49415
= 1 + -49805 = 1-49805
= log 31-48.
EXAMPLES. LXI. b. Page 190 (vii).
1. The amount for 10 years = (i^^)io x 100 = (f|)io x 100,
log (If )i^ X 100= 10 (log 26 - log 25) -f log 100
= 10 X (1-41497 - 1-39794) + log 100
= 10 X -01703 + 2 = 2-1703 = log 148;
.-. amount for 10 years = £148,
or compound Interest = £48.
2. The amount = {\\f x £1,
log \\\f X 1 -^ 8 (log 21 - log 20) = -16852 = log 1-474 ;
.-. compound Interest = £1-474 - £1 = £-474 = 9s. hid,
3. X — the no. of years ;
\\%%Y Is *1^® amount at end of x years, (T^f)* = 2;
.-. X (log 103 - log 100) = log 2, a; X (2-01283 - 2) = -30103,
— =23-4 years.
•01283
4. Let a? = the no. of years, {\%%Y, or (|f)* = 2,
X X (log 26 - log 25) = log 2, a; X (1-41497 - 1-39794) = -30103,
-30103 ,^^
^ = ^01703 = ^^-^-
ON LOGARITHMS. LXI b. 141
5. The present value = (H|)8 x £100,
log (iVi)^ X 100 = 8 (log 25 - log 26) + log 100
= 8 X (1-39794 -1-41497) + 2
= 8 (1-98297) + 2 = 1-86376 + 2
= 1-86376 = log 73-07;
.-. the present value = (IH)^ x £100 = £73-07.
6. Let a; = the no. of years,
(|oo6)a;^ population at end of x years, (1^W)* = 2,
X (log 1005 - log 1000) = log 2, a; X -00216 = -30103 ;
•30103 ^,. .
•'• ^ = •00216'^ ^®^^^"-^*
7. To find the amount for 1 half year we multiply £1000 by
lOli 203
100 ^^200'
the amount for 1 year =(%Uf x £1000,
log (U^^y^ X 1000 = 42 log (203 - log 200) + log 1000
= 42 X (2-30749 - 2-30103) + 3
= 42 X -00646 + 3 = 3-27132 = log 1868;
.-. the amount required = £1868.
8. Let X = the no. of years, (fW)^* = 3,
2x X (-00646) = log 3 = -47712,
•^7712 __
"^.0l292 = '^*^-
9. Interest at Id. for Is. per month is 8J per cent.
To find the amount for the first month we multiply Is. by - y^w or ^^ ;
the amount for 2 months = (ff|^)2s.,
144 months = (fff)i%.;
log (mV''= 144 X (log 325 - log 300)
= 144 X (2-51188 - 2-47712) = 5-00544
= log 101300 nearly;
.-. the amount = 101300s. = £5065 nearly.
10. The man puts by 2^d. at the end of the 2"** week,
2-d 4*^
2='^; 6'^
.- 226rf 52"d
log 2=^6 = 26 X log 2 = 26 X -30103 = 7-82678rf.
= log 6711000 nearly;
.-. 226d. = 6711000rf.= £27962.
142 ON LOGARITHMS. LXIb.
11. The velocity at the end of 1«* sec. = -001 ft. per sec.
2"d sec. = -001 X I ft
3'-*isec.= -001x(|)2 ft
25*hsec.= -001x(|P ft
log -001 X (1)24 := log -001 + 24 (log 4 - log 3)
= 3 + 24 X (-60206 --47712)
rr 3 + 24 X - 12494 = 1-99856 = log -9967 ;
•001 X (1)2* ft. per sec. = -9967 ft. per sec. = -679 miles per hr.
12. Let 2a; = the diameter of sphere,
l^r, a;3 = l c. yd., a;^ = 1 x | x ^j!^,
3 log a; = log 3 - log 4 + log 7 - log 22
= -47712 - -60206 + -84509 - 1-34242 =1-37773,
log a; = 1-79257 = log -622,
a;='622yds.; /. diameter = 2a; = 1-24 yds.
13. The required present value is
£125{r + r2+ r^^ where r=U|
= £125 i^^- £125 = £125 {1 - {mV^} x 51 - £125,
log (tUV^ = 13 (log 100 - log 102) = 13 X (2 - 2-00860)
= 1-99140 X 13 = 1-88820 = log -773;
.-. (K|)i3^-773.
Hence the required present value = £125 {1 - -773} x 51 - £250
= £125 {-227 X 51 - 1} = £125 {10-577} = £1322.
EXAMPLES. LXII. Pages 193, 194
1. cos 60° = 4; .-. ^ = 60°.
2. cos 120°= -J; .-. ^ = 120°.
3. sin30° = sin(180°-30°) = i; /. ^ = 30°, or 150^
4. tan 135°= -1; .-..4 = 135°.
5. sin 45° = sin (180° - 45°) = ] ; /. A = 45°, or 135°.
6. tan 120°= - ;^3 ; .. A= 120°.
7. sin (A+B + C) = sin 180° = 0.
8. cos(^ + 5 + O) = cosl80°= -1.
9. sinJ(4 + i^ + C')=si_n90°=l.
J. -f jB + (7 = 180°. LXn. 143
10. cosi(^ + l? + C) = cos90°=0.
11. tan(^+^) = tan(^ + JB + C-(7) = tan(180°-C)= -tanO.
12. coti(B + C) = cot{i(^+B + C)-i^} = cot(90°-4^) = tani^.
13. cos (A+B) = cos (180° -C)= ~ cos G,
14. cos (^ + 5 - C) = cos (^ + ^ + C - 2C) = cos (180° -2C)= - cos 2C.
sin A sin B - cos ^ cos B
15. tan^-cotJ5 =
cos A sin 5
cos (^ + J5) _ cos C
cos ^ sin B cos ^ sin JB '
sin J[ - sin JB _ 2 cos J (^ + jB) sin J (yl - B)
^^' Sn^ + sinl? ~ 2 sin i (^ + R) cos'i (^ - ^^ '
and cos J (^ + ^) = cos (90° - J C) = sin 4 C,
sinJ(^ + 5) = sin(90°-JC) = cosJC;
sin ^ - sin J5 _ sin ^ C . sin J (^ - B)
17.
sin A + sin B cos J O . cos J (^ — J5) *
sin SB - sin 3C _ 2 cos f (jB + C) . sin f (^ - C) _ cosf (I? + C)
cos 3C - cos '6B 2 sinf (B + C) . sinf (B - C) sin f (jB + C) '
.-. i(B + C) = 3-{180°-^}=270°-f^;
/. cot f ( JB + (7) = cot (180° + 90° - f ^) = tan f ^.
18. sin.4 + sinB = 2sini(^+^)cosi(^-J5),
but sin J (^ + jB) = sin (90° -^0) = cos J^ C,
and sinC~2 sin JCcos JO=2cosi(^ + i^) cosJC;
.-. sin ^ + sin £ - sin C = 2 cos J C cos J (^ - JB) - 2 cos J (^ +i?) . cos J C
= 2 cos i C {cos i (^ - i3) - cos i{A+B)}
= 2cos JC 2sin^^ . sin^L\
19. /. (sin A - sin i?) + sin Cr= 2 sin i C {sin J (^ - 7:^) + sin 4 ( J + ^)}
= 2 sin J (7 . 2 sin 4^ . cosJjB.
20. 2sin4^ . cos4^ + 2sin4i^ . cos .Ji? + 2sin4Ccos JC
= sin ^ + sin B + sin (7 = 4 cos ^A . cos ^B . cos J (7. [p. 192, Example 3.]
21. .*. (cos A + cos i?) + (cos C - 1) = 2 sin 4 C {cos J (^ - 1?) - sin J C}
= 2 sinj C {cosi (A-B)- cos J(^ + JB)}
= 2 sin J C . 2 sin J^ . sin J B.
144 ^+5 + (7= 180°. LXII.
22. cos2J^=J(l + cos^);
.'. cos^^A+coa^^B -Gos^iC = i{l-rco8A + co8B -cos G),
Now cos ^ + COS B =2 cos i {A+B) cos J (^ - 5) = 2 sin J C cos ^(A-B)
cosC = l-2sin2jC7;
.-. 1 + cos J^ + cos^-cosC = 2 sin^C {sin JC + cos J (^ --B)}
= 2 sinj C {cos J (^ +5) + cos J (^ - B)}
= 2 sin ^ O . 2 cos J ^ cos J5.
23. sin2j^=J(l-cos^);
.-. sin^J^ - sin2iJ5 + sin2i<7 = i {1 - (cos ^ -cos5 + cos (7)}
= i [1 - 2 cos iC {sin J (£ - ^j + cos JC} + 1]
= i [2 - 2 cos J C { sin i (B - ^ ) + sin J (^ + ^) } ]
= J { 2 - 2 cos I C . 2 sin J jB cos J^ }.
24. The expression = sin (90° - ^) + sin (90° -B) + sin (90° - C) - 1
= cos ^ + cos B + cos (7 - 1
= 4 sin 4 ^ sin i 5 sin J C. [See 21.]
25. (sin 2A + sin 25) + sin 20 = 2 sin C (cos {A-B)- cos (^ + 5)}
= 2 sin (7 . 2 sin ^ . sin jB.
26. (sin 2A - sin 2i?) + sin 2C = cos (7 {sin (^ + 5) - sin {A - B)\
= 2 cos A sin B cos C.
27. The expression = sin (180° - 2A) - sin (180° - 2B) + sin (180° - 20)
= sin 2A - sin 2jB + sin 2C
= 4 cos ^ sin B cos C. [Vid. 26.]
28. .*. cos2^ + cos2B + cos2C= - 2 cos (7 {cos (^ -5) - cos 0} - 1
= - 2 cos { cos {A-B) + cos (^ + jB) } - 1
= - 2 cos C . 2 cos A cos JB - 1.
29. sin2^ = J(l-cos2^), etc.
.-. sin2 A - sin2 B + sin^ c = J { 1 - (cos 2 A - cos 2JB + cos 2(7) }
= J [1 - 2 sin (7 {sin (B - A) - sin (7} - 1]
= - sin (7 {sin {B - A) - sin (B + ^)}
= - sin C . 2 cos B sin (- J^).
30. The expression
= cos (180° -2A) + cos (180° - 2B) - cos (180° - 2(7) + 1
= 1 - (cos 2A + cos 2J5 - cos 20)
= 1 + 2 cos cos (^ - 5) + 2 cos2 O - 1
= 2 cos O {cos (A-B)- cos (A-vB)]
= 2 cos . 2 sin A sin J5.
31. This is merely the expression of the fact that
sinj (^ +5 + 0) = sin 90°=1.
A+B + G= 180". LXII. 145
32. tan (A-\-B-\- C) = : — - — -7 = expanding numerator and denomi-
nator,
sin A cos B cos C + sin B cos G cos A + sin G cos A cos 5 - sin A sin J5 sin G
cos ^ cos B cos C - cos A sm i^ sin (7 - cos jB sin ^ sin C - cos G sin ^^in B *
dividing both numerator and denominator by cos A cos B cos (7, we have
, ,. „ ^. _ tan ^ + tan B + tan C - tan A tan J5 tan G
1 - tan £ tan C- tan C tan A - tan Z tan B
= because (^ + B + 0) = 180°.
33. Proceeding as in 32 we have
tunKA I B \ /^x ^^^^i^ + tan|Jg + tan^C-tan^^tan|J5tan^(; ^
^\ -^ -^ ) l-tan^^tanJ^-tanii^tanJC-taniCtanJ^*
.*. the denominator is zero.
EXAMPLES. LXIII. Page 195.
1. tan ^ = tan (90° - J5) = cot 5.
2. tanJB = tan(90°-^) = cot^; cos 0= cos 90°= 0.
3. sin 2A = 2 sin ^ cos ^ = sin (180° - 2B) = sin2B.
4. cos 2^ + cos2J5 = 2 cos {A+B) cos (A - J5) = 2 cos 90° cos (^ -^) = 0.
5. sm 2u4 = 2 sm ^ cos ^ = 2 .-.-=—;;- .
c c c^
6. cosec 2B = -.-- tt:^ = 7; . -r . = — ^— .
sm2B 2 a6 2a6
7. cos 2^ .= 1 - 2 sin2 ^ (See E. T. Art. 164.)
_ 2a2_62^a2-2a2
~ c2 ~ ^2 •
8. cos 2 J5 = cos2 j5 - sin2 B
, _ cos'-^ B - sin^ i^ _ sin'^ A - sin^ B
~ coB^B + sin2\B ~ sin2 A + sin'^ B '
9. sin2^5 = J{l-cos5}=i/'l--V
10. cos2 J ^ = 4 (1 + COS ^), as in 9.
11. (co8j^+sinJJ[)2 = l + sin^ = l + -.
c
L. T. K. 10
146 RIGHT-ANGLED TRIANGLES. LXIIL
12. Divide both numerator and denominator by c.
TVi a-b_/a b\ /a b\_sinA-smB
a-{h \c cj ' \c c) sinA + ^inB
_ 2 sin \{A-B) cos \(A-\-B) _ tan ^(A-B ) _ tan ^(A-B)
~2 sin^ (A+B) cos i{A-B)~ tanj (A + B) ~ tan 45° *
13. sin {A-B)-\- cos 2A = sin A cos B - cos A sin B + cos 2A
= sin A . sin A - cos ^ . cos A + cos^ ^ - sin^ A.
14. sin (A -- J5) + sin {2A -\-G) = sin (^ - J5) + cos 2A.
15. (sin^ -sin5)2 + (cos^ + cos5)2 = 2-2 sin ^ sin 5 + 2 cos ^ cos 5
= 2 - 2 sin ^ sin 5 + 2 sin A sin B,
_ 7sin ^ 4- sin J5 /sin ^ - sin B
~ V sin ^- sin £ "^ V sin^ + sinJ5 ^^®® ^^.J
_ (sin ^ + sin 5) 4- (sin ^ -sin 5) _ 2sin^
V'sin'^ A - sin"^ B J(co8^ B - sin^ B) '
EXAMPLES. LXIV. Page 203.
1, By iii. p. 203, sin^ = aA;, sini? = 6A;, sinC = cA:;
sin ^ + 2 sin jg afe + 2&^ sinC
a + 26 ~ a + 2f ~ ~ ~c~ '
sinM-msin^Jg _ k- (a^ - mb^ ) _ _ sin2 C
^' a^-m.b'^ ~ "a^-mb^ ~ ~ ~^^~ '
3. By iii. p. 203, a = d sin^, b = d sin 5, c = d sin C;
a cos A + b cos B -c cos C may be written
d sin ^ cos A-hd sin B . cos B-d sin (7 . cos C
= ^d (sin 2 J^ + sin 2J5 - sin 2C) = 2d sin (7 cos A cos 5 = 2c cos ^ . cos B.
See Examples LXII. 25.
4. (a + 5) sin J(7 = rf(sin v4+sin£) sin ^C
= 2c? sin i{A + B) cos J (^ -5) sin J C
= 26? cos J Ccos 4 (^ - B) sin J C = i sin C cos J (^ - 5).
5. (6-c) cos J^ = d (sin 5-sin (7) cos J^
= d sin ^ sin J (J5 - 0), as in 4.
- m, . sin A cos ^ + sin 5 cos B + sin (7 cos (7
6. The expression = ; -. — ^^ . ^
sin A sin B sin C
* _ I (sin 2 A + sin 25 + sin 2 0)
"" sin A sin 5 sin (7
2 sin ^ sin 5 sin (7 _ , ttttt^.-
= —■ — T—' — TT-- — 77 Examples LXII. 25.
sm A sm B sm C
TRIANGLES. LXIV. 147
7. a sin (B-C) + b sin (C-A) + c sin (A - B)
= d sin A sin {B - C) + d sin B sin(C -A) + d sin O sin (^1 - B)
= 0. [If sin (B - C) be written out in full.]
8.
a-b _d (sin A - sin JB) _ sin A - sin B
c ~ d sin C ~" sin (A + ^)
_ 2cos^ (^ + .B) sin^ (J^ -E) _ 2 sin ^ {A-B) sin ^ (^ +1?)
~2sini(^ + -B)cosJ(^+jK)~ 2sin2i(^ + i^)
_ cos B - cos A
""200^1(5 *
6 + c _ 2 sin ^ ( ^ + O) cos |(^ - 0) _ cos .6 + cos C .
^' a ~2sin4(J5+(7)cosJ(2? + C)~ T^BinH^ * ^ s in 8.J
10. \/(^^ sin B sin C) = (i sin B sin O
_ d^ sin ^ sin (7 (sin B + sin (7)
_ ___
_ Z>^ sin C + c^ sin ^
~~ 6 + c
11. Fromp. 237 E. T., a = ftcos(7 + ccos5,
6 = c cos ^ + a cos (7,
c = a cos B + b cos ^4 ;
.*. a + b + c=(b + c) cos A -i- (c + a) cos B + (a + b) coQ C.
12. As in 11, b + c-a={b-\-c) cos ^ - (c - a) cos 5 + (a - 5) cos (7.
a sin C sin A sin (7 sin ^ sin C
13,
6 ~ a cos C~ sinB- sin ^ cos (7 ~ sin {A + C) - sin A cos (7
__ sin A sin C
~ cos A sin C *
14. The expression = be (b cos C + c cos B) + ca (c cosA + a cos (7)
+ afc (a cos B + b cosA) = bca + cab + abc,
15. a cos (^ + J3 + C) - 6 cos (B'\-A)-c cos (^ + (7)
= -a + 6cosC + ccosJ5. [Art. 237.]
17.
18.
19.
2abc 2abc 2abc
tanJ5 sinJ5cosC bcosC a^ + b--c^ a^-h^ + c^
tan (7 sin (7 cos 1^ c cos B 2a ' 2a *
_«(«-c) «(s-6) _»(2«-6-c) _a«_
a a a ~ a~ '
^ /S {>-c){s-a) ) / Us-a){,-b) ] _s-a
10—2
148 TRIANGLES. LXIV.
«^ X 1.x IT. /f(«-^)(s-c s{s-h) 8-b
^ ^ V ( s(s-a) (s-c)(s-a)J s-a
21. c2 = a2 + 62_2a6cosC
= a2 (sin2 i C -t- cos^ J C) + 6^ (sin"-^ J (7 + cos^ J C)
-2a6(cos2iC-sin2JC)
= sin2JO(a2 + 2a64-62) + cos2jC(a2-2a& + fe2).
MISCELLANEOUS EXAMPLES. LXV. Pages 204, 205.
1, Let jPi be the length of the perpendicular from B on AG,
Then ap = hp-,^ by areas.
• A Pi ^Pi ^P
c OC DC
2. If 2 cos ^ sin = sin A, .. 2 sin ^ cos J5 sin C = sin2^ ;
/. sin2 A - sin2 B + sin2 C = sin2 ^ ; [Examples LXII. 29.]
.-. sin2 B = sin2 C; /. sin ^ = sin O ; .'. B = C.
sin 5 _ 6 _ sin 5 _ sin 5 _ 1
^* sin^ ~ a ~ sin SB ~ 3 sin JB - 4 sin^^ ~ 3 - 4 sin2^ '
- b 1
"a 3-4 sin2i? '
. sin.B sinC ,
4. — ?— = = «, suppose ;
c
.'. sinB
"W-
/ ( 7 • T» • ^1 fo2 sm J5 + c2 sm (7 , . , ^
.*. v/ 1 ?>c sm ^ sm (7} = may be written
^/ o-{-c
]cbc=—^^ '- , I.e. bc = b^ + c^-bc,
/. (6-c)2 = 0, i.e. b = c.
c IT. 1^ 14 / f^(^-M s(s-c) Z>c )
5. a cos i J5 . cos i (7 cosec hA=a. / -{ ^ . --— r- . ^ tt-^ :}- = s,
^ \ { ac ab (s-b)(s-c)]
similarly for the other expressions.
6. sin J^=|,Bin5 = ^
- cos C=cos(A'\'B)~ooqA cos 5 - sin J sin B =f . |f-f . ^\=if .
7. /. fc2 + c2 = a2; .-. ^=90^
TRIANGLES. LXV. 149
8. sin 2B - sin 2A + sin 2(7 = 4 sin A cos jB cos C ; [Examples LXII. 27.1
.-. 4 sin ^ cos jB cos (7=0,
i.e. sin^ = 0, orcosB = 0, orcosC = 0.
9. A = iof 180°, jB = I of 180°, a = f of 180°,
1 + 4 cos A cos B cos (7 = - cos 2 A - cos 2B - cos 2(7
= - cos 45° - cos 90° - cos 225° = ;
2 sin2^ + 2 sin2 (7-4 sin^B =: cos 2i + cos 2(7 - 2 cos 2B
= cos 45° + cos 225° =0.
10. The exp. = a cos J (B + C) sin ^ (5 - (7) + etc.
= ia (sin 5 - sin C) + 4 fc (sin (7 - sin ^) + Jc (sin ^ - sin B)
= 0; for asinJB = &sin ^.
11. In fig. 1 of p. 196 let D be the middle point of 5(7,
then AC^ = AD^ + CD^ - 2 AD . (7i) cos ADC,
also ^^2 = ^2)2 + B2)2 -2AD.BD cos ^DJ5.
Now cos^D(7 = cos(180°-^Z)5)= -cos^jDjB,
also GD = BD;
.-. by addition ^ C2 + ^ J52 =2AD^-^ CD^ + 51)2 .
/. b^ + c^ = 2AD^ + ia'' + ia^
and the result follows.
12. 6sin^ = asin J5;
.-. sin 35 = 2 sin5;
.-. 3sin5-4sin35 = 2sin5;
.-. 3-4sin25 = 2; /. sin 5=^.
Hence B = 30°, ^ = 35 = 90° and C = 180° -A-B^ 60°.
13. a&c (a cos^ +6cos5 + c cos C)
= i tt&cd (sin 2^ + sin 25 + sin 2C)
= 2ahcd sin A sin 5 sin C [Ex. LXII. 25. ]
= 2a26c sin 5 sin C = %S'K [iii. p. 203. ]
14. 26cos24C + 2ccos2J5 = 6 + 6cosC + c + ccos5
a2 + 62_c2 c2 + a2_52 1
= h + c-\ ^ + ^ — [ao + ac + a^) = a + h-hc.
2a 2a a. '
If then this =3a we have & + c = 2a. q.e.d.
15. By 11 above we have
4^D-^ = (2i>2 + 2c2-a2^,
45i<:-^=(2c*-» + 2a--fe--^),
4CF2^^2a2+-262-c2);
,*. by addition the result follows.
150 TRIANGLES. LXV.
16. In the figure of p. 230 let AD be the perpendicular from A on CB;
draw ^D'to bisect CB in D'; then
DD' GD-BD bcosC-ccosB
cot ADB =
AD 2AD 2AD
a^ + b^-c^-c^-a^ + b^ b^-c^ b^-c^ b^-
Aa.AD 2aAD 2acsmB 4:8
17. We have (i = c sin ^ = 6 sin (7, e=c sin^, /=a sin JB.
Also a = A;sin^, 6=/c sin5, c = /i;sin C;
.be ca . ah
/. 2d cos A = 2d -
2bc h
:. 2(cicosJ[ + «cosB + /cosC) = |{b^+c2-a2 + c2 + a2_52 + a2_|.52_c2|
1 ^ o lo o^ a . b c
= _{a2 + 62 + c^} = a.- + 6.- + c.j^
= a sin ^ + 6 sin 5 4- c sin G.
EXAMPLES. LXVI. Pages 208, 209.
1. 5 = 674-10, s-a = 321-85, s- 6 = 160-83, s--c = 191-42,
L tan J^ - 10 = i {log (s - 6) + log (s-c)- log s - log (s - a)} ;
/. L tan 1 .1 = 10 + 4 {log 160-83 + log 191-42 - log 674-10 - log 321-85}
= 10 + i {2-2063401 + 2-2819873 - 2-8287248 - 2-5076535}
= 9-5759748
9-5761934 9-5759748 /. D = i||^ of 60"
9-5758104 9-575810 4 =25-75'';
•0003830 -0001644
.-. 4^ = 20° 38' 25-75", ^=41° 16' 51-5".
2. a=484, 6 = 376, c = 522, s = 691, (s-a)=207, s- Z> = 315, «-c = 169.
The largest angles are opposite to the greatest sides and are therefore
A and C.
L tan 4 C - 10 = 4 {log («-«) + log (s-b)- log s - log (s - c)},
L tan 4 (7 = 10 + 4 {log 207 + log 315 -log 691 -log 169}
= 9 -8734581 = L tan 36° 46' 6" ;
.-. 4C = 36°46'6"; /. C = 73°32'12".
I, tan 4^ - 10 = 4 {log (s-c) + log {s-b)- log s - log {s - a)},
i tan 4 ^ = 10 + 4 {2-2278867 + 2-4983106 - 2-8394780 - 2-3159703}
= 9-7853745 = L tan 31° 23' 9" ;
.-. 4 /I = 31° 23' 9"; .-. ^ = 62° 46' 18".
SOLUTION OF TRIANGLES. LXVI. 151
3, s = 10142, s-a = 4904, s-h = 4460, s-c = 758,
Lt&n^A = 10 + i {log 4480 + log 758 - log 10142 - log 4904}
^ 10 + i {3-651278 + 2-8796692 - 4-0061236 - 3*6905505 }
= 9-4171366
9-4173265 9-4171366 /. D = ff-|f of 60"
9-4168099 9-4168 099 =38'';
•0005166 "^0003267
.-. i.4 = 14°38'38"; /. .4 = 29° 17' 16".
Lt&n ^B-10 = i {log(«--c) + log(s a) -logs -log (s-6)};
/. L tan J^ = 10 + i {log 758 + log 4904 -log 10142 -log 4480}
= 10 + J { 2-8796692 + 3-0905505 - 4-0061236 - 3-6d12780}
= 9-4564091
9-4565420 9*4564091
9-4560641 9-4560641 ^43/V.
•0004779 -0003450
.-. 9-4564091 = L tan 15" 57' 43" = L tan J B ;
.-. ijB = 15°57'43"; .-. 5 = 31° 55' 26".
4, s = 5875-5, 5-a = 1785-5, 6 = 3850, c = 3811,
L cos J J. = 10 + i {log s + log (s - a) - log b - log c}
= 10 + i {log 5875-5 + log 1785-5 - log 3850 - log 3811}
= 10 + i {3-7690448 + 3-2517599 - 3-5854607 - 3-5810389 }
= 9-9271526
9-9272306 9-9272306 D = Ht o^ 60"
9-9271509 9-9271526 =59";
•0000797 -0000780
.-. 4^ = 32° 15' 59"; .'. ^ = 64° 31' 58".
5, a = 7,6 = 8,c = 9, the greatest angle is opposite to the greatest side and
is (7, s = 12, s - c = 3,
LcosiC~10 = i(log9-logl4) = i(21og3-logl4),
L cos i C = 10 + J (-9542426 - 1-146128)
= 9-9040573
9-9040573
9;^040529 .-. D = g^-^^ of 60" = 2-8":
-0000044
.-. iC = 36°42'-2-8" = 36°41'57-2'';
.-. 0=73° 23' 54-4".
152 SOLUTION OF TRIANGLES. LXVI.
6, The smallest angle is opposite to the least side and is A^
L sin iA-10 = i {log 1 - log 8} = - ^ log 2,
LsiniA = 10 - -451545 - 9-548455.
Here D = ^%\% of 60" = 11'S'';
i^=20°42'17-3"; .-. .4 = 41° 24' 34-6".
7, Let a = 4, &=5, c = 6, s = 7-5, (s-c) = l-5,
in /\^(^-c)l /J5x3x3( /9 3
1/ cos J C - 10 = log 3 - log 4 = log 3 - 2 log 2,
jL cos J (7= 10 + -4771213 - -6020600 = 9-8750613 ;
9-8750613
9-8750142 .-. D = y^V of 60" = 25-35'' ;
-0000471
.-. ^0 = 41° 25' - 25-35" = 41° 24' 34-65" ;
.-. C= 82° 49' 9-3".
8. a = 2, &=;^6, c = l-fV3,
^^^ 2bc 2(1 + ^3)^6 ~2(1+V3)V6
_ V3(1 + V3 )_ 1 , .
(l + ^/3)V6~^/2' *• '
eo,^- (l + x/3)^ + 4-6 _ 2 + 2^/3 _1,
(7 =180° -60° -45° = 75°.
9. a = 2, 6 = ^2, c = V3-l,
co.J- ^ + (^^-^)'-^ - ^-^^^^ - ^.-^-135°
^^'^- 2^2(^3-1) -2V2(^"3^T)--;/2' "^-l^^'
^^'^- 4(V3-1) -4(V3-1)- 2 ' •• ^-^^ '
C = 180° - (^ + -B) = 180° - (135° + 30°) = 15°.
EXAMPLES. LXVII. Page 211.
1. (7= 180° -^-5 = 180° -53° 24' -66° 27' = 60° 9',
- ^^^^ - 338-65 X si n 53° 24'
^ ~ sinTo" ~ sin 60°!)' '
log a = log 338-65 + L sin 53° 24' - L sin 60° 9'
= 2-5297511 + 9-9046168 - 9-9381851
= 2-4961828 = log 313-46 ;
.-. a= 313-46 yds.
SOLUTION OF TRIANGLES. LXVIL 153
C = 180°-^-iB = 180°-48°-54° = 78°,
csin^ 38 X sin 48°
sine ~ sin 78° '
log a = log 38° + L sin 48° - L sin 78°
= 1-5797836 + 9*8710735 - 9*9904044
= 1*4604527 = log 28*8704 ;
.-. a = 28*8704,
csin^ _ 38 X sin 54°
~ sinO ~ ""sinr78° '
log & = log 38 + L sin 54° - L sin 78°
= 1*5797836 + 9*9079576 - 9*9904014
= 1*4973368 = log 31*4295;
.-. 6 = 31*43.
_ g sin (7 _ 1000 X sin 66°
^* ^~ sin .4 ~ sin 50° '
/. log c = log 1000 + L sin 66° - L sin 50^
= 3 + 9*9607302 - 9*8842540
= 3 -0764762 = log 1192*55;
.-. c = 1192-55.
4. ^ = 180°-5-C
= 180° - (32° 15' + 21° 47' 20") = 180° - (54° 2' 20") ;
.-. sin A = sin (180° - 54° 2' 20") = sin 54° 2' 20" ;
.-. d=^, of -000092 = -000031;
o\)
.-. L sin ^ =JL sin 54° 2' 20" = 9*908141 + -000031 = 9-908172.
, asini? 34° X sin 32° 15'
sin^ "sin (180° -54° 2' 20")'
.-. log 6 = log 34 + 1. sin 32° 15' - L sin (180° - 54° 2' 20")
= 1 -531479 + 9*9727228 - 9-908172 = 1-350535.
Here d = ^%\ of -001 = -0005;
.-. 1-350535 = log (22-41 + -0005) = log 22-415;
.-. 6 = 22-415.
C=180°-A-B
= 180° - 114° 18" = 65° 59' 42".
42"
Here d=-^,oi -0000563= -0000394
.-. L sin 65° 59' 42" = 9*9606739 + -0000394 = 9-9607133.
c sin A
a= . ,, ,
sm C
154 SOLUTION OF TRIANGLES. LXVIL
log a =:log c-\-L sin A~ L sin G
= log 24 + L sin 72° i' - L sin 65° 59' 42"
= 1-3802112 + 9-9783702 - 9*9607133 = 1-3978681.
•3978705 -3978681 .-. d = iff of -0001 = -0000862 ;
•3978531 -3978531
•0000174 -0000150
.-. -3978681 = log (2-4995 + -0000862) = log 2 -499586 ;
.-. 1-3978681 = log 24-996 ; :. a = 25 feet nearly.
9 -8249959 /. d = j\ of -0000234 = ^00001872 ;
9-8 249725
•0000234
/. L sin 41° 56' 18" = 9 -8249725 + -00001872 = 9 -82499122.
c sin B
.-. log h = log c + L sin B -LsinC
= log 24 + L sin 41° 56' IS" - L sin 65° 59' 42"
= 1^3802112 + 9-8249912 - 9^96077133 = 1^2444891.
Here d = |i|| of •OOl = ^00085 ;
/. 1^2444891 = log (17-55 + •0085) = log 17*5585 ;
.-. 6 = 17-559 feet.
EXAMPLES. LXVIII. Pages 213, 214.
1. 6 = 131, c = 72, 6-c = 59, 6 + c = 203;
^ + (7 = 180° -^ = 180°- 40° = 140°; i^ = 20°;
I,tan^(B-C) = log(6-c)-log(6 + c)+Lcot4^
= log 59 - log 203 + L cot 20°
= 1 -7708520 - 2-3074960 + 10-4389341
= 9-9022901.
9-9024195 9-9022901 i^or.frn- qa-
9-902160 4 9-902160 4 • - ^ = HI J of 60 = 30 ' ;
-0002591 -0001297
.-. J (jB - C) = 38° 36' 30". :. B-C = 1T 13'. Also jB + C = 140° ;
.-. 2B = 217° 13' ; ,\ B = 108° 36' 30" ;
2C= 62° 47'; .'. C= 31° 23' 30".
2. a-b = U, a + 6 = 56; ^ +^ = 180°- 50°= 130°;
tan i{A-B) = i^ cot J (7=i cot^ C ;
.-. L tan i{A- B) = log 1 - log 4 + L cot JC.
Butcoti(7 = tan4(^+5) = tan65°;
.-. L tan J(^ - jB) = I. tan 65° - 2 log 2
= 10-331327 - -602060 = 9-729267.
SOLUTION OF TRIANGLES. LXVIII. 155
Here D=|H of 60" = 49";
.-. iU-B) = 28°ir49";
:. A-B = 56° 23' 38", also A + 3 = 130^ ;
/. ^ = 93°ir49" and ^ = 36° 48' 11".
3. c-6 = l, c + 6 = 39, i^ = 4(60°) = 30°; + 5 = 180° -60° = 120°.
LtaiiJ((7-5) = log(c-6)-log(c + 6)+LcotJ^
= log 1 - log 39 + L cot 30°
= - 1-591065 + 10-238561 = 8-647496.
Here D = ifH of 60" = 34-6" ;
.-. J(C-jB) = 2°32'34-6";
/. 0-5 = 5° 5' 9-2", also O + J5 = 120°.
/. = 62° 22' 34-6" and J5 = 57° 27' 25-4".
4. a-6 = 124-610, a+^>=628-140, iO=39°13'; ^ + B = 101°34'.
L tan i (^ - 5) = log 124-610 - log 628-140 + L cot 39° 13'
= 2-0955529 - 2-7980565 + 10-0882755
= 9-3857719.
Here D = 4|*| of 60" = 47";
.-. i(^-5) = 13°39'47";
.-. ^-5 = 27° 19' 34" and ^+5+ 101° 34';
.-. ^ = 64° 26' 47" and 5 = 37° 7' 13".
5. a-6 = 30, a + 6 = 240, ^0 = 30°, ^ +5 = 180°- 60° = 120°,
tan i (^ - 5) = 1^ cot 30° = i ^/3,
LtanJ(^-5)-10=log^3-log8 = ilog3-31og2;
.-. Ltani(^-5) = 10 + ilog3-3log2
= 10 + -23856065 - -9030900 = 9 -3354707.
Here D = f ff| of 60" = 59";
.-. i(^- 5) = 12° 12' 59";
.'. A-B = 24° 25' 58", also A + B = 120° ;
.-. ^ = 72° 12' 59".
6. c2=a2 + 62 _2a& cos O
= 212+ 202 - 2 X 21 X 20 X cos 60°
= 441 + 400 - 2 X 21 X 20 X J = 421 ;
/. c = ^421 = 20-5.
7. c2 = (135)2 + (105)2 - 2 X 135 X 105 x cos 60°
= 18225 + 11025 - 2 X 135 X 105 X J = 16075 ;
.-. c = V15075 = 122-7.
156 SOLUTION OF TRIANGLES. LXVIII.
8. Leta = 5, & = 3, C=70°aO', ^ + 5 = 180° - 70° 30' = 109° 30',
tani(^-5) = |^cotiC = JcotJC;
.-. Ltani(^ -5) = logl-log4 + Lcot35°15'
= -21og2 + Lcot35°15'
= -•6020600 + 10-1507464
= 9-5486864 = L tan 19° 28' 50" ;
.-. J(^-jB) = 19°28'50";
^-5 = 38° 57' 40", also ^ + ^=109° 30';
.-.^ = 74° 13' 50" and B = 35° 16' 10".
EXAMPLES. LXIX. Pages 218, 219.
1. LsmA = ]oga-hLsmB-logb
= log 170-6 + L sin 40° - log 140-5
= 2-2319790 + 9-8080675 -2-1476763
= 9-8923702.
9-8924354 9-8923702
9-8923342 9^8923342 .-. D = ^^-^% of 60" = 21" ;
-0001012 -0000360
.-. 9-8923702 = 1. sin 51° 18' 21";
.-. A = 51° 18' 21" or (180° - 51° 18' 21"), i. e. or 128° 41' 39".
Since b is less than a each of these values is admissible.
When ^ = 51°18"21", (7=180°-^-^;
.-. 0= 180°- 91° 18' 21" = 88° 41' 39".
When A = 128° 41' 39", C= 180° - 168° 41' 39" = 11° 18' 21".
2. L sin ^ = log 6 + L sin ^ - log a
= log 119 + L sin 50° - log 97
= 2-075547 + 9-884254 - 1-986772 = 9-973029.
Here, D = |f of 60" = 56" ;
.-. 9-973029 = L sin 70° 0' 56'' ;
.-. B = 70° 0' 56" or 180° - 70° 0' 56" or 109° 54' 4".
Since a is less than b each of these values is admissible.
When 5 = 70°0'56", C=180°-^ -^ = 180°- 120°0' 56" = 59°59'4".
When B = 109° 54' 4", C = 180° - 159° 59' 4" = 20° 0' 56".
3. L sin 5 = log 97 + L sin 50° - log 119
= 1-986772 + 9-884254-2-076547
= 9-795479 = L sin 38° 38' 24" ;
.-. B = 38° 38' 24" or 180° - 38° 38' 24" = 141° 21' 36".
SOLUTION OF TRIANGLES. LXIX. 157
Since a is greater than fe, angle A is greater than angle B^ and only the
less value of B is admissible ;
/. J5=:38°38'24^
0= 180° -A- 1>^ = 180° - 88° 38' 24" = 91° 21' 36",
log c= log h + L sin G -L sin B
= log 97 + L sin 91° 21' 36" - L sin 38° 38' 24".
Now Lsin91°2r36"=Lsin(180°-91°2r36")=Lsin88°38'24";
.-. log c = log 97 + L sin 88° 38' 24" - L sin 38° 38' 24"
= 1-9^6772 + 9-999876 - 9-795479
= 2-191169 =log 155-3.
4. Lsin^=loga + Lsin C-logc
= log 24 + L sin 65° 59' - log 25
= 1-3802112 + 9-9606739 - 1-3979400=9-9429451.
Here D = ii| of 60" = 10";
/. 9-9429451 =L sin 61° 16' 10" or L sin (180° - 61° 16' 10")
or L sin 118° 43' 50";
.-. A = 61° 16 ' 10" or 118° 43' 50".
Since oa, :. C>A, and only the less value of A is admissible,
.-. ^ = 61° 16' 60".
5. Lsin-4=loga + Lsin C-logc
= log 25 + X sin 65° 59' - log 24
= 1-3979400 + 9-9606739 - 1-3802112 = 9-9784027.
Here i) = i*l of 60" = 48";
.-. 9-9784027 = L sin 72° 4' 48" or L sin (180° - 72° 4' 48") = L sin 107° 55' 1 2" ;
. ^ ^ 720 4' 48^' or 107° 55' 12".
Since c is less than a, both values of A are admissible.
When A = 72° 4' 48", 5 = 180° - J^ - C = 180° - 138° 3' 48" = 41° 56' 12".
When A = 107° 55' 12", B = 180° - 173° 54' 12" = 6° 5' 48".
We shall have the greater value of b when we take the greater value of J5,
i. e. when we take the less value of A. When A = 72° 4' 48", £ = 41° 56' 12",
a sin B
~ sin A '
.'. log 6 = log a + L sin ^ - iL sin ^
= log 25 + L sin 41° 56' 12" - L sin 72° 4' 48".
Here d = ^,of -0000234 = -0000030.
Id
.-. L sin 41° 56' 12" = 9-8249725 + -0000030 = 9-8249755 ;
.-. log h = 1-3979400 + 9-8249755 - 9-9784027 = 1-2445123.
Here d = |f f | of -001 = -00095 ;
.-. 2445123 = log (1-755 + -00095) = log 1-756;
.-. l-2445123 = logl7-56=log6, /. 6 = 17-56.
158 SOLUTION OF TRIANGLES. LXIX.
c mnA
6. sinC =
(a) sin C = f If X sin 30° = i Jf = 1 ;
.•. G = 90° and the triangle is not ambiguous.
(^) sin C = ^n>< sin 30° = U^ = |.
The triangle is possible, and there are two admissible values for G; since
c>a, .*. <7>^, and G may therefore be acute or obtuse; the triangle is
therefore ambiguous.
(7) sin C = U^ X sin 30° = Ht = A ;
«*. the triangle is possible; but since a>c, ^>(7, and G can only be acute.
The triangle is not ambiguous.
In the ambiguous case (^)
JL sin C - 10 + log 250 + log sin 30° - log 200
= 10 + logl000-log4 + logl-log2-logl00-log2
= 10 + 3-21og2 log2-2-log2
= 11 - 4 log 2 = 11 - 1-2041200
= 9-7958800 = L sin 38° 41' or L sin (180° - 38° 41'), i. e. L sin 141° 19' ;
.-. the angles of the obtuse-angled triangle are
^ = 30°, C=141°19', i^ = 180°-^-O = 8°41'.
Now log 6 = log c + L sin J5-L sin C
= log 250 + L sin 8° 41' - L sin 141° 19'
= log 1000 - log 4 + L sin 8° 41' - L sin 141° 19'
= 3 - -6020600 + 9-1789001 - 9*7958800 = 1-7809601.
Here d = ^ oi -001 = -00003 ;
/. -7809601 = log (6-0389 + -00003)= log 6-03893,
1-7809601= log 60-3893, .-. & = 60-3893.
MISCELLANEOUS EXAMPLES. LXX. Page 220.
1. ^ is obviously the smallest angle of the triangle, since it is opposite
to the least side. If we therefore apply the formula
■^VI'~^IJ^'i.
only the smaller value of A is admissible,
6=576-2, c = 759-3,s- 6 = 278-8, s-c = 95-7.
Now L sin J^ = 10 + J {log(s-6) + log(s-c) -log 6-logc}
= 10 + 4 {log 278-8 + log 95-7 - log 576-2 - log 759-3}
= 10 + 4 {2-4452928 + 1-9809119 - 2-7605733 - 2-8804134}
= 9-3926090.
L sin 14° 17' = 9-3921993, difference for 60" = 4959 ;
.-. D = *ff7 of 60" = 49-6";
.-. 4^ = 14° ir 49-6", .-. ^ = 28°35'39".
SOLUTION OF TRIANGLES. LXX. 159
2. Here B is the greatest angle and it is easy to see that h^^a^ + c^,
.-. b is ohtuse and therefore if we find B from the formula for sin JJB we
must take the obtuse angle. It is however best to avoid this investigation
by using either the tan ^B formula or the cos Ji? formula.
Here s = 10851-5, 5-a = 6850-5, s- 6 = 3208-5, s-c = 2909-5,
.-. L tan iB = i {log (s - a) +log (s - c) - log (s-h)- logs} + 10
= i {3-8357223 + 3-4638184 - 3-0380237 - 4-0354898} + 10
= 9-1130136.
X tan 52° 22'=: 10-1129282, difference for r-2613.
Here D = i^\\ of 60" = 19-6" ;
.-. 4^ = 52° 22' 19-6", .'. ^ = 104° 44' 39".
3. Here s = 10549, s -a = 1787-8, s - 7; = 2906, s- c = 5855-2;
.-. XtaniC=10 + i {3-2523189 + 3-4632956-3-7675417-4-0232113}
= 9-4624307.
itan 16° 10'= 9-4622423, difference for 60" = 4722.
Here D = ^|f of 60" = 24" ;
.-. J (7= 16° 10' 24", .-. (7 = 32° 20' 48".
L cot 43° 9' = 10-0280650, difference for 60" = -0002532 ;
SO"
.-. tZ = ^ of -0002532 = -00012666;
bO
.-. L cot 43° 9' 30" = 10-0280650 - -00012666 = 10-0279384.
Now Li8ini(C-B) = log 541 - log 10401 + L cot 43° 9' 30"
= 2-7331973 - 4-0170751 + 10-0279384
= 8-7440606.
8-7429222 =L tan 3° 10', difference for 60" = -0022845;
.-. D = Hfltof60" = 30";
.-. J(O-B) = 3°10'30", .-. 0-5 = 6°2r.
Also • G + B = 180° -A = 180° - 86° 19' = 93° 41' ;
.-. 2^ = 87° 20', .-. 5 = 43° 40'.
5. L tan J C - ^ = log (c - a) - log (c + a) + L cot 4 i?
= log 1109-3 - log 2637-7 + L cot 16° 29'
= 3-0440490 - 3-4212254+ 10-5288593
= 10-1516829.
10-1515508 = L tan 54° 48', difference for 60" = -0002682 ;
.•• 2) = ifliof60" = 30";
J (0-^) = 54° 48' 30"; .-. (7-^ = 109° 37',
+ ^ = 147° 2'; .-. 2(7=256°39'.
160 SOLUTION OF TRIANGLES. LXX.
6. L tan i(B- A)=log 52629 - log 125711 + L cot 54° 13' 30'' ;
/. log 125711 = 5 -0993698 + •0000035 = 5-0993733;
.-. L cot 54° 13' 30" = 9-8578031 - -0001332 = 9-8576699.
Now Ltani{B-A) = log 52629 - log 125711 + L cot 54° 13' 30"
= 4-7212251 - 5-0993733 + 9-8576699
= 9-4795217.
9 -4794319 = L tan 16° 47', difference for 60" is 4568;
••• D^AV^of 50" = 12";
.-. J(J5-^) = 16°4ri2"and5-^ = 33°34'24".
^ + ^ = 71° 33' ; .'. B = 52° 33' 42".
sin C = sin 108° 27' = sin 71° 33',
log c = log 89170 + L sin 71° 33' - L sin 52° 33' 42".
L sin 52° 33' = 9 -8997572, difference for 60" = -0000967 ;
.-. d = ^, of -0000967 = -0000676;
.-. log c = 4-9502188 + 9-9770832 - 9-8998248 = 5-0274772.
5-0274719 = log 106530, difference for 10 is 408 ;
.: d=^\ of 10 = 1-3;
.-. 5-0274772 = log (106530 + 1-3) = log 106531-3.
7. log c = log 3720 + L sin 62° 45' -L sin 74° 10'
= 3-5705429 + 9-9489101 - 99832019 = 3-5362511.
3-5362427 = log 3437-5, difference for -1 = -0000126 ;
.'.d = ^%x'l = 'OQ.
.-. 3-5362511 = log 3437-56; .-. c = 3437-6 yards.
8. ^ = 180 - 145° 18' = 34° 42'.
log 6 = log 1000 + L sin (180° - 100° 19') - L sin 34° 42'
= 3 + 9-9929214 - 9-7553256 = log 1728-2.
9. A = 180° - 138° 16' 20" = 41° 43' 40"
^ = log 9964 + L sin 41° 43' 40" = L sin 15° 9'.
Lsin41°43'=r9-8231138, difference for 60"=-0001417;
'*• ^"^W^^ -0001417 = -0000945;
.-. log a = 3-9984337 + 9-8232083 - 9-4172174
= 4-4044246 = log 25376;
.-. a = 25376.
SOLUTION OF TRIANGLES. LXX. 161
10. L sin i^ =^ log 1450 + L sin (180° - 100° 37') - log 6374
= 3-1613G80 4-9-9925013- 3-8044121 r=9-3494572.
9-3493429 :^ L sin 12° 55', difference for 60" = -0005505 ;
.-. ^ = UUof60" = 12'';
.-. 9-3494572=:Zsinl2°55'12".
sin B = sin 12° 55' 12" or sin (180° - 12° 55' 12"), i. e. 167° 4' 48".
Since oh^ :. C> B, and only the less value of B is admissible ; .
.-. B- 12° 55' 12".
A = 180° - 113° 32' 12" = 66° 27' 48".
11. L sin J^ == log 643 + L sin 52° 10' - log 872
= 2-8082110 + 9-8975162 - 2-9405165 = 9*7652107.
9-7651911 = 2. sin 35° 37', difference for 60" = -0001763 ;
.-. D = TVVTTof60" = 7";
.-. 9 -7652107 = L sin 35° 37' 7" =L sin 5,
sin i? = sin 35° 37' 7" or sin (180° -35° 37' 7") = sin 144° 22' 53".
Since c > 6, .'. C > B, and only the less value is admissible ;
.-. B = 35° 37' 7" ; A = 180° -B-C = 180° - 87° 47' 7" = 92° 12' 53".
12. L&mB = log 1000 + L sin 76° 2' 30" - log 2000.
L sin 76° 2' 30", 9*9869670+ -0000157 = 9-869827;
.-. L sin I? = 3 + 9-969827 - 3-3010300 = 9-6859527.
9-6857991 = L sin 29° 1, difference for 60" = -0002276 ;
.-. D = ^ff|of 60" = 40";
.-. 9-685927 = L sin 29° 1'40" = I. sin 5;
.-. sini^ = 29°l'40" or sin (180° -29° 1' 40") = sin 150° 58' 20".
a>b, .: A > 5, and only the less value of B is admissible,
.-. ^ = 29°1'40", C=180°-105°4'10" = 74°55'50".
13. L sin i>^ = log 873-4 + L sin 54° 23' - log 752-8
= 2-9412132 + 9-9100529 - 2-8766796
= 9-9745875.
9-9745697 = L sin 70° 35', difference for 60'' = -0000445 ;
.-. i)=Hf of 60" = 24";
.-. 9 -9745875 = L sin 70° 35' 24" or L sin (180° - 70° 35' 24") ;
i.e. L sin 109° 24' 36 " ; .-. B = 70° 35' 24" or 109° 24' 36".
Both values of B are admissible; for & > c, i.e. B>G; /. B may be
obtuse or acute.
^ = 180° -124° 58' 24" = 55° 1' 36" or 180°- 163° 47' 36" = 16° 12' 24".
L. T. K. 11
162 SOLUTION OF TKIANGLES. LXX.
14. LsmB = log 674-5 + L sin 18° 21' - log 269-7
= 2-8289820 + 9-4980635 - 2-4308809 = 9-8961646.
9-8961369 = L sia 51° 56', difference for 60" = -0000989 ;
.-. D^lt^of 60" = 17^
9-8961646 =rL sin 51° 56' 17" or L sin (180°- 51° 56' 17") =L sin 128° 3' 43";
.-. 5 = 51°56'17" or 128° 3' 43".
Since & > c both values of B are admissible.
15. L sin 5 = log 7934 + L sin 29° 11' 43" - log 4379,
.-. L sin 29° 11' 43" = 9-6880688 + -0001620 = 9 '6882308 ;
.-. L sin 5 - 3-8994922 + 9*6882308 - 3-6413749 = 9-9463481.
9-9463371 = jC sin 62° 6', difference for 60"= -0000669;
.-. i) = |if of60" = 10";
.-. 9-9463481 = L sin 62° 6' 10" or L sin (180° - 62° 6' 10") ;
.-. B = 62° 6' 10" or 117° 53' 50".
Since 5 > a both values of B are admissible.
16. Let jB, C be the angles at the base of the triangle, the sides sub-
tending them being b and c ; A being the third angle,
/. i(B-C) = 4(17°48') = 8°54'. &-c = 28-5; 6 + c = 182;
tan8°54' = 3\83^%cot4^;
.-. L tan 8° 54' = log 28-5 - log 182 + 1. cot J^ ;
.-. L cot iA = L tan 8° 54' - log 28 5 + log 182
= 9-1947802 - 1-4548449 + 2-2600714
= 10 -0000077 = L cot 45° nearly ;
.'. iA= 45° nearly ; ^ = 90° nearly.
.-. L tan 4 (C - J5) = log 1- log 9 + 1. cot 18° 39' 30";
.-. L cot 18° 39' 30" = 10-4717147 - -0002084 - 10-4715063 ;
.-. Ltani(C-B)=- -9542425 + 10-4715063 = 9-5172638.
9-5169097 = Ltanl8° 12', difference for 60" = -0004256;
••• D = |IHof 60" = 50";
.'. i{C-B) = 18° 12' 50" ; .\ C-B = 36° 25' 40".
C + B = 180° -■A = 180° - 37° 19' = 142° 41' ;
.-. 2B = 106° 15' 20" ; .. B = 53° 7' 40" ;
.-. log b = log 1000 + L sin 53° 7' 40" - L sin 37° 19' ;
.-, L sin 53° 8' 40" = 9-9030136 + -0000672 = 9*9030768 ;
.-. log 6 = 3 + 9 -9030768 - 9 '7826301
= 3-1204467 = log 1319-6 nearly;
.•.?> = 1319-6,
SOLUTION OF TRIANGLES. 163
2. cos A= —
EXAMPLES. LXX. b. Pages 220 (i), (ii).
1. When a = 6 = c; cos^=i; cosi^= ^-[|-^l =W3.
4 + 6-( l + ^3)2 ^ izl^^^ = \/^-\/^ ^ (\/3-l)
' 4^6 4^6" 4 2^2 *
6 + 4 + 2^3-4 2^3(^3 + 1) ^ 1
'''' 2V6(1+V^r ~2V6(v/3 + l) x/2*
coaC 4 + 2j3 + 4-0 _2(l + ^3)_
''^'^- 4(l + x/3) -4(1 + ^3)-*'
/. C=60°, jB=45°, ^ = 75°.
'^' '''''^- I672~ -'16V2 - 4^2 •
8 + 16-8^3-16 ^ 8(l-\/3) __J^
"""^^ 8V2(>y^-l) 8^2(^3-1)- ^2* ■
16-8^ 3 + 16- 8 _ _24- 8^/3^ _ 8 ^3(^3-1) _ ^3
'^''^ ^ - 16 (V3 - 1) ~ 16(v/3 - 1) ~ "16(^3 - 1) " 2 '
.-. C = 30°, ^=135°, ^ = 15°.
4. c2 = a2 + ^,-^-2a6cosl20°,
/. 19 = 4 + 62 + 26; .-. 62 + 26 + 1 = 16; 6 + 1= ±4, .-.6 = 3.
5. a2=16x7 + 36x 7 - 24 x 7 = 28 x 7 = (2 x 7)2.
a sin C 2 sin 75°
sin A sin 45
= 2V2xi^2(^3 + l)=V3 + l.
I
, 49 + 64-169 56 ^ ,_„,
7. cos .4 = zr^ = - ^7v = -h = 120"^
'• 2x56 2x56 ^
4 + 1-7
8. cos^= ^ =-i. .-.^ = 120°.
^ . a2 + 62-a2-a6-62
9. cos ^ = -— = - 4 ; '. A = 120°.
2a6
COB C = ~-t^^^- = f J = -8 ; .-. = 36° 52'.
11—2
164 SOLUTION OF TRIANGLES. LXX. b.
11. oos^=?^i^|^^i^=-tf=--65; /. .4 = 180° -49° 33'= 130° 27'.
ii X D X D
12. cos C=^^i^^^^= -If = --575 = 008 (180°-54° 54').
* 2x4x5 *" '
14. cBmA=a sin C; /. (^5 - 1) sin A = i (^5 + 1) (s/o - 1) ;
. ■ 1 v/5 + 1
•••«^^^=-j5ri = -4—
sin 54° = sin (3 x 18°) = 3 sin 18° - 4 sin318°
= i(V5-l){3-i(6-2V5)}=i(x/5-l)4(v/5 + 3) = i(^/5 + l).
.-. sin A = sin 54° ; .\ A = 54° or 180° - 54°.
15. • c2 = «2_^62_2a&cosC,
.-. 13 = a2 + 9 + 3a;
.-. a2 + 3a-4 = or (a- l)(a + 4) = 0; /. a = l.
sin ^ = - sin C, sin B—- sin C.
c c
16. (7= 180° -105° -45° = 30°,
csin^ ^2 cos 15° ._ , .
a = -. — ^ =^ J =^3 + 1,
sin <7 i
6 = 2^2 sin 45° = 2.
17. A = 180° - 75° - 30° = 75° = B.
.'. a = b» Also since the triangle is isosceles 4c=a cos 75°;
18. BinC=^gsin45°=^^x^2 = f =Bin60°;
.-. C = 60° or 120°, ^ = 75° or 15°;
sin 75° ,_. ^3+1 /_. . V^ + l
^=si^°^-/^'-JvW^'=^"V'v
sin 15° ,_ ^3-1 ,_ . V3-1
^^ "=sm20^^^'=V372^^' = '" V"-
SOLUTION OF TRIANGLES. LXX. b. 165
19. sin (7=.^^^ sin 30°=^ = sin 60°;
:. (7=60° or 120°; /. ^ = 90° or 30°.
When the triangle is right-angled a2=6-'» + c2=: 502(9 + 3) = (100^3)2.
20. sin (7 = 1, .: C= 90° and the triangle is not ambiguous ; see Fig. ii.
p. 216.
23. G = 180° - 54°, sin (7 = sin 54° = cos 36°
= l-2sin2 18° = l-i(6-2V5) = J(x/5 + l);
.-. c = J(V5 + l)V6x/2=iV3(v/5 + l).
24. h^ = a^ + c^-2acco8B;
.-. 4-2V3 = a2 + 4 + 2^3-2a(V3 + l)(V3 + l)iV2;
.. a2-a (2 + ^3)^2 + 4^3 = 0;
.-. {a-2J2)(a-^6) = 0;
.'. a = 2j2 or ^6;
^■^-'-^'^='^^-'="^'
or sin^=^=:^(^; .-. ^ = 60°.
When A = 90°, 0=75°; when A = 60°, C = 105°.
25. sin^ = i.5x| = J; .-. ^ = 30° or 150°.
See Figure iii. of p. 216.
26. sin A = '^^^ X ^^^L^ ^ J_ ; •• ^ = 45° or 135°.
2^2 4 fj2
When ^ = 45°, 1^ = 120°; when ^ = 135°, 5 = 30°,
Rin 120°
2> = ^^^^4^4(1+V3) = 2V6(1 + V3),
or, 6 = g|^°4(l + V3) = 2^2(l + V3).
27. a2=Z;2 + c2-2Z>ccos^ = 9{6 + 4 + 2^3-^6(V3 + l)x/2}
= 9{10 + 2V3-6-2^3} = 36;
.*. a = 6.
.-. C= 180° -60° -45° = 75°.
166 SOLUTION OF TRIANGLES. LXX. b.
28. sin B — \<^/ X \ which is greater than 1.
29. = 75°, 2c = 2asin75°-^sin45° = a(l + J3).
30. sin ^= J V^' sin 5 = 1,
sin C = sin (^ + 5) = sin ^ cos 5 + cos ^ sin 5 = J ^3 X f + J X I = ^ (3 ^3 + 4) ;
.'. the required ratio which is equal to the ratio of the sines is
iV3xlOV3:txlO^3:TV(3v/3 + 4)xl0^3 = 15:8V3:9 + 4V3.
EXAMPLES. LXXI. Pages 224-226.
1. ABC is a triangle in which ^ = 60°, (7 = 30 miles, 6 = 15 miles,
a2=:62 4.c2_ 26c cos ^ = 15^302- 2 x 15 x 30 cos 60°
= 225 + 900-450 = 675;
.-. a=V675 = 25-98...
2. Let A be the mouth of the harbour; let jB, G be the position of the
ships respectively after \\ hours; then AB = \ of ^ miles, AC=^ of 10 miles
and 5^0 = 45°;
/. BG^ = (-V-)2 + (15)2 _ 2 X -V- X 15 X J V2 = 15^ { A + 1 - 1 x/2}
= p^)2 {25 - 12^2} = (Y-)2 {25 - 12 X 1-4142...}
= (\^)2x(2-84...)2;
.-. J5(7= 10-6 nearly.
3. AB = c, BG=a, GA = b,
c sin B _ sin 120°
~~ sin G " sin 15°
= ^^ ^'n/?^ := _^^ Vid. E. T. Examples XXXIV.
^ x/18 + ^6 ^ J2 (3 + ^3 ) _ 1 -4142 x 4'73 2
3-1 2 2
= -7071 X 4-732= 3-346 miles.
4. From(3) ^C = i{^2(3 + v/3)}=iW2x/3(l + v/3)}.
The distance of the spire from the plane of A is
HEIGHTS AND DISTANCES. LXXL 167
5. Draw a quadrilateral ABCD^ so that AG and Bl) be the two diagonals
and I ABC = 120°, iBAC = ^5°, aAGB = W, z D^i? = 90°, ADBA = A5°,
L DAG = 4:5°. In the right-angled triangle DAB, DB = AB sec 45° = J 2 miles
In the triangle ABG, i?C= — ^— „— = . -.^^ = 7o -^ o" To = -Jo — i •
sm C sin 15^ ^2 2 J2 >/3 - 1
Now GD^ = DB^ + BG^-2DB .BG cos DBG
/. CZ)==-7^ = ^^^^^ii^=V3-l-l==2-732 = 2| miles nearly.
V o — 1 2
6. Angle GAB = 36° 18', CBA = 120° 27' ; .'. AGB = 23° 15'.
Let AB = c, BG=a, GAB = A, AGB = G.
Now loga = logc + -Lsin^ + I/ cosec (7-20
= log 1760 + L sin 36° 18' + L cosec 23° 15' - 20
= 3-2455127 + 9-7723314 + 10-4036846 - 20
= 3-4215287 = log 2639-5 nearly ;
.-. a =: 2639-5.
7. This is CasG III. of the Solution of Triangles ;
.-. L tan \(A- J5) = log 1346 - log 4934 + L cot 29° 8' 30"
= 3-1290451 - 3-6931991 + 10-2537194
= 9-6895654 = L tan 26° 4' 19" ;
.-. J(^-i?) = 26°4'19" and vi -5 = 52° 8' 38";
^ + ^ = 180°-- 58° 17'; .'. ^ + P = 121°43';
.-. 2^ = 173° 51' 38"; /. ^=86° 55' 49".
8. Let tlie distance from A to G = h,
,, ,, ,, ^ to i> = c,
GioB = a.
Let angle GAB = A, z GBA = B, aAGB = G;
.-. C = 180°-J:-J5 = 180°-61°53'-76°49' = 41°18',
b = c sin B cosec G ;
.-. log b = log c + L sin J5 + L cosec C - 20
= log 34920 + L sin 76° 49' + L cosec 41° 18' - 20
= 4-5430742 + 9-9884008 + 10- 1804552 - 20
= 4-7119302 = log 51515;
.-. & = 51515 feet.
168 HEIGHTS AND DISTANCES. LXXI.
9. Using same notation as in (8),
C = 180° - 72° 34' - 81° 41' = 25° 45;
5 = c sin ^ cosec (7 ;
/. log 6 — log c + L sin JB + i cosec (7-20
= log 37412 + L sin 81° 41' + L cosec 25° 45' - 20
= 3-5731038 -f 9-99540 .7 + 10-3620649 - 20
= 3-9305774 = log 8522-7;
.-. 6 = 8522-7 yards.
10. The height of the one above the other is 4970 x sin 9° 14',
i. e. 4970 X -1604555 = 797*5 yards.
11. In the triangle ABC, let A represent the point of intersection of the
railways ; at the end of an hour let the first train be at B and the second at
G then BG represents their distance apart; .-. J5C = 35 mis., ^J5 = 40 mis.,
u4(7 = a;mls., a^=h'^ + c'^ -2hc (to^A\
.-. (35)2 = a;2 + (40)2 + 2xa:x40cos60°;
.-. a;- -40a; -25 = 0.
From this quadratic a; = 25 mis. or 15 mis.
N.B. This is an instance of the ambiguous case.
12. Using same notation as in (8),
c2 = a2 + 62 _ 2ah cos C=82 + IO2 - 2 x 8 x 10 x J = 84;
.-. c = V84 = 9-165.
13. The height of B above A is the difference of the elevation of G above
B and the elevation of G above A ;
i. e. 10 sin 8° - 8 sin 2° 48' 24" = 10 x -1391731 - 8 x -0489664
= 1-391731- -3915312 = 1 nearly.
14. The sine of the angle which the tunnel makes with the horizon is
the height of A above B _ 1 _.i^q-,
the distance between A and B " 9*165""
= sin 6° 16'; .-. the angle is 6° 16';
tan i (i? ~ J ) = ^-^ cot 30° = ^ x ^3 = -192450 ;
.-. J5 - ^ = 2 X (10° 53' 36") = 21° 47' 12", 5 + ^ = 120° ;
.-. ^=i (98° 12' 48") =49° 6' 24".
HEIGHTS AND DISTANCES. LXXI. 169
15. Using the notation of (8) angle G = 180° -A-B
= 180° - 38° 19' - 132° 42' = 8° 59'.
In the triangle ABC a = c sin A cosec C;
.-. log a = log c + -L sin J^ -t- L cosec 0-20
= log 1760 + L sin 38° 19' + L cosec 8° 69' - 20
= 3-2455127 + 9-7923968 + 10*8064659 - 20
= 3-844375.
If h is height of mountain above B, h = asin 10° 15' ;
/. log 71 = log a + L sin 10° 15' - 10 = 3-844375 + 9-2502822 - 10
= 3-0946576 = log 1243-5;
.-. 7i = 1243-5 yards.
16. Using the same notation as in (8),
C = 180° - J[ - £ = 180° - 65° 37' - 53° 4' = 61° 19'.
In the triangle ABC, 6 = csin B cosec C;
:. log h = log c + L sin B-\-L cosec C - 20
= log 1000 + L sin 53° 4' + L cosec 61° 19' - 20
= 3 + 9-9027289 + 10-0568589 - 20 = 2-9595878.
If p be the perpendicular breadth of the river,
p^h sin A ;
.-. log p = log h -\- L B>m A - 10
= 2-9595878 + 9-9594248- 10
= 2-9190126 = log 829-87;
.-. i) = 829-87.
EXAMPLES. LXXII. Pages 227, 228, 229.
1, Let A be the position of the balloon when it was first observed, and
B the position of the man, so that the angle ^J5C = 60°; produce AE
horizontally to represent 1 mile, the distance travelled by the balloon in 10
minutes. Let BC = ^AE drawn parallel to it represent half a mile, the dis-
tance travelled by the man in 10 minutes. Produce BC to F; join EC ; then,
according to the question, angle ECF='dO°, Draw CD parallel to BA and
EF perpendicular to BF, EF represents the height of the balloon above
iro
HEIGHTS AND DISTANCES. LXXII.
the road. Now ABCD is a parallelogram; /. AB = BO=\ mile; /. BE
also = imile; /.DCF= I ABC^^O" and angle £Ci^=30°; .'. z D(7E^=:30°;
LBEC=LEGF=m''; :. i BCE = i BEC = SO°; .\ Z CD^ = 120° and
CB = BE = imile.
CE sin 120° ^^ sin 60° , .. ^3 ..
= -; ; .'. CE = —. — — TTT X A mile = -^ miles ;
BE sin 30° ' sin 30° ^ 2
i;i^=:Ci;sinJ5CF=iV3sin30°mi. = JV3m. = J^/3xl760yds. = 440^3yds.
2. In the figure let represent the foot of the tower,
and OB = x its height; z D^0 = 60°, z OAB = 90°,
LBB0 = 4:b'', ZDCO = 30°;
^=rtan60° = v'3; .-. OB = AO^^^x', .: ^0 = ~j.;
AO \/o
^ = tan45° = l;
OB = BO = x\
^ = tan30°=r_Lj ... GO = OB^'d = x^^.
Since 0^5 = 90°; ,\ AB'^=OB^- OA^-,
Also ^(72=002- 0^2^3x2- 'I = ^- , .-. AC
,_ 2x^6
3 '
.'. AC = 2AB,i.e. AC = AB + BC; .'. AB = BC.
3. Let X be the no. of miles the balloon travels per hour.
Then
in 20 mins.
Using figure E. T. p. 62, let Q be the position of balloon after travelling
1 mile, and P its position 20 minutes after; /. MQ = 1 mile, QP=- miles.
Z QOM= 35° 20', z POM= 55° 40'.
OM
-— = cot QOilf= cot 35° 20'; /. OM = cot 35° 20' miles,
MP = MO tan 55° AO';
1 + H = cot 35° 20' tan 55° 40', ~ = cot 35° 20' tan 55° 40' - 1
o O
__ sin 55° 40' cos 35° 20' - cos 55° 40' sin 35° 20 '
sin 35° 20' cos 55° 40'
_ sin (55° 40' -35° 20') ^
"sin 35° 20' cos 55° 40''
.-. x=S (sin 20° 20') (sec 55° 40') (cosec 35° 20').
HEIGHTS AND DISTANCES. LXXII.
171
4. Draw a figure of the form indicated in (2) and retaining the same
letters IDAO = SO°, L OAB = 90°, z 1)50 = 45°, ^J5 = afeet.
^=^tan30°=4Q; •'. AO = ODjS = x^S,
0B^ = AB^ + 0A^ = a^ + 3x^; .\BO = J{a^ + dx^),
OD X , _. v/5-1
BO ^{a'+'dx^)
x'^
= tan 18°=
^(10 + 2^5)'
" a^ + ^x^ 5 + ^5'
2a2
. .^.^ ^M3-v/5) _ a^(3-^5)(3 + V5)
^6-1 (V5-l)(V5 + 3) "1+^/5*
"2(1+^/5)'
V{2(l + V5)}
5. Using the figure and letters of (2) OD = x the height of the steeple,
ZD^0 = 45°, lOAB==90°, lDB0 = 15°,
AB^aft., ^ = tan45°=l, /. OD = OA=x,
OB^ = AB^ + OA^ = a^ + x^; .-. OB=^(a^ + x^).
0D_ X _^ i^o_ n/3-1
OB- sJia^ + x^)-^"""^^^ -^3 + 1'
.-. 2a; V3 = 2a2 - a^ ^3, 4a:2 ^3 = ^2 (4 _ 2 ^3) ;
/. 2a:3J = a(3i-l), a; = ^ (3J-3- i).
6. Let 5(7 be the inclined plane; BD the tower at its foot; zC5^=9°,
.-. Z C5D = (90° - 9°) = 81° ; let 5(7 be length
of line = 100ft., ZD05 = 54°; /. z5Da=45°.
BD =
BC sin C 100 X sin 54°
sin D sin 45°
= 100xi(l + ^5)xJ2 = 25(V2 + V10)
= 25 (1-4142 + 3-1628)
= 25x4-577 = 114-4 ft.
7. Let Z^5C = 47°; 5i) = 1000ft., zD5C=32°,
Z^5D = Z^5(7-zD5C = 15°; lADC = lT.
Since ^5(7 is a right-angled triangle and z ^50=47°;
.-. z5^C = 90°-47°=43°,
172
HEIGHTS AND DISTANCES. LXXII.
and since ADE is a right-angled triangle and z ADE — 77°;
.'. A DAE = 90°- 77° = W; :. z ^^D==43°- 13°=30°,
Z ^Di? = 180°- Z^i^D- z i>MD = 180°-15°-30° = 135°.
sin-B^D sin30° ^
.4(7-J[Bsin^5O = 4J5sin47° = 1414x'73135 = 1034ft.
8. Let OD be the chimney ; OAB the triangular area ; then OD = 150 ft. ;
L BDO = S0°; I ADO = 4:5°; I BDA = SO°.
BO
OD
=rtan30°
,, BO=^^=~^ = 50^3 = S6'Qn.
AO
DO
= tan45°=l; :. AO = DO = 150ii.
AB^ = AD^ + BD^--2AV.BD cos30°.
Now AD = OD cosec AD = 150 cosec 45° = 150 ^2,
BD = OD cosec OBD = 150 cosec 60° = 100 ^3 ;
/. AB^ = (150 J2)2 + (100^3)2 - 2 X 150^2 x 100 ^^3 x
x/3
= 45000 + 30000 - 45000 ^^2 = 75000 - 45000 x 1-414 = 11370 ;
.'. ^5=;^11370=:106-6.
9. Figure of E.T. p. 62. Let PQ = h (i.e. height of flagstaff), QM=x
(i.e. heij-^ht of tower), AP03I=a, z QOM=^, OM = yy i.e. distance of the point
of observation from the foot of the tower.
x + h
— tana, - = tau^,
y
x + h
- = tana-tanj3;
tan a - tan ^
Now x = y tsinp=
HEIGHTS AND DISTANCES. LXXIL
h tan /3 _ ^ sin /3 cos a
173
tan a- tan ^ sin a cos /3 - sin j3 cos a
hsinp cos a
~ sin(a-/3) '
10. Figure of E. T. p. 62. Let OM = h height of the cliff, OPM=p,
OQM = a, .'. FQ is the distance between the ships.
PQ = PM-QM,
FQ _ PM QM
OM" OM OM'
FQ = h {cot ^- cot a) feet
11, Draw a figure like that indicated in Ex. (8), putting the A, O^ B in
similar positions.
Let B be due S. of O and A due W. of B.
Then OB = h cot a,
OA'^ = BO^ + AB' = h^coi^a + d^;
OD^ h^
6 A
' h^cot^a + d'^
.-. ^2(i_cot3atan2^) = d2tan2^,
= tan2 18.
/. h =
cot'-^jS-cot'^a
d _ d sin a sin /3
;^(cot=^/3 - cot'-^a) ~ ^{sin^a cos^^S - cos^a sin^/S}
_ d sin a sin /3
~~ fj{ (s^^ * cos i^ + c<^s ^ sin ^ ) (sin a cos )3 - cos a sin jS) }
_ d sin a sin ^
~^-Jsin(a + /3).sin(a-/3)} *
12. Let AB be the height of the wall = h', AG the height of the man = BH\
lDCF=a, aDHG = P, z2^CD = 90° + a;
zDiIC = 90° -ft lCDH=^-a,
Since CA=HB, CH=AB = h.
In the parallelogram BG, GE = HB;
.-. ED-HB = ED-GE
= DG the required difference.
CH sin (90° 4- a) hcoQa
.'aO
Now Dif =
sin (/3 - a)
.-. DG = DHBmp =
~ sin (^ ~ a) '
/i cos a sin ^
sin (/3 - a)
174
HEIGHTS AND DISTANCES. LXXII.
,"yB
13. Let A be the point of observation, B the cloud, G the point on the
surface of the lake from which the light is reflected from the cloud. Since
from Optics the angle of inci-
dence = the angle of reflexion,
.*. BG and AG are equally in-
clined to the surface of the lake
and .'. also to the horizontal line
AD drawn through A,
lBAG = a + p, Z ^(7^ = 180° -2/3,
.-. lABG
= 180-(a + ^+180°
CB =
AG sin BA G _ ^O.sin ( ^ + a)
sin (iS - a)
sin ABG
Height of clond — GB sin/3
AG . sin ^ . sin (^ + a) _ h sin {^ + a)
sin [(3 - a)
sin (j8 - a)
14. Let A be the position of the spire which is nearer to the road and
B that of the spire which is farther from the road.
The height of A above the road can be found as in (13) to be — ; — — ^ ;
^ ^ ' sm (^ - a)
The
in the same manner the height of B above the road is
difference in height between B and A
- 7 ;jsin (7 + g ) _ sin(/3 + a)
~ (sin (7 - a) sin (/3 - a)
h sin (7 + a)
sin (7 - a)
= /i
= h
sin (7 + a) sin (/3 - a) - sin (j8 + a) sin (7 - a)
sin (7 - a) sin (^ - a)
cos (7 - /3 + 2a) - cos (/3 - 7 + 2a)
= }i.
2 sin (7 - a) sin (^ - a)
sin 2a sin (/3 - 7)
Now
' sin (7 - a) sin (§-a)'
the difference in height of the spires
= tan a;
the horizontal distance of the spires
.-. The horizontal distance = the difference in height x cot a
_ h sin 2a sin (fi-y) cos a __ 2/i sin a cos a sin (j3 — 7) cos a
~ sin (7 - a) sin (}i - a) ' sin a ~ sin (7 - a) sin (/3 - a) ' sin a
== 2h QQB^ Pt sin (i8 - 7) . cosec (/3 - a) . cosec (7 - a).
HEIGHTS AND DISTANCES. LXXIL 175
15. From note E. T. Answers, the point E in an unlimited straight line
\
V
/
t
\
/ //
B
\
'' / /
\
/ / /
'\
' *" y''^
-f
1^:"
GE, at which a finite straight line AB subtends the greatest angle, is on
the circumference of the circle of which GE is a tangent and ABG a secant.
Let lBEG = y and GE = c;
.*. ^G = c tan (a + 7), BG = ciainy.
From Euclid iii. 32, Z BAE=iBEG = y; and .-. in the right-angled tri-
angle ^(7£ a+2Y=j7r, .*. 7 = Jx-Ja.
AB~AG-BG = cidi.n{a + y)-cidiny
= c{tan (J7r + Ja)- tan(47r- Ja)}
(1 + tan J a l-tanjaj_ 2 tan J a
(1 - tan J a 1 + tan \aS~ * 1 - tan'-^ J a
= 2c tan a.
The height of the pillar = h-\-BG = h-\-c tan y = h-\-c tan (J tt - J a).
Or, let be the centre of the circle of which GE i^ tangent and ABG
secant ; draw OM perpendicular to AB ; .'. AM = MB and AB = 2AM. Draw
the straight line OE and (Euclid in. 18) OEG ia a. right angle; .-. MOEC is
a rectangle and MO=GE = c. Also angle ^ Oil/ = angle AEB = a (Eucl. in.
20); .-. ^Jf=il/Otana=:ctana; .'. ^B = 2c tan a.
Also GB .GA = GE^, i.e. GB {GB + AB)=:GE\
Let GB = a; .*. a (a + 2c tan a) = c2; .\ a^ + 2ac tan a- C' = 0.
From this quadratic a = c tan (J tt - J a) .
The height of the pillar =b + a = b + c tan (i tt - J a).
16. In the above fig. let A denote the object which is further removed
from the road, B that which is nearer to the road E, the point where AB
subtends the greatest angle, G the second point of observation. From note
E. T. p. 272 it is known that the point E is on the circumference of a circle
of which GE is tangent and ABG secant [cf. (15) above]. Let lBEG=^y
and GE = a ;
.-. AG = cUxi(a + y)^BQ^ci9,ny.
176
HEIGHTS AND DISTANCES. LXXII.
From Euclid in. 32 iBAE= A BEG = y; /. in the right-angled triangle
AGE a + 27 = i7r; /. 7 = j7r-Ja,
AB = AG-BG=:cta.n (a + 7)-ctan7
— a tan (jTr + ^a) -a tan (Jtt- 4a) = 2atana...cf. 15.
17. Let AB be the flagstaff and BG the tower, GE being the horizontal
plane. E the first point of observation and D the second. Since the angle
.Aa
/ / J
B
,^ .f / !
/ .'-'/ / /
^
^''''' / ,/ /
/'^^
' '' y'
-f*"
N !■:.'■
...--/)
AEB is equal to the angle ADB, the points A, B, D, E lie on the circum-
ference of a circle. Euclid in. 21. The centre of the circle is in the
straight line bisecting AB at right angles in M and in the straight line
bisecting DE at right angles in iV"; .*. MONC is a rectangle and MO = GN.
The angle AOM is half of the angle A OB ; hence by Euclid iii. 20 the angle
AOMis equal to the angle AEB; i.e. Z AOM=a,
AM
MO
M0 = CN=CD + DN=h-2k-\-k=h-k,
= tsLnAOM; .-. AM=M0 t&n AOM =(h-k) tana;
.-. ^i5 = 2^i¥ = 2(/i-A;)tana.
18. By comparing (15) and (16) above and note E. T. p. 272, it is
known that E the first point of observation is on the circumference of a
circle of which GE is tangent and ABG secant. The figure is like that on
p. 175 except that AGE is not a right angle. G is second point of obser-
vation and the straight line ABG makes with GE the angle AGE = p.
Let angle BEG = y; from Euclid iii. 32 lBEG= lEAG; .*. angle
EAG = y, and angle ABE = 1S0'' -y- a. But angle EBG = 1S0°- ^-y;
.-. /.ABE = p + y; .-. 180°-7 -a = |S-l-7; .'. 27 = 180° - a - ^3.
BE _ sinjS .j^^,_ asin/3 AB
EG'
Now — -^ =
BE =
sin(^-f-7) J5^
2a sin a sin ^
sin(/3 + 7)'
a sin a sin j8
~ sin 7 sin (j8 + 7) "" cos j8 - cos (27 + )3)
_ 2a sin a sin ^ _ 2a sin a sin jS
~ cos /3 - cos (180° - a) "" cos a -f cos /3
sma^
sin 7'
yds.
TRIANGLES AND CIRCLES. l77
EXAMPLES. LXXIII. Pages 239, 240.
, ab . ^ he . . ca . ^
1. A= — sinC = — sin^= - siiijB.
... ^ 10x4 . ^^_ 10x4 ^. .,
(i) A = — — -sin30°=-r — -=10 sq.ft.
^ ' 2 2x2
K V, on
(ii) A = -^ sin G0° = 25 ^3 = 25 x 1-732 = 43-3 sq. in.
C6§xl5 . ,^„,,. 200x15 ^^^^_
(ill) A = — ^ sm 17° 14 = -- — -- X -29626
Z o X Z
= 500 X -29626 = 148-13 sq. yds.
(iv) A=y/{s{s-a) (s-h) (s-c)}; s = 21, s-a = 8, s-h = 7, 8-c = 6
= ^/{21 X 8 X 7 X 6} =84 sq. chains.
a» 10x20 ,^^ ^ ,
(v) A = -^ = — — — = 100 sq. feet.
(vi) 5 = 1017, s- a = 392, s- 6 = 512, s-c = 113;
.-. A ^{1017 X 392 X 512 x 113} = (113 x7x8x8x3) = 151872 sq. yds.
2. s==21, s-a = 8, s-6 = 7, s--c = 6, A =^{21 x 8 x 7 x 6} = 84.
A 84
Radius of inscribed circle = r=- = 7— = 4,
s 21
A 84 ^.„ A 84
r=-^ = ^ = 14
3. 1H= -. — -r = -; „ = -; 7. .
Sin A sin B sm (7
The circle in which the first triangle can be inscribed is of diameter
2 4 4 v/3
.- ---^ = — — = — ^ . Diameter of circle in which the second triangle can be
sm60° ;^^/3 3 °
inscribed is ,-^^r. — —~- ; i.e. the diameters of the circles are equal ; there-
sin 30° 3 ^
fore the circles themselves are equal, and the two triangles can be both
^' sin^~bcsin^~ 2^' '' Js '
The area of triangle of (2) is 84 ((i), (iv)) sq. ft.;
^ 13x14x15 65 ^^ . ^
.'. R = — i — Z-, — = -— = 8i feet.
4 X 84 8 ^
L. T. K. 12
178 TRIANGLES AND CIRCLES. LXXIII
5. When the circle is the same the radius is constant; and when the
perimeters are all equal, 2s is constant and therefore s is constant; .*. r, and
s in the formula r = — , are constant; .*. ^S^ is also constant; i. e. the areas of
the triangles are all equal.
sin^ sinJ^ . ^ 6 sin ^
6. — ^— = ; .-. sin-B= .
a a
Since b>a; .-. B>A and only the smaller value of B is admissible;
.-. sinJB = V2x^^V3=-^; /. ^ = 45°; ^ = 60^;
Area of the triangle = = J x ^/3 x fJ2 sin 75°
n ,'^ aftsinC -^ c ^^ . ^ c
abc . . abc abc
... A = — ; or,from(4),iJ = — ; .-. ^=-^.
(ii) -,^=--- =2R; .-. a = 2i2 sin ^ , i = 2i^ sin i^.
sni ^ sm B
XT A absinC iR^ sin A sin B sin C _„„ . . . t> • ^
Now A = — = = 2R^ sm A sin B sm C.
q
(iii) ?•=-; .-. S = rs= A.
(iv) a = 2J2sin^, 5=:2iJsin5, c = 2Esin(7.
From (iii) S = rs — r^ {a + b + c)=rli {sin A -h sin B + sin C).
, , , . ^ , a^ftsinC , a^ 2ij sin 5 sin 6^
(V) ^=4.-'^BmC = i.--— = 4.-^2^-^^-^
a^gin^sinC , „ . ^d • ^ ^
= * . — . •, =ha^ sm -B sm C cosec ^.
^ sm ^ ^
(vi) /S = rs = r J (a + 6 + c)=ri? (sin^+sinJB + sinC)
= 4ri^ cos i^ cos J5 cos J O. E. T. p. 192.
a
Now 2B =
sin A 2 sin 4^ cos J^'
„ 2arcosi^ cosJ-B cosiC , , , ^, ,^
. S = ^ . , - — , , — ^^— = ra cosec i ^ cos ^ I> cos * (7.
2sm4^cosJ^ 2 2 2
>Sf2 -^^-<?
\/{s^s-a)(s-6)(s-c)} S'
TRIANGLES AND CIRCLES. LXXIII. 179
(viii) =hab8mC = ^—-^^^
_ i (gg - h^) 4R^ sin A sin B sin G
4ii2(sin2^-sin2^)
ija^- b^) sin ^ sin BsinC _ i (a^ - h^) sin A sin B sin 6"
sinM-sin^jB ~ sin (^ + J5) sin (^ - J5)
_ i(a^- b^) sin A sin B sin G
~ sin C sin (A-B)
= J (a^ - 62) sin ^ sin B cosec (^ - JB).
8. From (4) JR=j^and- = r;
. jR^_ and 6Rr=-p-- ,
Since a, 6, c are in a.p., a + c=26; .*. a + & + c = 3Z> = 2s;
3a6c 3«6c /, >„
2s do
9. ^ = IT sin (7; let a and 6 be constants being equal to 50 and 60 feet
respectively; the area will therefore be greatest when sin G has its greatest
value, i.e. when sin C = 1 ; /. greatest area =4 x 50 x 60 = 1500 sq. ft.
10. In the isosceles triangle ABG let b and c be the equal sides; 2a
the perpendicular and 2x the base, the perpendicular will bisect the base at
right angles ; .
now 2R = — — ^ =a;^/5-^ -7^=7^^;; .-. J2=: - . a;= - . 2a;=: - of the base.
smB V^ ,^5 2 4 8 8
11. iJ (sin ^ + sin ^ + sin C)
= J (2R sin ^ + 2J2 sin B + 2R sin G) = i(a + b-{-c)=s.
12. 4JJ2(cos^ + cosB.cos(7)
= 4:R^{-co8(B-\-C) + cosBcoaG}
= 4:R^ Bin B sin C = 2R sin B.2R sin G=^bc,
13" ^^ = ^ = sTnl)~0° = ''
\ab sin (7_a6a4-6 + c_ a6
~ s ~ 2 ~ 2 ~ a + 6 + c '
rt„ -, 2afe ac -f 6c + 6*2 + 2a^;
.*. 2ii + 2r = c + - — r — =
a-\-b-\-c a + b+c
[since = 00°; .•. c2 = a2 + fe2j
_ flc + 6c + a2 + 62 + 2a& c (a + 6) + (a + 6)2 _ (a -f 6) (a + 6 + g) _
a+6+c a+6+c ~ a+6+c ~
12—2
180 TRIANGLES AND CIRCLES. LXXIII.
14. r^r^-^r^r^-^r^r^
S j^ _^ A_ A ^
~s-b' s—c s-c's-a s-a's-b
_s{8-a)(s -b){s-c) s{s -a)(s-b)(s-c) s{8-a){s~b)(8-c)
~~ (s-b) {s - c) (s -c)(s- a) (s -a)(s- b)
= s(s-c) + s(s-a) + s(s-6) = 3s2-s(a + 6 + c) = 3s2-2s2 = s2.
15. From (4), E=g, r= ? ;
^ _ ^ abc S abc abc
^S ' s 2s a + b + c'
1 _ a + & + c _ 1 1^ 1
2rR abc be ca ab '
16. In fig. E. T. p. 234, r^ = I^F^ = AFi ta.n F^AI^
= 5tanJ^ [by ii. p. 234].
Similarly ? 3 cot J B = s = r^ cot J C,
Alsorcoti^coti^cot^(7 = g^/i , ^(^/^ , . , '^\-^^ , . ^ '^\~"K \
s\/ \(s-b)(8'^c) (s-c)(8-a) {s-a){s-b)f
17. ri + r.
_ a + & + c-(a + &) _ c J{s(s-a)(S'-b)(s~c)}
~ (8-a)(8-b) ~ {s- a) (s - b)
= ;y^'t"'\^,, =ccotiC. E.T.p.200.
18 1 + 1 + 1 := ^'2^3 +^3^ + ^1^2 ^ ^ (^2^3 + ^3^+ ^ 1^2)
ri rg rg rir2r3 rrjrars
The numerator of this fraction from (14) is rs^ = S8; the denominator is
(from (7), viii) S^;
.'. the expression is —-,= -= - .
S^ S r
19. By (vii) E. T. p. 235, JJi^asecJ^, but ri-r = IJi sin J^;
.*. — = tan i^ , similarly Ar-— = tan J B ;
/. their sum= tan J/1 + tan J B = sin J (^ + £) sec J^ . sec ^B
= cos J sec i ^ sec J-B = — , by (x) p. 235.
TRIANGLES AND CIRCLES. LXXIII. 181
20. By (x) p. 235, ri = 4R sin^A cos^B coa^G]
•'• ri + r2 + r3-r = 4E {sin ^^ cos J B cos JC + siniB cos J C cos 4^1
+ sin J C cos ^A cos^B -sin^A sin^B sin^C}
= 4R8ini(A + B + C) = 4:R.
21. From (7, I) vfe ohia.m ^ = 8 = ^J{s{s- a) {s-b){8- c)\
.'. ahc = 4R,J{s{s-a){s-b){s-c)}.
Now r = S-7'8 = J{s{s-a)(s-h)(s-c)\^s; /. ahcr = 4:R(s - a)(s - h) {s - c).
22. By (vii) p. 235, I/i = a sec4^ = 2i^ sin^ seci^ = 4Esin J^.
Similarly the other distances are iR sin J B, 4i2 sin J C.
23. By 17, r2 + r3 = acot4^=a; when J^=45°.
24. Each of the angles of the equilateral triangle is 60°; let each of
the sides be a,
^=2^ = 2sin60°=;^' /.SE^aVS,
s 43<x 4 3a 6^ ^
S J a2 gin 60° a'^jS 2 ajS _ ,.,
^ s-a ia 4 a 2
.-. 3E = 6r = 2ri.
r^ _ acos Ji?cos4Csec J^ _ 1 sin J. 4
Fc ~ he ~ 4R * smyS^n iC '
pL + ^ + !a = ^^ . , , ^ , p ■ ,^ {sin2 J^ + sinH^+ sin2A C)
he ca ah 4E sin J^ sin J5 sin JO * ^ 2 -r ^ r
l -2sin4^sin4Bsin4( 7 _^
= Tiisini^sini^siniC t^'^- ^^' P* ^^^'^
^ 111 [ix. p. 235.]
25.
4ii sin J^ sin JZ? sin J 2R r 2R'
EXAMPLES. LXXIV. Page 244.
1. The perimeter of a regular polygon inscribed in a circle of radius r
is 2nr sin - ; therefore the ratio of the perimeter of the polygon to the
n
diameter of the circumscribing circle is — - sin ~ = n sin - .
2r n n
When?i = 4, ratio is 4 sin J 7r = 4 sin 45° = 2^/2 = 2 x 1-4142 = 2-8284.
When 72 = 6, ratio is 6 sin|7r = 6sin 30° = 3.
When n = 8, ratio is 8 sin | tt = 8 sin 22^°.
182 REGULAR POLYGONS. LXXIV.
Now sin22i°=:i^(l-cos45°) = i^(2-v^2); ■
.-. ratio is 4 ^(2 - ^/2) = 4 V5858... =3-06148...
When n = 10, the ratio is 10 sin j\7r; that is, 10 sin 18°.
But sin 18° = J (^5 - 1). [Art. 93, Ex. XLII. 6].
Therefore the required ratio is \o. (^5 _ i) ^ 5 ^ 1-23606 = 3-09015.
When w=12, ratio is 12 sin ^^ir^Vi sinl5° = 12 x ^^^-^
= 3^2 y3 - 1) = 3 X 1-4142 X -732 = 3-10558.
When n=20, the ratio is 20 sin ^^ tt = 20 sin 9°, sin 9° may be thus found.
From E . T. p. 150, sin 9° + cos 9° = V(l + sin 18°) = W(3 + V^),
sin 9° - cos 9°= - ^[l - sin 18°) = - 1 ^(5 - ^5) ;
- .-. sin9° = i{^/(3 + ^/5)-^/(5-^/5)};
.-. the required ratio is 5 {sj i^ + ij^) - ij {0 - ^Jb)]
= 5 {iV10 + v/2-V(5- 2-23608)} = 5 (2-28825 -1-6625) = 3-1287...
When ?i=S0°, the ratio is 60 sin ^7r = 60 sin 3°,
sin 3° = sin (18° - 15°) = sin 18° cos le5° - cos 18° sin 15°.
Now cos 18° = sjl - sin^ 18° = J ^^(10 + 2 ^Ib) ;
.-. the required ratio is ^^ {(V^ - 1) (/v/3 + 1) - ^10 + 2^5(^3 - 1)}
^^ {1.236 X 2-732 - 3-8 x -732} = VV2 x -595...
= J X 15 X -595 X ;^2 = 15 X -595 X -353 = 3-1405.
2. Area of a polygon described about a circle is nr^ tan J tt
= 12xlxtanl5° = 12x'2679... = 3-215 nearly. [E. T. p. 120.]
3. Let r be the radius of the circle.
Then the area of the square described about it =4r2tan45° = 4r2.
The area of dodecagon inscribed in it
= 12r2sin 15° cos 15°=12r2 . ^^-"^ . ^±i = ^ 7-2 = 37-2.
4r2 4 .
The ratio of the areas is 5-» = „ •
4. (i) The perimeter of the described polygon is 27irtan-, 71=100,
18°
2r = 1 ; .-. the perimeter = 100 tan — = 100 x tan 1° 48'
= 100 X -0314263... =3-14263....
REGULAR POLYGONS. LXXIV. 183
(ii) The perimeter of the inscribed polygon is 27irsin-, 71 = 100,
18°
2r = 1 ; .-. the perimeter = 100 sin — - = 100 x sin 1° 48'
= 100 X -0314108... =3-14108... .
5. Let i\ be the radius of the circle inscribed in the equilateral triangle,
then the perimeter of the triangle is Qr^ tan J 7r = 6ri^3.
Let Tg be the radius of the circle inscribed in the regular hexagon, then
the perimeter of the hexagon is 12r^ tan lir — ir^^sj^.
Now by the question 6ri;^3 = 4r2x/3; .'. S)\=:2r^,
therefore r^ : rg :: 2 : 3,
i.e. the radii of the circles (and therefore their diameters) are in the ratio
of 2 : 3 ; but circles are to one another as the squares on their diameters,
therefore the given circles are to one another as 2'^ : 3^, i.e. as 4 : 9.
6. In the figure E.T. p. 242, bisect the angle at H by OH and the
angle at K by OK ; produce HO, KO to meet at ; it may then be shewn
that straight lines drawn from to the other angular points of the polygon
bisect those angles. Therefore OH, OK... are all equal, and the polygon is
divided into as many isosceles triangles as it has sides. Let n be the number
of sides of the polygon, then the angle HOK= — and the area of the polygon
is n times the area of the triangle HOK. In the isosceles triangle HOK
draw the perpendicular 03/, then OM bisects the angle HOK and the base
BK; .'. lHOM='^ and OM=ifil/cot'^ = ^ cot- .
n n 2 n
The area of the triangle HOK is OM x H3I= ^ . ^ cot - .
2 2 n
^, .... . a a ^TT na^ .180°
.*. the area of the polygon is n . - . - cot - = -:- . cot — -- .
7. Let a be the side of the pentagon, then area
.5^\eot36°=^«^ '-'^'
4 • 4 V(10-2Vo)*
Let b be the side of the decagon, then area
= -j-cotl8--2-.---^3^- .
Now ^' l + v/5 ^56^ ^/(104-2V5)
4 VaO-2V5) 2 • ^5-1 •
a2(V5 + l)(V5 - l) = 2i>2 ^(10 + 2^5)4/(10 - 2^5)
2a^=b^^S0, a' = b^^20, .: a = b */20;
. a_4/20
" b 1 '
184 REGULAR POLYGONS. LXXIV.
8. In fig. E. T. p. 242, OH=R and OM=r, OH=HM cosecHOMy
.-. JR = ^cosec-; .-. 2R = acoseG-, OM=HM cot MOH:
2 n n
.'. ?' = -cot— ; .*. 2r=acot-.
2 n n
JR + r=^ ( cosec- + cot- ) = ha . ( 1 + cos- Wsi
2 \ n nj ^ \ nj
sin--
n
a
= 2 cot 2^. [Art. 165.]
10. Let r be the radius of the circle, then from E. T. p. 283,
Side of Pentagon = 2r sin i tt = J ?' v/(10 - 2 ^5).
Side of Hexagon = 2r sin J tt = r.
Side of Decagon = 2r sin ^^^ tt = J r (^^5 - 1) .
If the triangle formed from one side each of these polygons is right-
angled, then the sum of the squares on the sides of the hexagon and the
decagon is equal to the square on the side of the pentagon.
Now r2+{ir(V5-l)}2=:r2 + ir2(V5-l)2 = Jr2(10-2v'5). q.e.d.
11. Let a, 5, c, <i ... be the n sides of an irregular polygon described about
a circle of radius r; then the polygon can be divided into n triangles of
which the areas are \ar^ ^br, \cr, ^dr,... Therefore the area of the polygon
18 ^r (a + h + c + d...), i.e. the product of the radius and half the perimeter of
the polygon.
12. Let ABODE F be an irregular polygon of an even number of sides
described about the circle HIKLMN, so that the sides AB, BC, CD, DE,
EFj FA touch the circle on the points H, I, K, L, M, N respectively. Let r
be the radius of the circle. Then from (11), the area of the polygon is
ir(AB-\-BC-^CD + DE-i-EF-^FA)
= ir{(AH+HN)-\-(BH+BI) + (CI+CK)
+ (DK+DL) + {EL + EM) + (FM + FN)}.
It may be proved by means of Euclid III. 37 that
AH=HN, BH=BI, CI=CK, DK=DL, EL = EM, FM=^FN;
,', the area of the polygon
= ir{2AH + 2BH + 2CK+2DK+2EM+2FM}
= r{AH + BH+CK+DK-hEM+FM}
=: {AB + CD + EF}.
In the same way if we begin with BC instead of AB we can prove that
the area of the polygon = r {BC + DE + FA} ; and so on for every other side;
therefore the area of the polygon is radius x the sum of every alternate side.
13. Area of a circle =:7rr2;
.*. area covered by the dome = 7r (54)2 = 3-1416 ^ ^54^2 gq f^^^
= 3-1416 X 824 = 1017-8784
^ 3-018 sq. yards,
REGULAR POLYGONS. LXXIV. 185
14. Area of the circle = Trr^ = 1 acre = 4840 sq. yards ;
, 4840 4840x7 ...^
.-. r=V1540 = 39-25...yds.
15. The area of the base of the cylinder is the length of the paper x its
thickness = 300 x 3 x 12 x yj^ — 72 sq. inches.
If r = radius of cylinder, 7rr2 = area of its base, and 2?-=: its diameter;
.-. irr'^— 72 sq. inches,
47rr2=:288 sq. inches,
OQQ
4r2 = ^^ = 91-673...sq. in.,
w
2r=,^91-673... inches = 9-575 inches.
16. Let I be the length of the carpet in feet.
Then area of the base is V- 1 sq. inches ; the area is also ?-2 where 2r is
diameter = 1447r sq. inches;
.-. J^Z=1447r; .-. Z = 967r = 96 x 31416 = 301-6 feet.
EXAMPLES. LXXV. Page 252.
. 27r
sm —
_ n ^5 . 27r n 2Tr n
2 n 27r n ^ir
n
. 27r
sin —
When n=:oo , — =0, and -j--l=l; [E. T. p. 247.]
n
:. the limit when n = oo of - i^^ gj^ — — -n-R^ when 9i = oo .
2 n
.IT . IT
tan - sin — .
2. wr* tan - = r^ .w . - tan - = ttH = irr^ — . — .
n T 71 W TT TT
— cos -
n n n
IT
sin —
When ?* = 00 , = ; .-. — =1 and — =1 ;
n TT TT
- cos -
n n
:. nr^ tan - = wr^ when n = op •
n
186 LIMITS. LXXV.
3. The circular measure of 10" is jgo^^eO = 6^0 "
TT a"
Now sin ^ < ^ ; /. sin 10" is less than ^.„„^ sin ^ > ^ - -r ;
64800 4
/. sin 10" is greater than ^^^ - ^ (
64800/
If 7r=:3-141592653589793 then ^-;t^^v^ = '00004848136811 ; sine 10" is
o4oUU
therefore less than this decimal. But 7^:o7^/^ is less than '00005, therefore
64800
a fortiori sine 10" is greater than -00004848136811 -^ J (-00005)3 ; i.e.
sin 10" is greater than -000048481368078...
The two decimals between which sin 10" lies correspond in their first
thirteen figures, therefore we have
sin 10" - -0000434813681
4. 2 sin (72° + ^) - 2 sin (72° - A)
= 2 {sin (72° + ^) - sin (72° - ^)}
= 4 sin ^ cos 72° = 4 sin A sin 18°
= {J5 - 1) sin A, [E. T. p. 59.]
2 sin (36° + ^) - 2 sin (36° - ^ )
= 2 {sin (36° + ^) - sin (36°- ^)}
= 4 sin ^ cos 36° = 4 sin ^ sin 54°
= (;^5 + l)sin^.
5. Fig. E. T. p. 251, rp2 = 2 . RO . TR nearly.
Let x = TP, ■^^% mile = TR,
x^ = 7914 X J^V% = 225 nearly ;
.-. a; = 15 miles nearly.
6. From E. T. Art. 300, TP^ = 2TR . RO;
/. TR =
2.R0'
TP TM
From Euclid VI. 8, ^ = y^= -025 ;
TP^ AAA^oK A TP^ -000625x3957 .,
.-. — = -000625, and ^-^-^ -^ ^ ^^^"^'
i.e. Ti? = -000625 x 3957 x 2640 feet
= 6530 feet nearly.
MISCELLANEOUS EXAMPLES. 187
EXAMPLES. LXXVI. a. Page 253.
1. The proof of Art. 107 is true for each of the four figures on page 96.
tan^=f; /. ^(l + tan2^) = .|; .-. cos^=|;
2. When ^ = 0, cos ^ = 1 and sec ^ = 1, .*. cos ^- sec ^ = 0, let X represent
cos 6 - sec 0,
As increases from to Jtt cos diminishes to and .*. sec increases
from 1 to 00 .
.•. X is negative and increases numerically from to oo .
As increases from Jtt to ir cos^ increases numerically from to 1 and
is negative.
.'. sec decreases numerically from oo to 1 and is negative ;
.*. X is positive and decreases from oo to 1.
3. See T. p. 104, (2 - sec ^ ) sin ^ = ;
.*. sin^ = and A=7ixlS0°, or cosA = j^ and ^ = 2n x ISO"" ± 60°.
4. (1) sin (A+B) . sin {A~B)=: (sin A .cos B-\- cos A . sin B)
X (sin a . cos B - cos A . sin B)
= sin^^ . cos^l? - cos^^ . sin^JB
= sin^A . (1 - sin^B) - (1 - sin2^) . sin^J? = sin^^ - sin^^.
sin A + sinB _ 2 sin J (a + JB) . cos i(A-B)_ tan J (a + B)
(2)
sin ^- sin B 2coai^(a + B) ,8m^(a-B) tan|(a-B)
5. cosM-cosJ.cos(60° + ^) + sin2(30°-^)
= coss ^ - cos ^ . sin (30° - ^) + sin^ (30° - ^ )
[since cos (60° + .4) = sin {90°- (60° + .-!)}]
= cos^ A - cos A X (J cos A -^^3 sin A )
+ J cos2^ - J^3 sin ^ . cos ^ + J sin2^
= cosMx(l-i + J-J)+| = |.
6. Greatest angle ■-= 78° 14', greatest side = 2183 . sin 78° 14'— sin 30° 22' ;
log (greatest side) = log 2183 + L sin 78° 14' - L sin 30° 22'
= 3-3390537 + 9-9907760 - 9-7037480 = 3-6260817 :
by the rule of Proportional Parts, d = -0001 x ^\ = -0000815 ;
.-. -6260817 = log (4 -2274 + -0000815)= log 4-2274815.
Hence greatest side = 4227-4815.
7. See Examples.XVI. 1.
8. See T. pp. 106, 107.
188 MISCELLANEOUS EXAMPLES. LXXVI. a.
9. sin30° = i, sin60° = W^» sin 90°= 1, sinl20° = sin 60°=:4V3»
sin 150° = sin 180° - 150° = sin 30° = J,
sin 180° = sin 7 x 30° = sin 210°= + sin (180° + 30°) = - sin 30°= - J,
sin 240°= sin 180° + 60°= -sin60°= - W^,
sin 270° = sin (180° + 90°)= -sin 90°= -1,
sin 300°= sin (360° - 300°) = - sin 60°= - 1^^,
sin 330°= sin (360° - 330°) = - sin 30° = - J.
2 sin^^ 1 - cos 2A
10. t&n^A = n-7 = ^ pTi •
^^' 2cos^^ l + cos2^
11. tan ^+^(l + tan^^) = 2, .-. tan^ = j, see 1 above.
When sin^ = |, then tan^=^ and sec^ = |^, the ratios are all positive
when A is less than 90°.
12. Let a, b, 1035*43 be the lengths of the three sides,
a = 1035-43 . sin 44°-^sin 70°, 6 = 1035-43 . sin 66°-^sin70°.
log a = log 1035-43 + X sin 44° - L sin 70°
= 3 -0151212 + 9-8417713 - 9 -9729858 = 2 -8839067
= log765-432; .-. a= 765-432 feet,
log & = log 1035-43 + L sin 66° -L sin 70°
= 3-0151212 + 9-9607302 - 9-9729858 = 3-0028656
=log 1006-6; .-. 6= 1006-6 ft.
13. See T. Ex. XVI. 2.
14. See T. pages 104, 106.
15. i3inS0°=^, ta.n60'' = ^S, tan 90°= re ^
tan 120° = tan (180° - 60°) = - tan 60° = - ^3,
tan 150° = tan (180° - 30°) = - tan 30° = - -j^, tan 180° = 0,
tan 210° = tan (180° + 30°) = tan 30°=-.- ,
tan 240° = tan (180° + 60°) = tan 60° = ^S,
tan 270° = oo , tan (360° - 60°) = - tan 60° = - ^3,
tan (360° - 30°) = - tan 30° = - 4v .
16. tan^+J(l + tan2^) = 3, .-. tan^ = |.
. _ tan ^ _ 4
When sin ^ = J then by 1 tan A = ^ and sec ^ = | , the ratios are all positive
when A is less than 90°.
MISCELLANEOUS EXAMPLES. LXXVI. a. 189
17. Let a =193, 6 = 194, c = 195.
2
^'^^" 194 X 195 ^{^(^" "^H^ -b){s- c)]
^ ^{291x96x97x98}
194 X 195
2
194 X 195
2
■ 194 X 195
Q.v/{3x 972x22x48x7'^}
^{32x972x22x42x72}
x3x97x2x4x7= ^^=^861538 ;
~ 97 X 2 X 39 X 5 ~ 65
/. A = sin-i . 86154 = 59° 29' 23''.
SimUarly it may be she;\rn that B = 59° 59' 23", /. G = 60° 31' 14".
18. (1) cosf ^ =cosyi cos^^ -sin^ sin^^
= cos^ cos J^ -2cos J^sin*J^
= cos ^ cos J ^ - cos i ^ (1 - cos ^ ).
(2) cos^-cos(^ + 5) = 2sin(^4-i5)sinJ5
= 2 sin ^cos JSsin J5 + 2cos^sin2J5
= sin sin 5 + cos sin 5 tan J 5.
19. See T. Art. 107, sec A = l-r-^/(l - sin2^) = i^|.
20. The expression = - "^ ^^ ^ = - tan (0 + 45°) ;
X — tan u
.'. the values of the expression are the same as those of tan a as a changes
from f TT to Itt.
21. Write out Art. 148 putting cot for tan 0j
tan2^ = l, tan^==tl, 0=n7r^iir.
22. See T. p. 118,
cos 5a = cos (4a + a) = cos 4a . cos a - sin 4a . sin a
= (2 cos22a - 1) cos a — 2 sin 2a . cos 2a . sin a
= [2 (2 cos2 a - 1)2 - 1] cos a - 4 sin a . cos a x (2 cos2 a - 1) . sin a
= [2 X (4 cos* a - 4 cos2 a + 1) - 1] cos a -- 4 sin* a . 2 cos' a + 4 sin2 a cos a
= 8 cos^ a - 8 cos^ a + cos a - 4 (1 - cos2 a) 2 cos' a + 4 (1 - cos2 a) cos a
= 8 cos^ a - 8 cos' a + cos a - 12 cos' a + 8 cos^ a + 2 cos a - 4 cos' a
^16 cos^ a - 20 cos' a + 5 cos a.
23. See Art. 179. If A lies between 540° and 630°, J a lies between 270°
and 315°,
sin 4 a is negative and is greater in magnitude than cos^a which is
positive ;
/. sin Ja + cos Ja = -/^(l + 8in^)
sin i a - cos J a = -fj(l - sin a)
= 2sin Ja= - ,^{1 + Bin A) - fj{l - Bin A).
190 MISCELLANEOUS EXAMPLES. LXXVI. a.
24. Length of string=;^(144 + 25)in. = 13in.,
the points of suspension of the ring are the angular points of a regular
hexagon inscribed in the ring ;
/. the side = 5 inches,
(13)2 + (13)2 -52 ^313
cos of angle required = -
2 X 13 X 13 338 '
25. 1° = one-ninetieth part of a right angle, 180°= tt radians;
• • ^°~ Tfto ^ '"^ radians = - — — . x A radians,
. \. .. 180° 18x7° .,^
.lradian = .-^=-^ = 5A°.
26. sin^=V(l-A) = l; tRnA = i-~i = i,
27. See the Figuree in T. p. 110. Let P^OM^ = A ; then in Fig. I.
P^OR = 180° -A; in Fig. II. P^OR = 180° + A ,
and the two triangles P^OM^ are equal in all respects and OM^ is of the
same sign in each ;
.-. cos (180° -A) = cos (180° + A),
In the figures in T. pages 104, 107 ROP' on p. 104 = 180°-^, ROP' on
p. 107 = 90° + ^ and OM'=ON\ P'M' = P'N'.
In figure on p. 104 OP' starts from OL and revolves in the negative
direction.
In the figure on p. 107 Oi' starts from O^and revolves in the positive
direction.
When OP on p. 104 crosses 0C7 on p. 107 it crosses OL; hence
M'P' on p. 104 always equal -JV^'P' on p. 107,
and - OM' on p. 104 always equals ON' on p. 107 ;
OM' ON'
•*• WP' = WP''^ •'■ «ot(180°-^) = tan(90°-^).
28. sin X (2 cos a: - 1) = 2 sin J x cos \x (4 cos^ ^ .r - 3)
= 2sin Jo; (4 cos^Jx - 3 cos ja:).
2 sin ^ - sin 2^ _ 2 sin ^ (1 - cos 6) _ 2&in2i (9_ ^^
^^' 2 sin e + sin 2^ ~" 2Tin'^(l + cos e)~2 cos^f 6> ~ ^^ ^ *
•*• 1^—- — ;r— -T^-A-;7 lias the same changes as tan-A^ and is always positive.
2sin^ + Bm2^ ^ '^ ^
As d changes from to tt tan^J^ changes from to 00 ,
TT to 27r 00 to 0.
30. 2tc cos 60° = 62^c2 - a2 = (ft + c + a) (6 + c- a) -26c,
.-. &c = (6-fic + a)(6 + c-a)-26c, or (6 + c + a)(& + c-a)=36c.
MISCELLANEOUS EXAMPLES. LXXVI. a. 191
a
31. T radians = 2 right angles, radians = - x 2 right angles,
TT
i radian = Jtt X 2 right angles = J x /^ x 180" = ^^ of 180°= 19^^
32. Let tan a = a:, tan/3=:y,
. , ^. tan a + tan /3 . ^ x i tan a + tan /3
tan (a + /3) = , — i~—r^-E » a + ^ = tan-i - — --^—- ,
^' 1 - tan a . tan /3 ^ 1- tan a . tan /3
or tan~^a; + tan~^i/ = tan~^ - — — .
1-xy
33. sin a; (2 cos a; + 1) = 2 sin J a; cos i a; {2 (1 -2 sin2Ja;) + l}
= 2 cos ^ a; (3 sin \x-^ sin^ Jx).
sin ^ + 2 sin J ^ _ sin ^ ^ cos J ^ + sin ^ ^ _ cos^^ + l
■ sin ^ - 2 sin J ^ ~ sin ^ ^ cos J ^ - sin \B~ cos J ^ — 1
= - -.-5T-^= -COt^J^.
Hence the expression is always negative and the numerical changes are the
squares of the changes of the cot of J ^. As ^ changes from ^ to 27r J ^
changes from to J tt and cot changes from oo to 0.
35. See T. Ex. 3, p. 192, and Ex. 19, p. 194. The equation may be
written
(4 cos 4^1 cos 45 cos 4 C) (4 sin J ^ sin JJ5cos4C)
=:12sin J^ cos J^ sin J^cos Ji?;
.-. cosmic = I, .-. cosJC = W3> .-.iO^SO".
7. sin 18°
36. ^='^i^r°=="^
sm 36° 2 cos 18°
37. See Art. 39. The minute hand has described (3x4 + 2) right angles
+ f right angle between twelve o'clock and 20 minutes to four or 14| right
angles.
38. If ^ is greater than 90° and less than 180°, cos A is negative,
cos A=- J{1 - sinM) = _ ^(1 - 1) = - f ^2.
39. cos ^ + cos 2^ = 0,
.-. 2cos2^ + cos^-l = 0; .'. (2cos ^ - l)(cos<9 + 1):^0 ;
.*. either cos ^ =- 1 ; whence ^ = (2n + l)7r,
or cos ^ = + 4 ; whence = 2mr ± } ir.
40. When a cos A = b cos B, then
,^ (62 + C2 - a2) = A (c2 + a2 _ 12) .
be ' ca^
... a2 (62 + c2 - a^) = 62 (^.2 + ^2 _ yi^^^ . ^2^2 _ ^4 ^ ^,2^2 ^ 54,
c2 (a2 - 62) _ (a2 _ 62) (^2 + 62) = ; .'. (a' - 62) (c^ -a^- 6^) = ;
/. either a2 = 62 or c2 = a2 + 62. q.e. d.
1^2 MISCELLANEOUS EXAMPLES. LXXVI. a.
41. Let A and B be the sides opposite 1 ft. and ^3 ft. respectively,
then sinB = sin30°x^3 = i;^3; .-.5 = 60° and C = 90°;
.-. the sides are 1, ;^3 and 2.
42. Let A = the greatest angle,
Logtan J^ = 10 + i{logl0-log3-log2}
= 10 + 1 {1-4771213 - -3010300}
= 10 + -1109243 - L 52° 14' 19-5".
^ = 104° 28' 39".
43. See Arts. 38, 39. The minute hand describes an angle of (4 + 3^
right angles or 630° between half past four and a quarter past six.
44. If A is between 180° and 270°, sin A is negative,
i ^ 1
.'. sinJ[= - - —
45. (i) sin 2 J[ = 2 sin A . cos A
2cos^ . „ , 2cot^ 2cot^
sin^ * cosec'-^^ l + cot*-*^ *
(ii) This is the same as Example 3, p. 192, for 2^ + 2^ + 20 = 180°
46. sin(? + sin2<9 = 0, sin ^(1 + 2 cos ^) = 0;
.-. sin^ = 0; whence, 6 = mr,
or cos^= -J; whence, ^ = (2?i + l)7r± J tt.
47. When 6 cos A = aco8B, then
^ (62 + c2 - a') = ^- (c2 + a2 - 62)
bc^ ' ac
.'. h^ + c^-a^ = c^ + a^-h^ or c2-a2 = 0; .'.c = a. q.e.d.
48. Let A be the angle opposite the side ^^2 ft.,
then sin^=^2sin30° = i^2; .-. ^ = 45°, or 135°.
5 = 30° and (7 = 105° or 15°,
also c = a cos 5 + 6 cos ^ = i X ^/2 X ^3 + J X ^2 = J {J2 + ;^6) .
180°
49. (1) A TT radians = ^V^ x = 90°.
TT
1 80° 7
(2) 5radians^5x~- = --x450° = 286°2r49"...
TT 11
In the third case the unit is f of 360°,
45° = 45 -f (f of 360) of | of 360° = | of the unit.
50. (sin 30° + cos 30°) (sin 60° - cos 60°)
= (4 + iV3)(W3-4)=|-i = i-sin30°.
MISCELLANEOUS EXAMPLES. LXXVI. a. 193
51. (1) cos2(a + ^)-sin2a = cos2a.cos2^ + sin2a.sin2/3
- 2 sin a . cos a . sin /3 . cos p ~ sin^ a
= (1 -sin^a) cos2^ + sin2a(l--cos2^) -2 sin a. cos a . sin/3, cos^
= cos2j8 - 2 sin^a . cos^jS - 2 sin a . cos a . sin ^ . cos j3
= cos p (cos ^ - 2 sin^ a . cos /3 - sin 2a . sin (3)
= cos /3 (cos p cos 2a - sin 2a . sin jS)
=:cos/3.cos(2a4-j8).
,^. ^ ^ ^ , - cos a cos ia
(2) 1 + cota. cotia = l+-; . - — ~
sm a sin ^ a
1 f . cos a. cos i a) cos A a T . , cos a)
= -. — -^sina+ ^.— i — ^h = . - -^2sm^a+ . , V
sin a I sin J a j sma ( smjaj
cos Ja |2sin-ia + l -2 sin^Jal _ 1 cosja
{-
sin ^ a j sin a ' sin | a '
62. (1) 5tan2a;-sec2a;=:ll,
/. 5tan2a;-(l + tan2a;)=:ll; /. 4tan2a; = 12.
tan x= ^ ^3 ; whence, x = mr^'^ir.
(2) sin5^-sin3^ = ;^2.cos4^;
/. 2cos4^.sini?=;^2.cos46/. cos4^ (2sin (9 - ,y2) = 0;
/. cos 4^ = 0; whence, 4^ = (27i±l) Jtt,
sin^ = J^2; whence, ^ = ?i7r + (- 1)»* J tt.
53. Area = i X 10 x 15 x sin 30° sq. ft. = J x 150 sq. ft. = 37^ sq. ft.
54. In this case (by Proportional parts) d = g^\\ of 60" = 21 -2",
sin-i (0-649300) = 40° 29' 21-2".
55. (1) 10' = ^^^ X ^ right angle = ^^^ of Jtt radians
= shj o^ V" i'adians = -0026... radians.
(2) J of a right angle =:yV o^ ^ radians = -314... radians.
In the third case the unit is ^ (5 x 180° - 360°) - 108°,
a right angle = -^%% of 108° = f of the unit.
56. cosa = f; .-. sina= ±f;y/48; .*. tan a= i ^^48= ±4 ^3,
cosec a = ± 7^J4tS = ± j^2 V^-
57. (1) cos2(a-^)~sin2(a + i3)
= cos2a.cos2^4-sin2a. sin2^ + 2cosa . cos^. sin a . sin/3
— sin^a . cos^/S-cos^a . sin^^S- 2 cos a . cos/3, sin a .sin/8
= C082a . cos2/3+ (1 - cos^a) (1 - cos^^) - (1 - cos^a) cos2/3 - cos^a (1 - cos^^)
= 4 cos2 a . cos2/3 + 1-2 cos^ a - 2 cos^ /3
= (2 cos^a - 1)(2 cos2/3 - 1) = cos 2a . cos 2^3.
(2) 1 — tan a tan J a = (cos a - sin a tan J a) sec a
= { 1 - 2 sin^ J a + 2 sin' J a) sec a = sec a.
L. T. K. 13
194 MISCELLANEOUS EXAMPLES. LXXVI. a.
58. (1) 5tan2a; + l + tan2:r=-7, /. 6tan2a: = 6;
.*. tan^— ±1, whence x — mr^^w.
(2) 2 cos 4:0 coQ 6 = ^2 cos 4:6, .\ cos ^ = J ^2; or, cos 4^ = 0,
whence d = 2mr^i7r; or, 4^ = 2n7r±j7r.
59. Area-V(7x4x2xl) = 2;^14 = 7-478... sq.ft.
60. Here D - ^|f| of 60" = 32-2^
.-. sin-i (0 -621500) = 38° 25' 32-2".
61. 76g= -76 right angles,
l-2«=(l-2 X 2-f-7r) right angles = l-2 x ^^ right angles
= -763 . . . right angles ;
.-. l-2« is greater than 76^.
62. See Arts. 92, 90. We have that 2 sin ^ = sin B + cos B,
.'. l~2sinM=l- J(sin5 + cosJ5)2=i (sin B-cos£)2;
.♦. cos2^=cos2(B + 45°).
63. (1) tan2^-sinM=tan2^(l-sin2^-^tan2^)
= tan2.4 (l-cos2^) = tan2^ .sin2j.
/o\ 4. 4 4. o ^ COS A cos2yl
(2) cot A - cot 2A = - — 7 r-^—
sm A sm 2 A
2cos2^-cos2^
sm2A sin 2 A
sin {x + Sy) + s m {Sx -^y) __2 sin2 {x + y) . cos (x - y)
sin 2x + sin 2y ~ 2sm.[x + y) cos {x - y)
sin (a; + 1/)
64. See Art. 178. When A = 240°, IA = 120° and then sin J J[ is greater
than cos J A and is positive ;
.-. sini^ + cosJ^= +^(l + sin^), sin J^ -cos J^= +>y(l -sin^).
Hence the formula is true.
65. See Art. 154.
cos (^ + B) = sin {90 - (^ +B)} = sin {(90 - ^) + (- JB)}
= sin(90°-^).cos(-5) + cos(90-^). sin( -£)
= cos ^ . cos ^ - sin ^ . sin B.
[Since sin (-i^) = - sin J5 and cos ( - j5)^cosB.
sin A . cos {B + G)~ sin B . cos {A + C)
= sin ^ . cos B . cos C - sin ^ . sin J5 . sin (7 - sin J5 . cos A . cos G
+ sin B . sin ^ . sin C
=006 C (sin ^ , cos ^ -- sin I^ . cos^)=cos G , sm(A-'B).
MISCELLANEOUS EXAMPLES. LXXVI. a. 195
66. Let a, b be the two sides,
^-_ sin 70° 30' - ,„ sin 78° 10'
" = ^^"^5^3^20" ^ = ^^'^^i5^3]F20"
log a=:log 102 + L sin 70° 30' - L sin 31° 20'.
log c = 2-009 + 9-974 -9-716 = 2-267 = log 185.
log b = log 102 + L sin 78° 10' - L sin 31° 20'
= 2-009 + 9-990 - 9-716 = 2-283 = log 192.
Area of triangle = J (192 x 185) . sin 31° 20' sq. ft.
= 17760 X -520016 sq. ft. = 9235-48416 sq. ft.
= 1026-lG49sq.yds.
67. 2-3« = (2'3-^7r) of 180° = 2-3 x^'^ of 180° = 131t8t°,
.-. 2-3<' is the greater.
68. (1) cotM-cos2^ = cot2.4(l-cos2.4tan2^) = cot2^(l-sin2^).
(2) tan A + cot 2 A = ^^^^ + ^^^-f = cosec 2 A (2 sin2^ + cos 2 A )
sm 2A sm 2A ^
= cosec 2^.
^ 2sin2(^-^ ) Bin(x + y) ^^^.^
^ ^ 2 sm (a; + 2/) COS (a; -2/) ^ iff
69. See Art. 178. When ^^ = 150°, cosj^ is greater than sin^A and
is negative ;
.*. sin J^ + cos J^ = -;^(l + sin^), sin J^ -cos J^= +;^(1 -sin^).
70. sin 30° + sin 120° = J + 4^3 = ^2 . cos 15°.
71. (1) l + cos^ + sin^=2cos2j^ + 2cosJyl sini^
= 2cos J^ (cosi^ + sin^^)
= V{4 cos2i^ (1 + 2 cos iA sin J^)}
= ^{2(l + cos^)(l + sin^)}.
1 sin J
(2) cosec 2^ =
2 sin A cos A 2 sin2^ cos A
cosec2 A cosec2 A
2cot^ 2^(cosec2^-l)*
(3) sin f TT + sin ^ TT - sin f «• = 2 sin I TT cos f TT - 2 sin I TT cos ^ TT
= 2 sin f TT (cos I TT - cos f tt) = 2 sin f TT sin I TT sin f tt,
and sinf 7r = sin(7r-|7r) = sin^7r.
72. I^et ^, 5 be the angles opposite the sides whose lengths are 185,
192 feet respectively.
sin^ = sin31°20'x^§f ;
.-. L sin ^ = L sin 31° 20' + log 185 - log 102
^9-716 + 2-267-2-009 = 9-974
= i sin 70° 30', ^ = 70° 30'.
13—2
196 MISCELLANEOUS EXAMPLES. LXXVI. a.
L sin B = L sin 31° 20' + log 192 - log 102
= 9-716 + 2-283 -2 -009 = 9-990
= L sin 78° 10'; .-.5 = 78° 10'.
The area = J (192 x 185) sin 31° 21' sq. ft.
= 17760 X -520016 sq. ft. = 9235-48416... sq. ft.
73. Let the angles be a - 2j3, a -ft a, a + ft a + 2ft
Their sum namely, 5a = 27r ; .-. a = | tt, also a + 2^ = 6 times (a - 2^).
.-. 14^=5a; .'. p = -f^ of f of 7r = |7r; .-. a + 2/3=(| + |)7r; etc.
74. See Art. 75 and Art. 141, Ex. 4,
sin (180° + ^) = sin (90° + 90° + ^) = cos (90° + ^)= -sin^,
cos (180° + ^) = cos (90° + 90° + ^)= -sin(90°+J)= -cos^.
75. (1) See Art. 107. cotM = -^, = ^^^- =^-,-1.
(2) cotM + cot2^ = (cosec2^ - 1)2 + cosec2^ - 1
= cosec^yl -2cosec2^ + l + cosec2J -1.
When ^ = 30° the above statement becomes (^3)* + (^3)2 = 2-1- 2^,
that is 9 + 3 = 16-4; which is true.
76. See Art. 147. 2 cos^ - coss 6 = 0; .: cos2 (9 (2 cos (9 - 1) = ;
.-. either, cos^ = 0; whence, d — nir + ^Tr^
or cos ^ = J ; whence, = 2mr ± J tt.
77. sin2jB = sin.4 cos^,
.-. l-2sin2^=sin2^-2 sin A cos^+cos2.4^(cos^ -sinA)^:
.-. cos2B = 2cos2(^ + 45°).
78. See 48 above. Let the two triangles he ABC j^, ABC^ in Fig. iii.
on p. 216. Then c = ;^3, 6 = 1, &2 = ^2 + a2 - 2ca cos 30° ;
... l = 3 + a2-3a; .-. a = i(V3 + l) or i(s/S-l);
.'. BC^ = i i^S + l), BC^^i(s/S-l),
and the areas of the triangles are in the ratio BG^^i BC.^.
79. Let the angles subtended by each be
a, a + /3, a + 2/3, a + 3/9, a + 4ft a + 5/3 respectively,
then a + 5/3 = 6a; whence, a = p; also 6a + 15/3 = 27r;
.-. a = /3 = T2T^ = T*iV radians.
80. Art. 75. See Ex. 4, p. 107.
tan (90° + 90° + ^)= - cot (90° + ^) = tan J .
81. See Art. 148. sec36>-2 (tan2 6/ + l) = 0; .-. sec^i?- 2sec2^=0;
.-. either sec ^ = ; which gives no solution,
or, sec — 2; whence = 2mr =*= J tt.
MISCELLANEOUS EXAMPLES. LXXVI. a. 197
82. Art. 154.
sin {A 4- B)=coa [90 - {A+B)]= cos [90 - ^ + ( - B)]
= cos (90 -A) .cos{-B)- sin (90 - ^) . sin ( - B)
= sin ^ . cos B + cos A . sin 5,
since sm(-B)= - sin B and cos (-B) = cos B.
cos A, (ios(B+ C) -cosB . cos(^ + C)
= cos A . cos B . cos C - cos A . sin B . sin G
- cos ^ . cos A . cos G + cos 5 . sin A . sin C
= sin G (cos JB . sin ^ - cos ^ . sin B) = sin (^ - 5) . sin G.
83. (1) 1 + cosyl -sin^ = 2cos2J^-2sin4^cos4^
= V{4 cos2i^ (cos J^ - sin J^)^}
= ^{2 (1 + cos ^) (1 - sin A)}.
(2) sec 2A = g-^axry = 2^:^~22 '
(3) cos f TT + cos f TT + cos f TT + 1 = 2 cos f TT cos |7r 4- 2 COS^f TT
= 2 COS f TT (cos f TT + COS f Tt) = 4 COS f TT COS f TT COS f TT,
and cos f TT = COS (tt - ^ tt) =; - cos ^ TT. q. e. n.
84. The shorter diagonal = ^(5^ + 3^ - 2 . 5 . 3 . cos 60) in.
=x/(25 + 9-30x4)in.=V(19)in.==4-35...in.,
the longer diagonal=^(52 + 32 + 2 . 5 . 3 . cos 60°) in.
= V(25 + 9 + 30x J)in.=:^49in. = 7in.,
area of parallelogram = 5 x 3 x sin 60° sq. in.
" ¥- X ;^3 sq. in. = ^f x 1*732 . . . sq. in. = 13 sq. in. nearly.
85. (1) sinf ^ = sin^cos J^4-cosy4 sin J^
= 2sin J^ cos2J.4+cos^ sin ^^
= 8in Jy4 (1 + cos^ +cos^).
(2) sin(^ + 5)-sin^ = 2sinJ5cos((9 + Jo)
= 2 sin J 5 (cos ^ cos J 5 - sin d sin J 5)
= cos sin 5 - sin 2 sin^ J5
= cos^ sin 5{l-tan^2sin2J5-^(2 sin ^5cos J5;}.
86. (1) sin 100 + sin 50° r= 2 sin 30° cos 20° = cos 20° --sin 70°.
(2) ^S = tan 60° ; .*, we have to prove that
tan 60° + tan 40° + tan SO^ = tan 60° tan 40° tan 80°,
which is true by Ex. LXII. 32, since 60° + 40° + 80° = 180°.
(3) sin A-sinB cos C = sin (B + G)- sin B cos C = sin G cosB,
Also sin J5 - sin ^ cos G = sin (A + G)- sin A cos G = sin G cos A ,
and the result follows.
198 MISCELLANEOUS EXAMPLES. LXXVI. a.
87. SeeT.p. 56. sin 18°= J (^5- 1).
4 sin 18° cos 36° = 4 sin 18° (1 - 2 sin2 18°) = (^5 - 1) {1 - i (>/5 - 1)2}
= i(V5-l)(2 + 2V5) = i(V5-l)(V5 + l) = l.
88. Let a, b he the other sides of the triangle,
a = 1006-62 X sin 70°-^ sin 66°.
log a = log 1006 -62 + X sin 70° - L sin 66°
= 3 -0028656 + 9-9729858 - 9*9607302 = 3-0151212 = log 1035-43.
log & = log 1006-62 + L sin 44° - L sin 66°
= 3-0028656 + 9 -8417713 - 99607302 = 2-8839067 = log 765-4321.
89. Whole circumference = 27rr = 167r ft. ,
.-. length of arc which subtends at the centre an angle of 50°
= /A X IBtt ft. = V- X 3-1416 ft. = 6-9813 ft.
90. Fig. p. 106. Let P01/=^; i^OP' = ^ - 180°.
-sini?OP'= -sin(^-180°) = sin^, sin30° = J, [Art. 92]
sin 2010° = sin (5 x 360 + 180 + 30) = - sin 30° = - J .
91. Art. 144. The general value of cosec-i(y^2) is 2w7r±|7r.
92. (1) cos2^ ^ cos^P - 2 cos A . cos B (cos A . cos B - sin A . sin B)
= cos^A+cos^B -2cos2^ . cos2P4-2cos ^ . sinP . sin^ .cos^
= cos2^ (l-cos2^) + cos2p (1 - cos^^) + 2 COS ^ . siu P . sin^ . cos J5
= cos'^A . sin^^ + cos^P . sin2^ + 2cos^ . sin jB .sin^ . cos I?
= sin2(^+P).
(2) cos2^ 4-sin2^ , cos 2P = cos2^ + (1 - cos2^) (2 cos2J5 - 1)
= cosM + 2cos2p-2cos2^ . cos2p- l + cos2^
= cos2p+2cos2^-^2cos2^ . cos2jB-l + cos2p
= oos2p+ (1 - cos2jB) (2 cos2^ - 1)
= cos2p + sin2Pcos2^.
93. a^ COS 2B + h^cos2A=a^l-2 sin^B) + b'^l-~2 sin^ A)
= a^ + b'^-2a sin Ba sin B -2b sin Ab sin A
= a2 4- fe2 _ 4tab sin A sin B [for a sin B = b sin A],
94. c = a sin 135°-^ sin 15° 43';
.-. log c = log a + L sin 45° - L sin 15° 43'
= 2-0899051 -J log 2 -9-4327777 + 10
= 12-0899051 - i X -30103 - 9*4327777
= 2-5066124 = log 321-1207.
95. Art. 60. Let r = the radius of the circle,
, 27rr 360° 36x12.. 216 „, _ _„ „^
12 50° 5x27r 5x3-1416
MISCELLANEOUS EXAMPLES. LXXVI. a. 199
96. See Ex. XVL 4. cos ^ + 1 = 5 sin ^ ;
/. cos2^ + 2cos^ + l = 25- 25cos2^; /. 26 cos2^ + 2 cos ^ --24;
.-. cos2^ + tVcos^=4-|; .-. (cosJ[-H)(cos^ + l) = 0;
.', either cos ^=^ or cos^=-l.
This last value makes sin A=0, which satisfies the equation in a partial
sense only ; so that A = (2n + 1) tt.
97. See Art. 147. The angle whose secant is -- 2 is (2n + l) tt^^tt.
98. (1) sin2^ + sin2jB + 2sin^ . sin5cos(il+5)
= sin^A + sin^B + 2 sin A . sin B . (cos ^ . cos B - sin A . sin B)
= sin^A + sin^B + 2 sin A . sin B . cos A . cos B -2 sin'^A . sin^B
= sin^A (1 - sin2^) + sin^^ (1 - sin^^) + 2 sin ^ . sin £ cos ^ . cos B
= sin^A . cos^B + sin^B . cos^^ + 2 sin ^ . cos ^ . sin JK . eos B
= 8m^{A-{-B).
(2) sin^^ - cos2^ cos 2S = sinM - (1 - sinM) (2 cos25 - 1)
rrr sinM - 2 cos^ i? + 2 siu^ ^ cos^i? + 1 - sinM
= sin2i? - cos2£ (1-2 sin-^).
99. See Art. 178. 1200° = 3 x 363° + 120°.
sin 120° is greater than cos 120° and is positive,
.-. sin 1200° + cos 1200° = -h ^(1 + sin A ).
sin 1200° -cos 1200°= - J {1 - sin A.) .
cos 2A cos 2P _ 1 - 2 sin^^ 1-2 sin^i? _ 1 1
100. — ^- ^2 - a'' ~ 62 -a2~P'
since h^ sin^A = a- sin-^ B.
ocit^ A — 1
101. (i) 2 cot 2^ = — — =cot^-tan^.
^ ' cot ^
(ii) sin (sin~i f - sin~^ j%) = sin (sin-^ |) cos (sin~^ 5^)
- cos (sin-i f) sin (sin~i r\)
-fxl|--4x3:\=U. Q.E.D. [See 106 (iii)].
(iii) cot (^ + 15°) -tan (^-15°)
_ cos (A + 15°) cos {A - 15 °) - sin (A - 15°) sin (A + 15° ) _ 2 eos 2 A
sin (A + 15°) cos (A - 15°) "" sin 2A + sin30° *
102. When cos (2x + Sy) = i and cos (3a; + 2y) = i ^3,
we have 2x + 3?/ = 2xir ± J tt and Sx + 2y = 2xir =t ^ tt ;
.*. solving these equations we obtain the required values.
103. Sin X increases from to 1 while x changes from to ^ tt ;
.-. sin (tt s!n x) changes from to 1 and from 1 to as tt sin x goes from to tt.
sin (tt sin x) goes from to 1 and from 1 to again as x goes from J tt to tt.
sin (tt sin x) goes from to - 1 and from - 1 to as oj goes from tt to | tt,
and repeats as x goes from | tt to 27r.
200 MISCELLANEOUS EXAMPLES. LXXVI. a.
104. tan (tan-i a + tan-^ h) = ^~— ,
tan (2 tan~i a) = ^ ;
" l-x[l-x)~ 1-x-k-x'^ '
.'. Ax~4:x'^ = l; .'. x=\.
105. Let A and B be the two positions of the ship (7, D that of the
lighthouses.
AB = 2Q,BAC = 2'2Pm\BAD = i^°', :. BD = 20;
5(7 = 20. tan 22° 30', .-. log i? (7 = log 20 + L tan 22° 30'- 10
= 1 + -30103 + 9-6172243 - 10
= •9182543 = log 8-2842;
(7D = 20-B(7 = 11-7157.
-i.x« /n . ^n /2sin2i<9 /I
(ii)
-cos 6
cos 6
tan 56 + tan 3^ _ sin 50 . cos Sd + sin 3^ . cos 50
tan 56 - tan 36* ~ sin 56 . cos 3^ - sin 36^ . cos 56
sin 8^ 2 sin i6 . cos 4^
(iii)
^n 26 sin 2^
1^3 .
= 2 cos 2^. cos 4^.
sin-^TV=/8» .-. sin^=TV; cos^=ijf,
sin (a + j8) = sin a . cos j8 + cos a . sin /3
_3 15 4 8_45 + 32_77
~5'^17"^5^ 17~~85" ~85'
a + j8=sin-i|t, or sin-^ f + sin-i ^^ = sin-^ |^.
107. 2 sin2 6* - (1 + ^3) sin 2(9 + 2 ;^3 cos2 ^ = 0,
(sin ^ - ;^3 cos 6) (2 sin 6> - 2 cos ^) =0,
sin^-;^3cos^ = 0; tan^ = ;^/3; .-. ^ = W7r + j7r,
or 2sin^-2cos<9 = 0; tan ^ = 1; .'. 6 = nx + i'7r.
108. (i) See Ex. LXIV. 7.
(n) c(cos^+cos5) = c |-^,y- + ^^^ (
=:-L{a(624.c2_a2) + 6(a2 + c2_52)j
MISCELLANEOUS EXAMPLES. LXXVI. a. 201
= J^{a + b) 4. (s - a) (s -b) = 2(a + h) 8m^C.
109. The diameter of the circle circumscribing AEF is(F£-r-sin A) and
the angle FEB = FCB ; .: A ABC is similar to AFE;
.-. FE : AE = a : b;
.'. diameter of the circle = -: — - x -r- = -. — ^ cos A = a cot A .
sm A b sm A
Similarly for the other diameter.
110. Let X be the distance from the second point of observation to the
top of the tower,
then a; = 50 X sin 10 -^ sin 5°
= 100 X sin 5° cos 5°H-sin 5° = 100 cos 5°,
loga; = 2 + Lcos5°-10
= 11-9983442 - 10 = 1-998342 ;
.-. height of tower = a; . sin 15,
log (height) = \ogx+L sin 15° - 10
= 1-9983442 + 9-4129962-10
= 1-4113404 = log 25-7834.
111. sin^ -4 = sin 5 . cos 2^,
or 1 - 2 sin2 ^ = 1 _ 2 sin 5 . cos 5
= {cos2B - 2 sinB . cos -B + sin2 ^}
= 2 { (cos J5 - sin B)-r-sJ2} {{cos B - sin B)-^J2}
= 2 . sin (45° - 5) cos (45° + B) .
112. (i) sin^i (cos2^+cos4^ + cos6^)
= sin A (2 cos 4^ . cos 2 A + cos 4^) = sin ^ . cos 4a (2 cos 2A + 1)
= sin ^ . cos 4a (2 - 4 sin^ ^ + 1)
= cos 4a (3 sin A -4 sin^ A) *
= sin SA . cos 4a.
(ii) sin-ij = 30°; .'. 2sin-ii = 60°,
and co8"ij=:60°; .'. 2 sin~^ J = cos-^ J.
113. Let AB and BC produced meet in E.
Draw BFy GG perpendiculars to AE ; then,
cos (^ + P) = cos (180° -E)=- cos E, and
AB cos A-BG cos (A+B) + GD cos D = AB x AF-^AB + BG x cos E+DG
= AF-hFD + DG = AD,
[Notice that ^i^=-F^.]
202 MISCELLANiEOUS EXAMPLES. LXXVl. a.
114. SQ0°-(2A + 2B) = 2C; /. tan 2C= - tan (2^ + 2B),
and proceed as in LXII. 32.
Let tan2^=x, ta.n2B — y^ tan 2(7 = ^, then, since x + y + z=^xyz, we have
2(7=: 360° -2A- 2B. We have to prove that
(l-y^)(l-.^) (l-.^)(l-x^) (l-a:2)(i_y2)
2^/22 22;2a; 2ar2z/ ~ '
that is, that, cot 45 cot 40 + cot 4(7 cot 4^1 + cot 4^ cot 4i? = 1,
that is, then tan 4^ + tan 4B + tan 4(7= tan 4lA tan 4J5 tan 4(7,
and this is true since 4^ + 45 + 4C = 2 x 360°.
115. Let a and h be the distances the two ships are respectively from the
beacon, 6 = 1 mile since the triangle formed is isosceles,
then a = sin 75° 9' 30" -f sin 52° 25 '15",
log a = 9-9852635 - 9*8990055 = -0862580 = log 1-219714.
116. Let ^,5, (7 be the three angles
cosa = f, .-. sina = f, cos5 = Jf, .-. sin 5 =3?^,
cos 0= - cos (A+B)= - (cos A .cosB- sin A . sin B)
= -(fxH-|x3-y=-(tf-|0)^_i«.
sin(7 = v/il-(if)'}=\/960-r35, tan (7= -^960-35.
117 {'\ ^^° ^ + 2 sin 3^ + sin 5A _2 sin 3^ . cos 2a 4- 2 sin 3^
cos A -2 cos 3^ + cos 5A ~ 2 cos Sa . cos 2a ~ 2 cos 3a
_ sin 3^ (1 + cos 2a) _ sin 3^ x 2 cos^ A
~ cos 3^ (cos 2a - 1) ~ cos 3^ x 2 sin^ A
__ (4 sin^ A- 3 sin ^ ) X cos^ ^ _ 4 sin ^ - 3 cosec A
"" (4 cos^ ^ - 3 cos A) X sin'^ A~ 4: cos ^ - 3 sec ^
(ii) Let cot-i 3 = a, cot-^ | = /3,
cota = 3, cot^ = |,
_ , ^. cota. cot/3-1 3x|-l ^
cot(a+^)= ^^t^^J^^ =-3fF=^'
a + /3 - cot-i J ; or, cot-i 3 + cot'^ f = cot-i J.
118. Area = V(1452 x 1210 x 240 x 2) sq. yds.
=^(112 X 22 X 3 X 112 X 10 X 2^ X 3 X 10 X 2) g^^ y^^
=sj{ll* X 26 X 32 X 102) sq, y(jg, ^ 112 X 2» X 3 X 10 sq. yds.
= 121 X 4 X 10 X 6 sq. yds. = 4840 x 6 sq. yds. = 6 acres.
119. Fig. p. 232.
ad^^ + bd^^ + cd^^ = ar^ cosec2 ^A + br^ cosec2 J 5 + cr^ cosec2 J C.
a- 9 o^ A S (s-a) (s-b)(8-c) , , ,
Smce ar^ cosec^ i ^ = — ^ / V^ / \ — ^^ x abc = abc x(s-a)s;
^ s^(s-b)(s-c) ^ ^ '
.*. ad^ + 6c?2^ + ct!32 _ (ibc x{s-a + s-b + s- c)-t-8 = a6c.
MISCELLANEOUS EXAMPLES. LXXVI. a. 203
120. I^et A be the foot of the slope, jB, C the two points of observation,
D the object, then BDA = a, CDA=p, 7 = angle.
BD _ sin (7-/3) BD _ sin^ ^
c ~ sin (jS - a) c "" sin a '
sin 7 sin 7 cos /3 - cos 7 sin /3
" sin a ~ sin (/^ - a) '
sin 7 _ sin a tan 7 cos )3 - sin a sin /3 ^
** cos 7" sin (/3-a) '
sin a sin j3
/^ sinacosi3\
.-. tan 7(1 — : T 1 =
sin (jS - a) '
2 sm a cos 8 - sin /3 cos a ^ ^ ^
.•. cot 7 = ; — ^—~. — — ^- = 2 cot iS - cot a.
sin a sm /3
121. tan-i (i tan 2A) + tan-^ cot A
- A tan 2^ + cot ^ , , tan^^(l - tan2^)+cot^
= tan-i /— ,-; ^^, 7—: =tan-^ :; — - — 7- — ^ — -^r--
122.
1 -4 tan 2^ cot^"~ 1 - l-7-(l -tan^^)
i + l-tan2^ ~:
- tan^^ - tan A ~
l-tan^o; 1 + tanJa;
, . tan^^ + l-tan2^ , .^ .
= tan-i-- — - — - — 5— — 7 - = - tan"*icot^^. q.e.d.
tan A - tsLii^A - tan A
l + tan^ar 1- tan J a;
.'. tan^ ^x = 2; .\ ^x = tan-i ± J .^3 = wtt ± |^ tt.
123. When A lies between 90° and 180°, then J^ is between 45° and 90°;
in which case sin J^^ is greater than cos^^ and positive;
/. ^(l + sin^) = cos J^ + sin4^ = l-2sin2J^ + 2 sin J^ cosj^
= l + 2sin J^ (cos J^ -sinj^4)
= 1 + 2 sin J ^ ;y/(l - sin J^),
for ^A lies between 22 J° and 45° and in that case cos J^ is greater than
sinj^ and is positive, and /. cos J^ -sin J^= +^(1 - sin .J^).
124. asin\4 + Z> sin^ + csin C=0, acos ^ + 6cosJ5 + ccos C=0.
a sin A cos C + & sin 5 cos C + c sin C cos C= 0,
a cos A sin (7 + 6 cos B sin G + c sin G cos G = 0;
:. asm(G-A) = h8m{B-G); :. a-^sin (B - G) = b-i-8m{G -A).
Similarly a-^ sin (B — G) = c^ sin {A - B) ;
.'. a:h : c = sm(B-G) : 8m(G-A) : sin (A-B).
125. Let AB^ BG be the two parts of the ascent, GD = a the height of
mountain,
then AB = AC.i^-^\; BC=AC .^^^l AC=-^ .
sm (p - a) sm (/3 - a) sm 7
,. AB+BC=-A- X «in(^-7) + Bin(7^)
Sin 7 sin (/3 - a)
^^2sin4(^-a)co8{4(a + / 3)-7} ^^ cos { ^ (a 4- j^) - 7}
2 sin 7 cos J (j3 - a) . sin J (/3 - a) ' sin 7 cos J (/3 - a) *
204 MISCELLANEOUS EXAMPLES. LXXVI. a.
n«« ^ / / i o i\ r. i /sin 3^ 4- sin ^^
126. cosec 2^ (cosec A + cosec 3^ ) =cosec 2 A { . , ,
^ \ sm ^ sin 3^ y
_ 2 sin 2^ cos A _ sin (3^ - A )
~ 2 sin ^ cos A sin A sin 3.4 sin^ A sin 3^
_ sin A cos 3^ - cos SA sin ^
"~ sin^^ sin SA
127. 2cos4^ cos<9 + ^2(sin^ + cos^)cos^=0,
.-. either cos ^ = and 6 = mr + ^7ry
or cos4^ + cos(^-i7r) = 0, that is 2 cos (|<9- i tt) cos (f ^ + i7r) = 0,
whence, either cos (f ^ ~ ^ tt) = and f ^ - 1 tt = 7i7r + i ir,
or cos(f (9 + i7r)=:0 andf<9 + |7r=:n7r + j7r.
128. See Ex. LXII. 19, we have
sin B + sin C - sin ^ =4 cos ^A sin JJ5 sin J C,
sin C+sinA -sin 5 = 4 sin J^ cos JjBsin JC,
and the result follows by multiplication.
129. 10i-5=10^ = ^/(1000) = 31-622,
10'^ = 104 = ^/10 = 2-154, 10^*^ = lot = 4/10000 = 21-534.
130. Let AB, AC he the two chords,
a, p the angles which the chords make with the tangent. Join J5(7,
AB _ sin J^CjB _ sin a
**^®^ IC~sIn:i5C~sm^*
131. Let n be the number of degrees in a polygon of x sides; and let x
be the number of grades in a polygon of y sides ; then
a; X n = 90 X (2a; - 4), 2/ x n = 100 x (2?/ - 4) ;
.-. i^^"^ or xy-20y + lSx = 0; /. a: = 20-18-;
18ic
.*. since x and y are positive integers — is an integer and less than 20,
let the integer be X, also y = — - - 18 ;
20y .
.'. — - IS an integer, u say ;
X
.-. XAt = 20x 18 = 2x2x2x3x3x5;
.-. X may be either 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, or 18,
and then x is 19, 18, 17, 16, 15, 14, 12, 11, 10, 8, 5, or 2,
and y is 342, 162, 102, 72, 54, 42, 36, 22, 18, 12, 6, or 2 respectively;
but 35=2 and y = 2 gives n = which is not a 'polygon.'
Therefore there are only eleven solutions ; of these solutions only a; = 5, 8,
10, or 12 give integral values for n.
MISCELLANEOUS EXAMPLES. LXXVI. a. 205
132. cos 11^4 + 3 cos 9^ + 3 cos 7^ + cos 5^
= 2 cos 8^ . cos 3^ + 6 cos 8^ . cos A
= 2 cos 8^ (cos 3^ + 3 cos A) = 2 cos 8^ (4 cos^^)
— 8cos3^ X cos 8^ = 16 cos^^ J (008^4^ - sin2 4^)
= 16cos^^ xco8(4^+j7r) cos (AA - Jtt).
133. sin-^a; + sin~iJa; = sin-iJ^/2,
.', x^ {A- x^) = 2 -2~J2x sj{l - x') + x'^-x*;
.: 3.7:2_2=-2^2a;V(l-a5');
.-. 17a:* -20x2 + 4 = 0; .', x = ^{^ (5-2 J2)].
134. Jlog2-i(Lcos^-10) = l-log2 + 4{2LcosB-LsinC7-10}.
135. The first distance = ^jB sin 45° = AB x J ^2,
the second distance = AB+ AB xiJ2 = AB x ^ {^2 -•- 1) ^2,
the third distance = ^Z) = ^2 AB^ (>y2 + 1) ^2.
136. The expression = ^ { cos 2 (a + j3) - cos 2 {a + y) + cos 2 (/3 + 7)
- cos 2 (^ + a) + cos 2 (a + 7) - cos 2 (7 4- /3) } .
137. a = log|f|f = 21og5-101og2 + log41 = 2-2&-106 + log41.
138. sin 81° + sin 39° - sin 21° + sin 99°
= 2 sin 90° cos 9° - 2 sin 30° cos 9° = 2 cos 9° (sin 90° - sin 30°)
= 4 cos 9° cos 60° sin 30° = cos 9° = sin 81° = sin 908.
139. sinw+1 .^ + sinn-l . ^ = sin2^,
2 sin n^. cos ^ = 2 sin ^. cos ^; /. cos^ = 0; whence, ^ = n7r±j7r.
sin 71^- sin ^ = 0, 2 sin 4(n- 1)^. sin J (?i + l)^ = 0;
.-. sin J (n - 1) ^ = ; whence (n - 1) ^ = 2m7r^
sin i (/I +1)^ = 0; whence (n + 1) ^ = 2m7r.
140. A. circle can be circumscribed about the quadrilateral
sm (a 4-/3 + 7) sm (7-/8) sm (a + j8 + 7)
141. cos 2 A + sin 2B = sin (90° - 2^ ) + sin 2B
= 2sin{45°-(^-5)}cos{45°-(^+jB)}.
cos 2 A - sin 25= sin (90° -2A)~ sin 2B
= 2 sin {45° - (^ + i?) } { cos 45° -(A-B)},
142. (i) cos 55° + cos 65° + cos 175°
= 2 cos 65° . cos 120° + cos 65° = 2 cos 65° x ( - J) + cos 65°.
(ii) sin224° - sin2 6° = (sin 24° + sin 6°) (sin 24° - sin 6°)
= 4 sin 15° . cos 9° . cos 15° . sin 9° = sin 30° . sin 18° -\{J5-1),
206 MISCELLANEOUS EXAMPLES. LXXVI. a.
143. cos2^-cos2L' + cos2(7-cos2I>
= 2 sin (A-{-B)sm{B-A) + 2 sin (C + D) sin (D ~ G)
= 2 sin {A+B){Bm (B - ^) + sin (D - C)}
= 2Bin(A + B)2smi(B-A+D-C)cosi{B-A-D + C)
=4 sin (A-[-B) sin i (180° - 2A - 20) cos 4(2B + 2(7- 180°)
= 4 sin (^ + 5) cos (A + (7) sin (B + C).
144. log 5 + log 7 = a, 2 log 5 + log 13 = 6, log 5 + 2 log 7 = c, and the result
follows by solving these equations.
145. Ill the figure E.T. p. 216, let B be the train in its first position;
A the town ; CgC^ the positions of the train 18 miles from A ; then
AB = 20, ABG = i5 and 5- = a^ + c2-2ac cos45°;
.-. (18)2 = (24)2 + c2-^2x24c; /. c = 6(2 ;^2±1) ;
.*. time required = -^^ of 6 (2 ^^2 --t 1) hours.
146. 4 sin A cos A sin {A~B) = (4 sinM - 3) cos (A~B)
.*. 4 sin^^ cos ^ cos jB - 4 sin A sin B cos^A
= (4 sin^^ - B) (cos ^ cos B + sin ^ sin B) ;
.-. 4 sin ^ sin -B ( - cos^^ - sin^^) = - 3 (cos A cos i? + sin ^ sin B) ;
.*. sin A sin B = S cos ^ cos B. Q. e. d.
147. tan (tt . cot 6) = cot (tt . tan 6). Let tan ^ = a,
tan - = , or 1 - tan - . tan tt . a = 0.
a tan tt . a a
(^^+..„)=_
tan - + tan tt . a
tan ( - + TT . a ) = =oo ;
tan - . tan wa
a
.-. - + 7r. a = W7r+-; .'. 2a2- (2?i+ l)a= - 2,
16a2 - (2n + 1) a + (27i + 1)^ = 4^2 + 4/i - 15 ;
.*. 4a or 4tan^ = 2/i+l±^47i2 + 4n-15.
148. ^2 + ^2^2 + 2 cos a.
2a; = 4cos2a-2 + 2cosa=(ic2-f2/2_2)2_2 + j:2_|_^2_2
= (a;2 + 1/2)2- 3(^2 + 2/2)^
149. 2EsinC = c and (c- & cos^)2 = c2 4-Z>2cos2^ -2&ccos^
r=a2 - 62 ^ ^,2 cos2^ = a2 - 62 sin2^. Q. E. D.
ah
150. tan <tan-i r; — + tan-^ -
6 + c a-\-c\ ^ ah
{f>-Vc){a-Vc)
a^^ac-^h^^hc {a^-\-\P-)-\-ac-^hc . , . o .o ox x i
= -I T 5^ 7 =^ -^- — , ^ , — = 1 (since a2 + 62-^2) = tan jTr.
ah-\-ac-\-hc-\-c^-r(?.h c{a-^h-\-c) ^ ' *
MISCELLANEOUS EXAMPLES. LXXVI. a. 207
151. 60° = § right angle, 50k=-.J right angle, i7r = f right angle; the
fourth angle = (4 - 1 - i - f ) right angle = f right angle ;
.-. the angles are 60°, 45°, 135°, 120°.
152. cos(sin-iwi + sin-^?i) = cos^7r = 0;
.-. ^(1 - wi2) ;^(1 - n2) - mn = ; .-. 1 - m^ - n^ + m^n'^ = mV ;
.*. m^~l- n^ ; .*. m = ± ^(1 - n^), or sin-^ m = =t cos~^ n.
. „, 1-cosa l-ecos3~co8 3 + e
153. tan2Aa=- = -
^"^^ ^ 1 + cosa l-ecosj3 + cos/3-e
_ (l4-6)(l-cos^) ^ 1 + e
~(l-«)(l + cos/3) 1-e ' '^^'
154. 2 cos ((9 - J tt) cos (2<? - J tt) = cos ^ + cos {Sd - J tt) .
155. ^252^2 _ 452^2 ^ ^2^2c2 (1 _ sin2 ^)^
a262c2 COS A - 4&c6f2 = a252c2 / cos ^ - ^ sin2^ j .
We have to prove that
27, ^r^ { A sini^sinCsin2^\2
C0S2jy C0S2 C = COS A 7— n-j
\ sm^^ J
= (sin J5 sin C - cos B cos C - sin B sin C)^.
156. cos2 (^ + 5) + cos2 (^ - ^) - cos 2A cos 2 JB
= 2 cos2^ cos2jB + 2 sin2^ sin2 5 - (cos2^ - sin2^) (cos^^ - sin25)
= cos2^ cos2£ + sin2J^sin2jB + cos2^ sin2jB + cos2jBsin2^
= (cos2^ + sin2^) (cos25 + sin2^) = l X 1.
157. Let x° and 2c° + 10° be the units.
Then 4/ca;° = Sk (a:° + 10°) ; .-. a;° = 30°.
158. cos7°30'=V{i(l + cosl5°)} = ^|i^l + ^^^^-)j.
= i>y{2 (4 + ^/6 + ^2)}= J V{2x/2 (2^2+ ^3 + 1)}.
Now (-l + V2 + x/3Px(2 + ^2) = 2 (3 -^2-^3 + ^/6) (l+V2)s/2
= 2^2(1 + 2^2 + ^3),
and the result follows.
159.
sin i ^ + cos I J5 - sin i C _ 2 cos ^(A + C) , sin J (^ - G) + cos ^B
sin J^ + cos^(7-sin J^ " 2cos J(^+P) sin J (/4 - JB) + cos J C7
_ 2cos j: (^ + C){sin ^(^ - C) + sin ^{A- C)}
~2cosJ(^ + B){sinJ(^-i?) + sini(^ + ^)}
_ cosi(^ + (7)(2sini^ .cos:^(7) _ cos (45° - :| ^) . cos | C
~ cosi(^+5)2sinJ^.cosJ£ "" cos(45°- jcj . cos J^
_ (cos J 5 + sin J 5) cos J G
"" (cos J C + sin J C) QoTj B '
208 MISCELLANEOUS EXAMPLES. LXXVI. a.
160. a2{6^ + c2-26ccos(J5-(7)}
= a%^ + a^c^ - 2a?hc cos B cos C - 2a'^hc sin B sin C
= a%^ + a'c'' - J (a2 + fe2 _ ^2) (^2 ^ ^2 _ ^,2) _ 26^02 sin2^
=:4{2a262 + 2a2c2-a4 + 54^c4_262c2_4^2c2 + (52 + c2-a2)2}
[since 262c2 sin2^ = 262c2_ 262c2cos2^]
= i(264 + 2c4-462c2). Q.E.D.
161. cos 30 + sin 30 = cos + sin 0, sin 30 - sin = cos - cos 30,
2 cos 20 . sin = 2 sin 20 . sin ;
.*. either, sin = 0; whence, = wtt,
or tan 20 = 1; whence, 20=W7r±j7r.
162. a; = 3cos0 + cos3</> = 3 cos0 + 4cos30- 3 cos 0=4cos3 0,
2/ = 3 sin - 3 sin + 4 sin^ = 4 sin^ ;
•*• {i«)^ = cos20 and (Jy)3 = sin-0;
.'. (Jx)§ + (j2/)^ = cos204-sin20 = l; or, a;3 + 2/? = 48.
163. cos218° sin2 36° - cos 36° sin 180°
= TV(10 + 2V5)xTV(10-2V5)-i(l + V5)i(l-v/5) = xVxfJ-A = A.
164. 2 (4 cos3 - 3 COS 0) + 2 cos - 1 = 0,
2 cos (4 cos2 - 1) - (2 cos + 1) = ;
/. (2 cos0 + l)(4cos20-2cos0-l) = O;
.*. either cos = - J ; whence, = 2mr ± § tt,
or 4 cos- 0-2 cos 0-1 = 0,
that is, (4 cos - >y5 - 1) (4 cos + ^^5 - 1 = 0, ^
whence = 2n7r±i7r or = 2n7r±f tt.
165. Ii^ this case the angles at A and C are right angles
.-. BD'^ = a?-\-d} = h^ + c^;
.', (s - a){s- d) = s2 ~ (a + d)s + ad
= s^-(a + d)^{a + b + c + d) + ad
= s2 - J (a- + d- -\- ab + ac ■]- db + dc + 2ad) - ad
= s^-^(b' + c^ + ab + ac + bd + dc + 26c) - be
= s^~^(b + c){a + b + c-{-d)-bc = {s-b){s-c).
The area =sf{(s - a) {s -b)(s - c) (s - d)} =^{{s - a)^s - d)^}.
166. sin(^+5)cos^ = 3cos (^+5) sin^,
/. sin A cos A cos ^ + sin ^ cos2^ = 3 sin A cos ^i cos 5-3 sin^^ sin B ;
.-. sin 2A cos B = sin 5 (1 + 2 sin2^) = sin B{2- cos 2A ) ;
:. sin(24+5) = 2sin J5;
.-. sin (2^+i^)cosB = 2sinBcosB;
/. sin (2A + 2B) + sin 2^ = 2 sin 2B, q.e.d.
MISCELLANEOUS EXAMPLES. LXXVI. a. 209
167. sin 18° + cos 18° = ^(1 + sin 36°) = ^(1 + cos ^4°) = ^(2 cos^ 27°).
, «^ tan 2^ + tan ^ , ^^ ^ ^ r^
168. 1 — Ki — TTE-^ n + tan 26 + tan ^ = 0,
1-2 tan 26 tan 6
.'. either, tan 2^ + tan ^ = 0,
whence tan ^ = ; and, 6 = mr,
or tan^=±Ay3; and ^ = n7r±j7r,
or, 2 - tan 26 tan ^ = 0, whence, 6 = mr^ tan~^ J fj2.
169. 4 cos3 2^+4 cos3 2B + 4: cos^ 20
= cos 6^ + cos 65 + cos 60 + 3 (cos 2A + cos 2B + cos 20)
= 2 cos 3 (A+B) {cos 3 (.4 - B) - cos S(A-^B)} + 1
+ 6 cos (^ + £) {cos (^ - 2?) - cos (A + B)}+3
= 4 sin 30 sin SA sin 3J5 + 12 sin 20 sin 2A sin 2B + 4.
170. ^^ cos^ ^A + ca cos^ J B + «& cos'^ J
= s(s-a)+s(s-5) + s(s--c) = 3s2_2s2=:s2.
171 . sin 7^ = sin 6^ cos 6 + cos 6^ sin 6
= 2 sin 3^ cos 36 cos <9 + (4 cos^ 2^-3 cos 2^) sin 6
= 2 (3 sin (9 - 4 sin^^) (4 cos^^ - 3 cos 6)cos6+...
= sin ^ { 3 - 2 (1 - cos 26>) } {2 (1 + cos 2(9) - 3} (1 + cos 26) + . . .
= sin ^ { 1 + 2 cos 26} { 2 cos 26> - 1} (1 + cos 26) + ...
= sin ^ {8 cos3 2(9 + 4 cos2 2(9 - 4 cos 2(9 - 1}.
Now 7^ = 7r is satisfied if either of the above factors is zero.
But 8 cos 6 cos 26 cos 3^ - 1 = 4 cos 26 {cos 46 + cos 26} - 1
~ 4 cos 2(9 {2 cos2 26" - 1 + cos 26} - 1
^ 8 cos3 2^ + 4 cos2 2(9 - 4 cos 2^ - 1.
Hence the statement of the question is true.
172. sin2 ^{A+B)- sin2 i {A - B)
=:{smi(A+B) + sm^(A-B)}{smi{A-{-B)-smi{A-B)}
= 2 sin^J[ cos JjR x 2 cos \A sin JJB = sin^ sin B,
Hence when ^4 + jB is given sin A sin B has its greatest value when A = B,
Now suppose (A-\-B-\-C) given, then
sin A sin B sin has its greatest value when A = B = C.
For suppose we keep [and .-. (A+B)] unaltered, then the value of the
expression is increased by making A = B.
Similarly by keeping (B + O) unaltered the value is increased by making
B = G.
Hence the greatest value is when A—B — C
Now cos A cos J9 cos = sin (90° - ^ ) sin (90° - B) sin (90° - 0),
and since (90° - ^ + 90° - 5 + 90° - 0) is constant this expression has its
greatest value when A = B = C and then A=B — G= 30°.
L. T. K. 14
210 MISCELLANEOUS EXAMPLES. LXXVI. a.
173. cot-i3 = sin-VTV = cos-VA' sin-Hv^5 = cos-i|^5;
= (f + l)i^/2 = ^^/2 = sini7^.
174. sin SA + sin SB + sin 3 C
= 2 sinf (^ + B) cos f (^ -^) + 2 sin f Ccosf C
=:2sinf (71+5) {cost (^-B) + COS 1(^+5)}
[For I C= 270° - f (^ + B) and sin (270° - (9) = - cos ^
and cos (270° -d)=- sin ^]
= -4cosf Ocosf ^ cosf J5. q.e.d.
a S
175. We have ^^-^ — -= ; and a=c, suppose; .*. 4:8'^ = a^b (s - a) ;
£1 sm A. s — a
.-. i{(2a + b)b{2a~b)}=a%; .'. b^ = 2a^. q.e.d.
176. sina; = l- sin2a: = cos2a;.
/. cos'*a; = sin2a; = l-cos2a:. q.e.d.
177. sin 10° sin 50° sin 70° == J (cos 40° - cos 60°) sin 70°
= i cos 40° sin 70° - J sin 70°
= I sin 110° + J sin 30° - J sin 70° = J sin 30° = ^
for sin 110° = sin 70°.
-. «« . L 1 ^ cos ^ , a; - sin ^1
178. tan -^tan-i :, r—^ - tan-^ —- 1
{ 1 - a: sin ^ cos ^ J
id a: - sin ^] (^ x cos d a; - sin ^]
X -
_ \ X cos a: - sin ^1 j
~ (1 - a; sin ^ cos ^ J 1
sin ^ (1 - 2a; sin B + x^) , ^
( = tan d.
1 - a; sin ^ cos
cos ^(1- 2a; sin^ + a:*^)
179. Since ^ + B + C + D = 360°,
.-. J(^+5) = 180°-i((7+i)), /. cos4(^+J5)+cosi(C + D) = 0;
.-. cos J (^ + 5) + cos i (C + D) + cos J (^ + C)
+ cosi(B + 2)) + cosi(JB + (7) + cosi(^+i)) = 0,
expanding, the statement is seen to be true.
180. Let AB be the line, C its middle point ; let D be the foot of the
tower and h its height,
then AD = BD and DC is perpendicular to AB.
Also AD = h cot a, CD = h cot /3,
and a^ = AD^- GD'^ ; :.a^ = h^ (cot^ a - cot2 /3) ;
.'. h = a sin a sin j3-f-^ [sin^/S cos^a - sin^a cos^/S}
= asinasin/3^;^{sin (^ + a) sin (/3-a)}.
181 . sin (a - jS) cos 2^ + cos (a - /3) sin 2^3
= sin{(a-/3 + 2/3)} = sin(/9-a + 2a)
= sin (j3 - a) cos 2a + cos (/3 - a) sin 2a.
MISCELLANEOUS EXAMPLES. LXXVI. a. 211
182. sin(? = 3»^; /. cos^=H.
sin2^=:2xAxH = H*; 2cosH^-l = H; •*. 2cos2i^ = f«;
/. 8in2^.cosi(?=l§txJ|f,
and tan ^ = ^^^ x ^ = f^,
183. If the angle 26 is in the first quadrant sin 26 is positive and so
are cos 6 and sin 6.
If 26 is in the second quadrant sin 26 is positive and 6 would be in the
first quadrant and both sin 6 and cos 6 are positive.
If 26 is in the third quadrant sin 26 is negative, cos 6 would be in the
second quadrant and is negative sin 6 would be positive, .*. 2 sin ^ . cos 6 is
negative.
If 26 is in the fourth quadrant sin 26 is negative and cos 6 would be in
the second quadrant and is negative sin 6 would be positive, .*. 2 sin 6 . cos 6
would be negative ; .*. sin 26 has always the same sign as 2 sin 6 . cos 6.
184. 2xiaxccos^ = c2 + ia2-(^Z))2.
Also a6cosO=Ja2 + 62_(^i))2.
.. a(&cosO + ccos^) = c2 + 62 + Ja2-2^Z)2;
/. a2 = c2 + 62^ja2_ 2^2)2.
.'. 4J^i)2=2c2 + 262-a2=c2 + 62^..26ccos^.
185. r=Zsini^
.*. —3- = cosec \A cosec \B cosec \ G
^^^ and rs = S; [Ex. LXXIII. (7)iii.]
~ (s-a){s-b){s-c)
Imn ahc
186. Let -; = k. then a cos 6-^b cos d> = k sin (6 + d)) — c,
sm <p
.*. sm (^ + 0) = - sin 0.
Again, c2-.26ccos0 + 62cos2</>=a2cos2^ = a2-a2sin20 = a2- 62sin20;
.-. 26ccos0=&"^ + c2-a2;
where s = \(a + h-\-c).
2 2
Similarly sin ^ = — J {s is - a)(s -h){s - c)} = — S\
.*. cos (6 + </)) = cos 6 cos <p - sin'^ sin <p
= { (a2 + c2 - 62) (62 + c2 - a2) - 165f2} -^4a6c2
= (2c* - 2a2c2 - 262c2) ^^abc^;
.-. tan (^ + 0) = - sin X 2ab^(c^ -a-- 62),
.-. tan {6 + <p)x (c2 - a^ - 62) = 4tbcS, q.e.d.
14—2
212 MISCELLANEOUS EXAMPLES. LXXVI. a.
187. This is another way of stating that when A-{-B + C = 180°;
sin A cos A + sin B cos B+sinC cos G = 2 sin A sin B sin C.
[See Ex. LXII. 26.]
188. By Ex. XLIII. 9, we have
sin A sin B sin G
= i{sm{B-\-C-A)-}-sm(G + A-B) + sm(A+B-C)-sin{A+B + C)},
Let A=^-y, B = d-p, C=d-y,
sin(^-/3)sin(^-7)sin(j3-7) = J{sin2(^-j3) + sin^ + sin2(j3-7)-sin2(^-7)
and two similar statements.
By addition we have
sin (6 - p) sin (6 - 7) sin (p-y) + sin {0 - 7) sin [6 - a) sin (a - 7)
+ sin {6 -a) sin (6- §) sin (a - /3) = J {sin 2 (/3 - 7) + sin 2 (7 - a) + sin 2 (a - /3) }
= sin (j3 - 7) sin (7 - a) sin (a - ^). [See Ex. XLIII. 7,]
189. 2 sin2^ + 2 sin2 jB + 2 sin^ C = 3 - cos 2^ - cos 2B - cos 2C
= 2 - 4 sin ^ sin J5 sin G, [Cf . Ex. LXII. 21.]
This expression has its least value when A=:B = G hy Ex. 1.^ '^ and there-
fore the least value of sin'^ A + sin^^ + sin^ 0= 3 sin^ 30° = j.
190. The perimeter of a regular circumscribing polygon of m sides
IT
= 2nr . tan - , where r is the radius of circle.
n
The perimeter of a regular inscribed polygon = 2nr . sin - ,
but here 2nr . tan - — inr . sin - .
n n
:. 1 = 2 cos — , or cos - = i = cos i tt ; .*. w = 3,
191. msin2^=:nsin^ (1),
^.cos2^=g . cos ^ (2).
From (1), 2m.cos^=n.
Substitutiiig in (2) for cos 6, we get
^^ * ^2 -i^=^ • ^ ' 2n2p - 4m2p = 2mng'.
192. See Ex. XLin. 20.
tan 4:A + tan A 2 tan 2 A + tan A - tan^ 2 A tan A
193. tan 5 A =
1 - tan 4:A tan A 1 - tan^ 2^-2 tan 2 A tan A
_4 tan ^ (1 - tan^^) +tan^ (1 - tan2^)2- 4 tanM
~ (1 - tan2^)2 - 4 tan2^ _ 4 tan2^ (1 - tan2^)
4 tan ^ - 4 tan^^ + tan A -2 tan^ A + tan^^ - 4 tan^ A
^ 1-2 tan2^ ^ tan*^ - 4 tan2 ^ - 4 tan2 ^ ^ 4 tan*^" *
MISCELLANEOUS EXAMPLES. LXXVI. a. 213
194. cos 36° = sin 54° = sin 3 x 18° = 3 sin 18° - 4 sin^ 18°,
also by XLII. 6, 4 sin218°=l - 2 sin 18°;
/. 4 sin 18° cos 36° = 12 sin2 18° - 16 sinn8°
= 12 sin2 18° - (1 - 2 sin 18°)^ = 8 sin^ 18° - 1 + 4 sin 18°
= 2 - 4 sin 18° - 1 + 4 sin 18° = 1.
Again, 2 cos 36° - 2 sin 18°= 2 sin 3 x 18° - 2 sin 18°
= 6 sin 18° - 8 sin^ 18° - 2 sin 18°
= 4 sin 18° - 2 sin 18° ( 1 - 2 sin 18°)
= 2sinl8° + 4sin218°
= 2 sin 18° + 1 - 2 sin 18° = 1. q.e.d.
195. r=g=:^^^-^)<^;^)<^-^) = .tan^^taniBtaniC.
196. cos(^ + i7r) = cos(a + j7r), /. ^ + J7r = 2n7r=t(a + Jtt).
This is the complete solution.
197. >os(^ + i5)cos(^ + C)(cos^+D)
= 2{cos(5-C) + cos(2^ + i? + a)}cos(^+D)
=cos (360° - 20) + cos (2B - 360°) + cos (360° + 2^) + cos (360° - 2D)
= cos 2A + cos 2jB + cos 2(7 + cos 2D ;
.*. cos (j; + 2^) - cos 2^ + cos (a: + 2D) -cos 25 + etc. =0;
.-. 2sin (4x + 2^)sin Ja; + etc. =0;
divide by sin J x and multiply by cos ^ x, which may be done provided sin J x
is not zero ; then
2sin(Ja; + 2^)cos Jic + etc.=:0; /. sin (ic + 2^) + sin2^ + etc. =0;
.-. sin (x-\-2A)+ etc. = - (sin 2A + sin 2D + sin 2C + sin 2D)
= 4 sin(^ + D)sin(yl + C)sin(^+D), as above.
198. sin2^ + sin2D + sin2 C - 2 sin ^ . sin D . sin (7 - 1
= sin2^ - 2 sin ^ . sin D . sin C - J (1 - 2 sin2D) - J (1 - 2 sin^C)
= sin2^ - sin A {cos (B - C) - cos (B + C)}-^ (cos 2D + cos 20)
= sin2^ - sin A {cos {B-C)- cos (D + 0)} - cos (D + 0) cos (D - O)
= {sin ^ + cos (D + 0) } {sin ^ ~ cos (B-C)}
= {cos (90° - ^) + cos (D + 0)} {cos (90°-^)- cos (D~0)!
=2 cos J (90° -^ + D + 0). cos 4 (90°-^ -D-0)
x2sin4(90°-^-D + O) .sin i (90°-^ +D-0).
199. 2i2{cos24^ + cos24D + cos240}
= 2D{2sinJ^sinJDsinJO + 2} [Ex. LXIL 21.]
abc
,,, ^A
sabc
=4D-^ +4D -+4R=r + 4:It. [Ex. LXXIII. 4, 7 (i".)]
214 MISCELLANEOUS EXAMPLES. LXXVI. a.
200. Ij6t ic be a side of the square ABGD ; let the diagonals intersect in O.
Let and xp be the least angles the diagonals of the quadrilateral make
with a side of the square ; then one of the angles between the diagonals is
90° - - xf/.
Twice the area of a quadrilateral whose diagonals AB, CD intersect in O
is
{OA xOG+OBx 00+ OBxOD + ODx OA) ain AOB=AB x OD sin AOB ;
.'. 2C=Mcos(0+^).
Now cos0 = -; cos^ = -;
••• «i^ V = -p- ; sm^ xp = -p^ .
20= /ife {cos cos ^ - sin sin ^} ;
.-. 2G=^x^ - ^{Wk'^ - x^(h'^ + k'^)+x^}\
:, {20-xy=hV-x^(h^-^k^) + x^;
^_ hV-4:0^
MISCELLANEOUS EXAMPLES. LXXVI. b. Page 269.
1. 2cos^-cos2^ = rt (i),
2sin^-sin2^ = 6 (ii).
Square (i) and (ii) and add, then we obtain
4 (cos2 d + sin2 0) + (cos2 26 + sin^ 20) - 4 (cos 6 cos 20 + sin sin 20) = a^ + h\
i.e. 4cos^ = 5 -a2-62^
cos2^ = 2cos2(9-l=:3-%(5-a2-62)2_i^i(5_^2_52)2_i.
Substitute for cos and cos 2^ their values in (i),
then J (5 _ a2- 62) _ 1 (5 _ a2_ ^,2)2^1^^.
.-. (a2+62)2_6(a2 + 62)^32^12_8a; /. (a2 + 62_3)2^12_8a.
2, * hco8 0-\-ksin0=l (i),
Z cos ^ + m sin ^ = 1 (ii).
Multiply (i) by I and (ii) by h and subtract,
. , l-h
.'. sm^ = - .
Ik - mh
Multiply (i) by m and (ii) by k and subtract,
m-k
.-. cos 0~-
mh - Ik
Now sin2^ + cos2^ = l; .-. }^-^'^'+p'fJ^. = l,
{Ik-mhy {mh-lky
:. (l-hY'\-{7n-kY={mh-lk)K
MISCELLANEOUS EXAMPLES. LXXVI. b. 215
3. The diagonals of a rhombus bisect each other at right angles ; there-
fore the length of the side of the rhombus with diagonals 2a, and 26, is
Ja^ 4- ly^' Denote the angle subtended by diagonal of length 2a, A and that
by diagonal of length 26, B ;
, (a2 + 62) + (a2 + 62)-4a2 62-a4 ^-.^ m . . ^.^ .
.-. cos A = "^ ^ / „ . , ,, ' = -2-^. [E. T. Art. 240.]
„ (a2 + 62) + (a2 + 62)-462 a2-62
''''^^"" 2(a2 + 62) ~a2 + 62*
4. 2 0082^^1 = l + cos^; .-. 2 cos i ^ = >y2 + 2 cos A.
Also 2cos2JJ=:l + cos4 J;
.'. 2cosJ^ = ^2 + 2cosi/ = V{2 + ^(2 + 2cos^)}.
Similarly, 2cosi^ = V[2 + \/{2 + V(2 + 2cos^)}].
A
In the same way we may proceed for 2 cos ^^ .
5. If 8a; = log,3, .*. ^8^ = 3 and e^'^^si^^S,
Also ^-'^=3-i = i-;
6. A radians = degrees.
TT
Now we are given that tan = tan A°;
IT
.-. ^l?2! = nxl80° + ^°; .-. A (^^°- l) =nx 180°;
nxlSO^XTT
•*• ~ 180° -TT '
.'. A IS some multiple of r—^- .
loU — TT
'y -\- CL
7. Since a, /3, 7 are in A.p. , /3-a=:7-/3 and 7 + a = 2j8; .*. ^— — =^.
. 7-a 7+a
«- . 7 + ct "V — a
Now Sin a + sin 7 = 2 sin --_- cos ^-^ = 2 sin j3 cos (/3 - a).
216
MISCELLANEOUS EXAMPLES. LXXVI. b.
8, Let ABCD be the circle of which the centre is O; AD the
nearer to the centre, and BG the parallel
chord further removed from the centre.
Join OA, OD, OB, OC ; and from draw
the perpendicular OPP' bisecting AD in
P and BG in P'; OBG and OAD are
isosceles triangles;
O5C=J(180°-72°) = 54°,
and OAD = J (180° - 144°) = 18°.
PP'= distance between the chords.
PP'=AP'- AP= OP sin OBG - OA sin OAD
= radius x sin 54° - radius x sin 180°
= radius x (sin 54° - sin 18°)
= radius x {J (^5 + 1) - i (v/5 - 1)}
= half of the radius.
9, 4 sin {d - a) sin (rnO - a) cos (6 - md)
= 2 cos {e - md) {cos {6 - mS) - cos {6 + md - 2a) }
= 2 cos2 (6 -m0)-2cos{d- md) cos (<9 + md - 2a)
= 1 + cos 2 ((9 - md) - { cos (2(9 - 2a) + cos (2m(9 - 2a)}
= 1 + cos (20 - 2me) - cos [26 - 2a) - cos (2md - 2a).
10, x* + y*i-z^- 2yh^ - 2z^x^ - 2x-y^
= - {AxY -x^- ^^V -y^ + 2'i/ V ^ 2^^0:2 - z^)
= _ {ixY - {x^ + VY + 2 (a;2 + 2^2) ^2 _ ^4}
= -[4a:y-{(a;2 + ?/2)2_2(a;2 + 2/2);sH2^}]
= _ 1(2x2/)^- (0:2 + 2/2-^2)2}
= - {2a;2/ + (a:2 + 2/2-2;2)} [2xy - [x^^ + y^ - z^)]
= -{{x + y)''-z}{z''-{x-y)^
= -{x + y + z)(x + y-z)(z + x-y){z-x + y).
:. log of the first exp. = log (a: + ?/ + 2;) + log (2 - a; - 2/)
+ \oQ{z + x-y) + log (z + y- x),
sin 3(9 + ^/3 cos 3^=1,
sm ^^ + \^ cos 3^ ^ j^ gin (3^ + 60°) = sin 30° ;
chord
U. (i)
e=-
-M
M = mr-lTr-\-(-
■irTV^ = TV{6n7r-27r+(-l)
(ii)
(iii)
2 2
3^ + j7r=7i7r + (-l)^i7r;
71 TT
¥ 9
sin m^=:cos?i^; .*. cos ( J tt - m^) = cos w^ ;
.'. Jtt- m^ = 2r7r±n^; /. ^ (m±7i) = Jtt- 2r7r
cos(j8 + a:) _msinj8^
cos [a- x) n sin a *
cos (/3 + a;) + cos (a-a:) _ m sin^ + nsina
cos {a-x)- cos ()3 + a;) ?z sin a - m sin p '
i)N
MISCELLANEOUS EXAMPLES. LXXVI. b.
217
2 cos ^ (g + /3) C08 ^ (j3 - g + 2x) _ m sin /3 + ?i sin a
2 sin i (g + iS) sin i (j8 - g + 2a;) ~ w sin g - m sin /3 '
^ , m sin /3 + n sin a
.'. cot A (g + iS) . cot i (/3 - g + 2x) = — . — ?- r— - ;
2 ^ '^^ '^^^ ' n sm g - m sm ^
, , / « V - -. / ^, m sin /? + ?i sin g
.-. coti(/3-g + 2a: =tan|(g + ^) . — z—^ r— r;
^ ^'^ ^ ^ ^ "^^ n sm g - m sin j8
, . (, ■, ■ . m sin j3 + w sin g)
/. a; - i (g - fi) = cot"! -^tan i (g 4- /3 . — ^-^^ r-- V .
(iv) tan m6 = cot 7i^ ;
/. cot (i7r-m^) = cot7z^; .*. \ir -md = rTr + n6,
or, since tanm^ = cotn^, /. md and nd are complementary; therefore
771^ + n^ = rTT 4- J TT.
(V)
(vi)
(vii)
tan ^ + tan2<9 + tan 3^ = 0;
= 0:
sin d sin 2^ sin 3^
cos ^ cos 2^ cos 3^
sin ^ cos 3^ + sin 3^ cos ^ sin2^_
cos (y cos 3^ 008 2^"" '
sin 4^
cos 6 cos 3^
2 sin 2^ cos 2^
sin2^_
■*'SS^2^-"'
sin2^_
■^cos2"^~ '
cos d COS 3^
/. sin2^(2cos2 2(9 + cos^cos3^) = 0;
.-. sin 2^ (4 cos2 2(? + 2 cos ^ cos 3^) = ;
sin 26 (4 cos^ 26 + cos 4(9 + cos 2(9) = ;
.-. sin2^(6cos3 2(9 + cos2^-l)=0;
•. sin26>(3cos2(9-l)(2cos2(9 + l)=0.
cos 8^ - cos 5^ + cos 3(9 = 1,
cos 8^ + 2 sin4(9 sin (9 = 1,
l-2sin2 4^ + 2sin4^sin^ = l,
2 sin 4^ (sin ^ - sin 4(9) = 0,
4 sin 4^ . cos ^ 5^ sin J 3^ = 0.
cos ^ . cos 3^ = cos b6 . cos 16 ;
.-. cos 4^ + cos 2^ = cos 12^ + cos 29;
. cos 4^ -cos 12^ = 0; .'. 2 sin 4^ sin 8^ = 0.
12.
tan
A-h
a-h C
— i cot ^ .
a + h 2
XT 14. ^^ ^ ^-^ tan 0-1 /^t\
Now let - = tan0; /. ,=, ^ — - = tan 0-.-);
6 ^ a + 6 tan 0+1 V 4/'
/. if = tan-i - , tan J (^ - J5) = tan (0 - J tt) cot -^^ .
218
MISCELLANEOUS EXAMPLES. LXXVI. b.
13. 62 + c2-26ccos(60° + ^)
= c2 + 62 _ 26c (cos 60° cos A - sin 60° sin A)
= c2 + 62 - 6c cos A - 26c sin 60° sin A
= c2 + 62- J (62 + c2-a2) - 2ac sin 60° sin B
=c2 + a2-J(a2 + c2-62)-2acsin60°sinB
= c^+a^-ca cos B - 2ca sin 60° sin B
= c2 + a2 - 2ca cos (60° + B),
Let Oj, Og, O3 be the centres of the equilateral triangles described on BC,
GA and AB respectively.
Then
Now
0^0^^= O^C + O2C2 - 2O1C . O2C COS O^GO.^
^^_S a2sin60° a
^1^-7= fa -73'
_^ -S 62 sin 60°
V3'
.-. 3OiO22 = a2 + 62-2a6cos(60°+C)
= a2 + 62 - J (a2 + 62 - c2) + 2a6 sin 60° sin C
= i(a2 + 62 + c2)+25fV3,
3O3O32 and 3O3O12 are reducible to this expression, therefore
O1O2 =0203 = 0301.
14. Let be the centre of the base of the tower ; from A draw AM and
from B draw BN to touch the circular base of the tower, then the angle
0AM = a, OBN=p. Let r be the radius.
Then r = O^ sin a = OB sin ^ ;
AB = OA-OB = -
sma
.*. the diameter = 2r
sin j3 ' ' * ~ sin /3 - sin a '
2a sin a sin ^
sin )3 - sin a
MISCELLANEOUS EXAMPLES. LXXVL b. 219
15. Let OF — h be the height of the tower,
05P=j7r~a and 0JP = j7r-fa.
Angle ABP=0AP-0BP=iTr + a-lir + a=^2,a,
AB = OB- OA = hcot (45° -a)-h cot (45° + a)
sin 2a
j cos (45° - g) COS (45° + «) ^ _ , _
(sin (45"° - a) sin (45° + a) i ~ si:
(45° -a) sin (45° + a) ^ sin (45° + a) sin (45° - o)
,2 sin 2a ^,, _
_ ji — =^2h tan 2a.
COS 2a
16. Through C let the horizontal plane A'B'G be drawn and let AA\
BB' be vertical lines.
Then B'CA' = e, &nd ACA' = \ BCB' = fi, AB = a = AG. ljeiBC=x.
Draw BK horizontally to cut AA' in K,
Then AB^=BK^ + AK^ = (B'A')^+{AA' -BB')^
= {B'(P + CA'-' - 2B'G . CA' cos 6} + (a sin X - a; sin fif,
or a^ = (a;2 cos^/x + a^ cos^ \ - 2aa; cos fx cos X cos 0) + (a sin X - a: sin fx)-
= a;2(cos^;u + sin^^x) + a^ (cos2X + sin^X) - 2ax {cos ix cos X cos ^ + sin X sin fi} ;
.*. x^ = 2ax {cos /JL cos X cos ^ + sin X sin /x } ,
or x = 2a cos 6 cos X cos /x + 2a sin X sin /x
=acos^{cos {X - fi) + cos {\ + fx)} +a{co8(X-fx) -cos(X + /Lt)}
= acos(X-)u) {1 + cos d}+a 8m{\ + fx){cosd-l}
= 2a {cos (X - m) cos2 J^ - cos (X + />t) sin^^}.
17. b^ = c^ + a^-2caco8B; .\ a^-2caco8B + c^-b^ = 0.
If aia2 are the two values of a, we have from the theory of Quadratic
Equations in Algebra,
(i) ttj ■^a2=2c cos B. (ii) aia2=c^ - b'^.
(iii) From (i) (aj-f a2)^ = ic^co8^B.
From (ii) 4aia2 cos^P = 4 (c^ - feS) cossp ;
.-. (a^ + flg)^ - 4aia2 cos^ JB = 46'^ cos^P,
i.e. ttj' + 2aia2 + ^2* - ^a^ag cos^B = 46^ coa^jB ;
.-. aj^ - 2a^a^ (2 cos2B _ 1) + aj^ = 462 cos^ B ;
.-. aj2 - 2aiaa cos 2B^a,^=W cos^B.
220
MISCELLANEOUS EXAMPLES. LXXVI. b.
(iv) Vide fig. iii. E. T. p. 216. If a perpendicular be drawn from
the middle point of AB, the centres of the two circles lie in this perpen-
dicular; centre of circle circumscribing ABGo being the point where the
perpendicular from the middle point of BC^ intersects the perpendicular
from the middle point of AB, and the centre of the other circle being the
point where the perpendicular from the middle point of BC^ intersects the
perpendicular from the middle point of AB. The length of the distance
between the middle points of BC^ and BC^ is
2 *
Let X be the distance between the required centres, then
%-
2x
-2 = sin^,
2sin.B
(v) The diameters of the circumscribing circle is
Sin B sm G
which is the same for both triangles. Therefore the circles are equal.
(vi) E. T. p. 216, fig. 111.
262 * ■ 2&3 '
but &2 ^ ^2 4- aj^ - ^2a^c = c^ + a^^ - J2a^c ;
cosC,^C, = ^^±^^--^^l-
.-. 2&2zz: 2c2 + a^2 + ^^2 _ c^2 (u^ + a^)
= 2c2 + ai2 + a22-2c2 ^
.'. cos G.
^1 + «2'
2 •
.see (i)
18. AI2 bisects the exterior angle at A^ so also does AI^, .*. 12^-^3 ^^ ^
straight line. I^A bisects the angle A and the bisectors of the angle A and
of the external angle at A are perpendicular.
MISCELLANEOUS EXAMPLES. LXXVI. b. 221
The angle GI^B = 180° -I^GB -1^30= BG'I-hC'BI=iB + ^0=90° -^A;
.-. the angles of I^I^I^ are 90° - i^ , 90° - iJ5, 90° - J C, so also are the angles
of each of the triangles IiBG, l^GA^ l^AB.
ABainBAI^ __ AB cos ^ A
n BI^A cos BI^A " cos J G sin J G
2cGoahA 2a cos i^ , . ,.,
= — ^ — A- ~ — f— = a cosec hA (i).
sm C sin ^ ^
(iii) Similarly I^I^ = c cosec ^ C and Iil^ = b cosec ^B.
Area of IiIoI^=^{I^I^ x I^I^ sin -fgVs)
_6c cosec J C cosec i^ sin (J TT- J ^) _ be cos J^
"^ 2 ~2sinJCsini5
- y r-^t
abcsjs
^ / j (g-a)(s-b)(g-c)(s-a) |
' \/ ( ab ,ca f
Now
j (g-a)(s-b)(g-c)(s-a) ] 2V{(s-a)(s-Z>)(s-c)}
( ab ,ca
abc8 _ abcs _ sa
' 2S ~~ be sin ^ ~" sin A '
abe^Js _ abe,J{8(8-a){8-b)(s-c)}
2j{(s-a)(s-b){s-c)} 2^{(8-a)(s-6)(S"C)(s-a)(s-6)(s-c)}
sin J^ sin JJ5 sin J O '
(iv) The radius of the circle circumscribing Iil^^
I^I^ _ c cosec ^C ^ c ^ c
~2&mI^I^I^ 2sin(j7r-JC) 2 sin J C cos JO sin C"" *
19. In 18 we proved that ABG was the pedal triangle of I^I^L^, and
.*. we have that in the above figure, making the necessary alterations in the
letters, ^ = 90° - i FDE or FDE = Tr-2A,
and BG=EFBeo{90°-iFDE) or ^F=acos^.
(iii) ADxa=:2S = BExb=GFxc;
1 1 I_^ ^ e _a + b + c 8 _1 1
'''AD'^BE'^CF~2S^2S^2S~ 2 S~S~r'
(iv) JD = ^, BeJ4^ GF=^;
a c
AD^ 45f2 be be
•• BE.GF a2 ^^452 a2*
(v) The triangle AEF is similar to the triangle ABG, and its sides
are respectively acosil, bcoaA, ccos^; therefore the radius of the circle
oircums' lUing AEF is R cos A.
222 MISCELLANEOUS EXAMPLES. LXXVLb.
(vi) Badius of the circle circumscribing DEF
FE FE
see (ii)
. see (i)
2sinFD£ 2sin(7r-2^)
a cos A 2R sin A cos A
~2sin2]i'
_ R sin 2^
~'2sin2^
20. a2 = 62_|.c2_26ccos^=:62^_c2_26c(l-2sin2J^)
= &2 ^ c2 - 26c + 46c sin^l = (6 - c)2 |l + ^^^ sin2 |l .
Let tan2 d = -pr^^ sii^^ 4 »
then a2=:(6-c)2{l + tan2(?} = (6-c)2sec2(?; /. a = (6 - c) sec ^.
21. a2 = 62 + c2 - 26c cos ^ = 62 + c2 - 26c (2 cos2 J^ - 1)
= (6 + e)2-46ccos2|=.(6 + c)2|l-^^cos2^J.
46c A
Let j3 be an angle such that sin2^= cos2— ,
then a2=(6 + c)2(l-sin2^) = (6 + c)2cos2^; /. a = (6 + c)cos^.
2J(6c) . A 2^/(347x293) sin 19° 51'
= 7 sin - = -^^^-^ -f ;
6-c 2 54
.-. L tan 19 = log 2 + i (log 347 + log 293) + L sin 19° 51' - log 54
= -30103 + \ (2-5403295 + 2-4668676) + 9-5309151 - 1-7323938
= 10 -6031457 = L tan 75° 59' 51" ;
.-. 6> = 75°59'5r.
Now a = (6 - c) sec ^,
.-. log a = log (6 - c) + L sec ^ - 10
= log 54 + L sec 75° 59' 51" - 10
= 1-7323938 + 106162489 - 10
= 2-3486427 = log 223-17 nearly;
.-. a = 223-17 nearly.
^ „ , . ^ 2^hc A 2^(347x293) ^._^,
Or, from above, sin fl= 7^^^^ cos — = ^^ ^,^ cos 19° 51';
'^ 6 + c 2 640
.-. L sin /3 = log 2 + J (log 347 + log 293) + L cos 19° 51' - log 640
= -30103 + \ (2-5403295 + 24668676) + 9-973398 - 2-80618
= 9-9718466 = L sin 69° 35' 30";
s./3=69°35'30".
MISCELLANEOUS EXAMPLES. LXXVI. b. 223
Now a=(b + c) cos ^ = 640 x cos 69° 35' 30" ;
/. log a = log 640 + L cos 69° 35' 30" - 10
= 2-80618 + 9-5424624 - 10
= 2-3486424= log 223-17 nearly;
.-. a = 223-17 nearly.
22. Now
a _ 6 _ c
sin A ~" sin B ~ sin G '
a h-c
sin A sin B - sin G *
and sin B - sin 0=2 cos h(B + G) sin i(B-G)
= 2 siniA sin i(B - G);
_ (b-c) sin A _(b-c) cos ^A
•'• ^~2sini^sini(^-C) ~ STfpB^Op
h -\- c 6 + c
If = tan~ir tani^, then tan0=r tanj^.
o-c o-c
h-c
But since i2in\(B-G) = - — cotj^,
.-. iQ,n\A = ^^ Goi\(B-G)\
.-. tan0 = J-±-^^%oti(B-C);
^ b-c b+c ^^ "
.-. tan = cot J (5-0),
therefore and \(B-G) are complementary angles and
sin \{B-G) = cos ;
ih - c) cos i A
.-. a=' ^^ — — ^— ;
cos
.*. loga=log(6-c)+IiCos J^ -Lcos0.
23. Since BD = GD, the triangle ABD = triangle .4 GD ;
therefore 2 x area ABD = area -4BC7 ;
.-. 2 X (7 X iiZ) X sin BAD = 6c sin ^ ;
. ^,^ dsin^ 6sin^ n << 0^^
••• ^"»^^^ = ^:i^ = VP^T2?r^^-) Ex. 11. p. 205.
h sin ^ b sin J[
^ x/(62 + 62 ^ c2 - a2 + c2) ^(62 + 26c COS ^ + c2) •
224 MISCELLANEOUS EXAMPLES. LXXVI.b.
24. From E. T. p. 235 (vii.) 11, = -^ = ^^^^ = 4iJ sin M = ^.
similarly AR sin iB = y; and 4^RsmiC=z,
xyz = 64i23 sin J ^ sin J B sin \ (7,
d (x2 + 2/2 + -^2) = 32^3 (sin2 \A + sin^ J 5 + sin^ \ G)
= 321^3 (1 _ 2 sin i ^ sin 4 B sin i G) [LXXIII. 25.]
= 321^3 - 642^3 sin i ^ sin J i? sin J C ;
.-. xyz + tZ (x2 + 2/2 + z2) = 327^3 ^ 4^3^
Qrt qIj Q/*
25. The sides , , are in the proportion of a, &, c; i.e. the
^^ s-a s-a s-a
two triangles are similar and they are therefore equiangular. The area oi
the triangle ABG is Ja6 sin G. The area of the second triangle
= iJ/ i\ -7 r\C sm(7=p^ r^smC.
^ {(a-b) (a~b)) (s- a)^
Badius of circle escribed to side BG (i.e. a) of ABG
_ S _ab sin G
~s-a~2{s-a)'
Radius of inscribed circle of the second triangle
area of the triangle
~ half its perimeter
^s^ab . _, . \ sa
(s - a)2 ^ / o _ ^
^ js^ab
(.-a)2
Therefore the circles are equal.
26. Let QA=x, QB = y, QG = z; angle AQG=SLngle AQP wangle BQC
= 120°, cos 120°= -i, sinl20°=W3,
a^=BQ^-^QG^ + BQ,QG,
b^ = AQ^+QG^ + AQ.QG,
c^ = AQ^ + BQ-^ + AQ.QB;
.'. a^ + b^ ■\-c^=2x^ ^2y^ -\-2z^ + xy + yz + xz.
Let the area of triangle ABG=A\
.'. 2 A = yz sin 120° + zx sin 120° + xy sin 120°,
A = i^S(xy + yz + zx),
d(xy + yz-\-zx) = A J3A.
Now 2(x^-{-y^ + z'^) + {xy + yz + zx) = a^ + b^-{-c^,
2(x^-\-y^ + z^)-\-A(xy-{-yz + zx) = a^-hb^ + c^ + 4:jSA;
.-. x + y + z = J{i(a^-{-b^ + c^) + 2^3A}=d;
,', y + z = d-x, y^+2yz+z^=(d-xf; but y^ + z^ + yz=^a^;
.'. yz = (d-x)^-a^
MISCELLANEOUS EXAMPLES. LXXVI. b.
225
also yz-\-(xy-{-xz) = i,JSA; .'. yz = ^jSA -x(y+z),
but x + y-{-z = d; .'. yz = ^A,jS-x(d-x);
.-. (d-xY-a^=^A^S~x(d-x);
.-. d^-dx-h^=iA^S;
/. dx = d^--a^-iAj3 = i(a^ + h^ + c^) + 2^3A-a^~-^As/'S\
.'. dx = i(h^-\-c^-a^)-\-^A^S;
_ JS{b^-hc^-a^)+4.A 5^2
. ^. 4V2A+^6(62 + (j2_tt2)
. . v^ — .
2 {12 ^3A + 3 (a«+ 62+ c2)}*
27. When sin A and cos A are both known then the different possible
values of A differ by 360°.
4 SfiO°
Hence the different possible values of — differ by .
a A
So that if - be the least positive value of — the different angles are
n ^ n
a m360°
n n
In the figure let P^OR be ""; let P^OPq = ^^=zP^OP^ = P^OF^, etc.
Then all possible values of A are given by the lines OPo, OP^, OP^... and
these angles in general have different values for their sines ; and there are
n of them and no more.
28. Let AB be the given arc ; R any point in it ; O the centre of the
circle; OA and OB the bounding radii so that angle AOB = a. From R
draw RC, RD perpendiculars to the radii; join OR; let angle AOR = ^ and
angle i^ Oi^ = a - /3. Therefore
CD : OG = 8in a : cos/S,
and OG = ORGOBp; /. CD : rcos)3=rsina : rcos/S.
L. T. K. 15
226 MISCELLANEOUS EXAMPLES. LXXVI. b.
29. Let be the centre of the smaller circle and B the centre of the
larger ; E the point in which a tangent touches the smaller circle and D the
point in which the same tangent touches the larger; produce BO and BE to
meet at A ; from the point E draw EG parallel to ^J5 to meet BD in G.
If 6 is the angle between the tangents it may be easily shewn that angle
BAD=ie. Jjet AB = x, OB = EG=a + h, GD = BD ^^ EO^a^^b.
Then ED^ = EG^- GD'^ = {a + b)^- (ar>^b)^ = 4:ab,
^ ,^GB a-^b 1/ /a /b\
30. E. T. p. 234, fig.
Let AI bisect EF at right angles in K ; then z FEI= 90° - FEA = ^A ;
.-. EF=2EK=2EIco8KEI=2rcosiA.
Similarly, FB = 2r cos J B and BE = 3r cos J (7 ;
.-. EF : FB :BE :: 2rcosi^ : 2rcosiB : 2rcosiG
:: cos J^ : cos JJ5 : cosJC
31. (i) BB : D(7 = c : b,
,\ BB '.BB + BG = c :b + c\ :.BB=-^.
6 + c
AB sin 5 , ^ ac sin B be sin ^ 2bc , ,
-rr^ = . ., y , :.AD — ^ r^ — - = — — — = :; COS^^.
BB smj^ (6 + c)smJyl (h + c)8m^A b + c ^
/••v ^T^ <^^ a sin 5 ,
(ii) GB = r — = - — ; — - , as above.
^ ' b + c sm ^ + sm C"
(iii) aBFB : AABB = BF : BA = a : a + fc,
AABB : AABG=BB : ^C=c : & + c;
/. ABFB : A^^C^ac : (a + 6)(6 + c).
.-. Area of triangle BBF= 77 ttt ^ •
(6 + c)(& + a)
Area of CDJB =-^ ^^^— tt ; area of AFE = ^^^
(c + a){c + b) ' ~ (^a + b){a + c) '
A DJei'^zr aABG-{A BBF+ A CDJ^J + A AEF)
" \ ~ (b + c)(b + a) "~ (c + a)l(cTfc) ~ {a-^b)(a + c)J
_ 2a6c)Sf
Now
MISCELLANEOUS EXAMPLES. LXXVI. b. 227
abc _ ^ ^ ^
{a-\-b)(b-[-c){c + a) ~ b + c * c + a ' a + b '
a _ sin^ _ sin J^ cos J^ _ sinj^
6 + c ~ sin ^ + sin C ~ sini(B + C)cos J(5- C) ~ cos ^ (B -G) '
abc _ sin J A sin \B sin J C
(a ^^6) (6 + c) (c + a) ~ cos \(B - C) cos i (C - ^) cos ^ {A - B) '
2abc 2 sin ^^ sin J jB sin 4 (7
•'• '^7 Tvn — : — w — : — v^'J)-
{a + b)(bi-c)(c + a) cos i(B - C) co8 i(G - A) cos i{A - B) '
32. Since a, /3, 7, 5 are the angles of a quadrilateral inscribed in a circle
a + 7 = 7r, /3 + 5 = 7r, a + 7 + jS+5 = 27r;
.-. 7 + 5 = 27r-(a + j8) and 5 + a = 27r- (/3 + 7) ;
cos(a + ^) .cos(/3 + 7) . cos (7 + 5) cos (5 + a)
= cos(a + j8). cos(j8 + 7) . cos{27r-(a + ^)} cos {27r-ip + y)}
= co82(a + /3).cos2(^4-7)
= cos2(a + ^) cos2{7r - (a - j3)} =cos2{a-V/3) cos2(a - p)
= {cos(a + ^)cos(a-^)}2 = (sin2a-cos2/3)2 [Ex. xxxv. (25).
= (l-cos2a-cos2j3)2.
33. Since 9-^ and 62 ^^^ values of d, we have
a cos $1 + bain 0-^^ = (i),
a cos d2 + b sin 6,2 — c (ii).
Multiply (i) by sin di and (ii) by sin $2 and subtract ;
.*. a (cos di sin $2 - cos ^o sin 0^) = c (sin ^g - sin ^j) ;
.*. a sin (^^ - ^2) = <^ (sin ^1 - sin ^g) ;
.-. 2a sin J ( ^j - ^2) cos J (^^ - e^) = 2c sin ^ {d^ - d^ cos J (^1 -f ^2) 5
.-. - . cos J (^1 - ^2) = - cos H^i + ^2) •
Now multiply (i) by cos 62 and (ii) by cos d^ and subtract ;
.-. b sin (^1 ~d^ = c (cos ^2 - cos ^j) ;
.-. 2b sin J ((?! - i^g) cos i (6>i - ^2) = 2c sin J ((^^ - 6.^ sin ^ ((^^ -f d^) ;
.-. -cosi(^i-^2) = ^sini(^i + ^2);
.-. - . cos i(^i + ^2)= - . sin 4(^1 + ^2)= ^ cos J((9i - d.y
34. Solving the first two equations, assuming that cos^, cos</), cosx
are none of them zero, we have
cos^ + cos^cosx cos^ + cos0cosx l-cos^x*
228 MISCELLANEOUS EXAMPLES. LXXVI. b.
Substituting in the third equation we have,
1 - cos^ X = cos (cos + cos 6 cos x) + cos 6 (cos 6+ cos cos x) ;
.-. cos2 e + cos^ (f) + cos^x + 2 cos ^ cos cos X - 1 = 0,
and the required result follows.
From E. T. Ex. XLIII. (20) the expression may be put into the form
^+0+x 4>+x-^ x+<9-0 o+<p-x
8 cos — ^ — ^ . cos - — ^ . cos ^— ^r — ~ . cos — - — - .
a Z a 2i
In order that this expression may be zero one of the four cosines must be
zero ; therefore one of the four compound angles must be some odd multiple
of a right angle, i.e. (^ ± ± x) = (2w + 1) tt.
35. E. T. p. 234 fig. The first circle drawn in the manner indicated in
the question will have its centre and the point of contact with the inscribed
circle of the triangle ABC in the straight line Al. Let r^ be the radius of
this circle, then the length of ^I=r + ri + ri cosec J^; and this whole dis-
tance also is r cosec \A ;
.-. ri(l + cosec J^) = r (cosec 4^ - 1);
_ r (cosec J^ - 1) _ r (1 - sin J^)
^"~ l + cosecj^ ~ l + sinj^
Let rg be the radius of the second circle drawn in the manner indicated
in the question ; then, as above, it can be shewn that
^ ^r,(l -8in^^) ^ / 1-sin^^ y
^ 1 + sinJ^ \\-^^m\Aj
By repeating the process we have, where r^ is the radius of the 7ith similarly
described circle,
Nowri + r2 + r3+ r„, when w=go
[1 + smJ^ \l + smj^/ \l+smj^/ J
_ Jl-sin^^ / l-sinj^\|
~^ jl + sinjj:'^ V ~l + sinJ^/J
l-siniJ[ , , 1 , 1N
-'*• ^ ==Jr (cosec J^-1).
• 2sini.l
36. FromO draw OF perpendicular to T^C; then angle 5 OP = angle A
and angle (7BO = 90°-^; angle ^i^O = B- (90°-^) = ^ +B-90° = 90°- (7.
In the same way it may be shewn that the two angles into which AT> divides
the angle A are 90° - B and 90° - G ; and the two angles into which GF
divides G are 90° - A and 90° - B,
BD _ sin (9 0° -0) __ cos C
AD ~ ainB ~" sin B '
Similarly — — = -, — - .
AD sm G
MISCELLANEOUS EXAMPLES. LXXVI. b. 229
.■.BD + GD = BC=a=AD('^ + ''-^).
\sin B sin CJ
1 «in C cos G + sin B cos ^ sin B cos 5 + sin G cos (7
Similarly
AD a sin jB sin G 2B sin ^ sin B sin C
1 sin G cos C -f- sin A cos ^
i^£ ~ 2E sin ^ sin i? sin C
1 _ sin ^ cos A + sin B cos ^ ^
C^- 2i^ sin .4 sin £ sin C '
111 _ sin 2A + sin 2B + sin 2(7
• * ZD "^ ;RE "^ Ci^' ~ 2ii sin A sin i^ sin G
4 sin ^ sin jB sin C 2 ,.„ -r^r^T /«^. -.
= 2ie sinXsin £ sin C = R ' L^^" ^^"- (25)-]
37. Draw the figure ABGD and let the diagonals of the quadrilateral
intersect at ; let
OA=a, OG = c, .'. a-\-c = h; OB = b, OD = d, ,\b + d=k.
The area of the quadrilateral
= ^AOD-{-ABOG + AAOB + AGOD
— J ad sin d + ^hc sin d + ^ah sin ^ + ^^ c<i sin ^
= J (a(Z + 6c + a6 + cd) sin ^ = J (a + c) (6 + d) sin d = ^hk sin ^.
38. If
X _ y _ z
sin X ~ sin F "" sin Z *
a; y cos Z _ z cos F
" sin Z ~ sin F cos Z~ sinZ cos F *
a; _ i/cosZ + 2CosF _^y gosZ + z cob Y
" sin Z ~ sin F cos Z+sm Z cos F ~ sin (F+ Z)
_ y cosZ + 2;cos Y _y cosZ + 2;cos F^
sin (180° -A') sIEz '
.'. x=y cosZ + zcosY,
39. If -S and r are respectively the radii of the circumscribed and in-
scribed circles of a regular polygon of n sides, each of which is a, it may be
seen from E. T. Arts. 283, 284 that
R — \a cosec - and r = A a cot ;r- .
Let X and y be the length of the sides of the given polygons of n and 2n
sides respectively.
I. In the first polygon r = Jx cot - .
In the second polygon Jt = \y cosec — .
From the question R = r, :. 2y cos ^r- = x cos - .
^ 2n n
230 MISCELLANEOUS EXAMPLES. LXXVI. b.
II. Now 3 + J3:4iJ2 = hycot-—:hxcoseG- = ycos---:x
IT TT i. -r
= a;cos- cos 77- : x from I.
n 2n
.-. cos - cos -- = -7—/^ = ^ vo • ^ = cos 15° . COS 30° ;
n 2n 4^2 2^2 2
/. -=30°andn = 6.
n
It will be interesting to the student to observe that this is an instance of
the Trigonometrical solution of a Cubic Equation.
For assume cos 7r- = x. then
2/1
cos^r- cos- = cos ^7- ( 2cos2^ — 1 ] = x(2x^- 1);
2n n 2n\ 2n J
and the equation becomes 2x^-x=-—r-,^ •
40. Since AI^^ bisects the angle A, therefore the points A, Ij I^ lie on
the same straight line; produce AI to I^ cutting the circumscribing circle
at V ; join 1^0 cutting the circumscribing circle at the point K and produce
it to meet the circle again at H, Now I^G bisects the exterior angle at G
of the triangle ABC and IG bisects the angle AGB, therefore the angle ICIi
is a right angle ; and a circle with 11^ as diameter can circumscribe the
triangle IGI^. But from E. T. Art. 282, we know that VG=VI; therefore
V is the centre of the circumscribing circle of the triangle IGIu hence
Now I1O2 - OK^ = I^K . I^H (Eucl. II. 6)
=I^A .I^V. (Eucl. III. 36, Gov.)
But ^i^ = slifi:i' • (see fig. E.T. p. 234)
and I^V=IV= VG=2R sin J^. (E. T. Art. 282.)
Therefore LO"^ - 0P^= .~-\-- . 2JR sin ^A = 2Rr. ;
.-. Ii02-OP2 = 2i?ri, i.e. 1102=2224.22^^.
41. Assume a;=tan J^, y = tB,n ^B, then z = ta,n^G from Ex. (33) p. 194
and4^ + iJ5+JC=^0°; .'. ^+^ + (7=180°; and from Ex. (32) p. 194
tan A + tan B + tan (7= tan A . tan B . tan C,
2 tan 4^ 2tan|JB 2tan|C
^'^' l-tan24^'^l-tan24J5"^l-tan2iC
2 tan 4^ 2 tan 4 E 2ta.niG
~ 1 - tan2i^ • I - tan2i^ * 1 - tan^J G '
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