THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA LOS ANGELES PLANE TRIGONOMETRY, MENSURATION, AND SPHERICAL TRIGONOMETRY. PLANE TRIGONOMETRY, MENSURATION, SPHERICAL TRIGONOMETRY. EEV. J. F. TWISDEN, M.A., Late Scholar of Trinity College, Cambridge; and Mathematical Professor, Sandhurtt College. LONDON AND GLASGOW: RICHARD GRIFFIN AND COMPANY, PUBLISHERS TO THE UNIVERSITY OF GLASGOW. 1860. Sciences Library QA \ TKIGONOMETRY. TBIGONOMETEY is the art of measuring triangles. The meaning of the word, however, has been much extended, so that it embraces the determination of the situation and distance of all the points in a given space in which the situation and distance of some pouits are given. The surveyor measures one or more lines and angles, and finds from these all the other points to be settled by calculation. If we imagine the various parts of the space to be surveyed connected by straight lines, besides the length of the lines and angles which they include, those angles also are to be considered which the various planes to which they relate make with each other. If the geometer has chosen some points of mountains which, for the purpose of the survey, he considers as connected in triangles, they must, as they lie in various planes, be reduced to the horizontal plane ; so that a plan may be drawn on which all these various elevated objects shall appear in one plane. But if we consider the apparent celestial sphere, in the centre of which the observer seems to stand, the various points of the same may be regarded as con- nected by arcs drawn from this centre ; and thus we shall have spherical triangles' as we had before plane ones, which again serve to ascertain the various points on the surface of the sphere. Trigonometry is divided into plane and spherical ; and, in general, teaches to find from three given parts of a triangle (of which, however, in plane triangles, one at least must be a side) the three remaining parts. The application of Plane Trigonometry to Navigation and Astronomy will be found in the treatise on this subject, in the " CIRCLE OF THE SCIENCES," by Professor Young. MENSTJBATION is the art of ascertaining the contents of superficial areas or planes, of solids or substantial objects, and the lengths, breadths, &c., of various figures, eithe r collectively or abstractedly. The mensuration of a plane superfices, or surface lying level between its several boundaries, is easy : when the figure is regular, such as a square or a parallelogram, the height, multiplied by the breadth, will give the super- ficial contents. In regard to triangles, their bases, multiplied by half their heights, or then* heights by half then* bases, will give the superficial measure. The height of a triangle is taken by means of a perpendicular to the base, let fall from the apex or summit. Any rectangular figure may have its surface estimated, however numerous the sides may be, by simply dividing it into triangles by drawing lines from one angle to another, but taking care that no cross lines be made : thus, if a triangle should be equally divided, it may be done by one line, which must, however, be drawn from any one point to the centre of the opposite face. A four-sided figure will be divided into two triangles by one oblique line connecting the two opposite angles ; a five-sided figure (or pentagon) by two lines, cutting, as it were, one triangle out of the middle, and making one on each side ; a six-sided figure (or hexagon) will require three diag- onals, which will make four triangles ; and so on, to any extent, and however long or short the several sides may be respectively. The most essential figure is the circle, of which mathematicians conceive it impossible to ascertain the area, with perfect pre- cision, except by the aid of logarithmic and algebraic demonstration. PLANE TRIGONOMETRY. 1. On representing Lines and Angles by Numbers. If we have a line of any length, -we can represent it numerically by the number of times it contains a given line, which we take to represent unity. Thus if we take a line a foot long to be the unit of length, a line seven feet long can be represented by 7. Of course, the same holds good of any other line. And so when we speak of a line 8, 5, or whatever the number may be, we mean that the line in question contains 8 or 5 of the given unit, as 8 feet, or 5 feet. And of course, if we can represent lines by num- bers, we can generalize the numbers by letters, and thus we can represent lines by algebraical symbols : so that a b c, x y z, &c. may be understood to represent lines. In the same manner as before, if we speak of a line a, we mean a line containing as many units of length (e.g. feet) as a contains units of number. On the same principle we may express angles by numbers or by letters. This is done by dividing the right angle into 90 equal parts, each of which is called a degree, and dividing the degree into 60 equal parts, each called a minute, and the minute into 60 equal parts, each called a second. An angle is then expressed as being so many degrees, with odd minutes and seconds, e.g. 36 degrees, 57 minutes, 31 seconds, (which is usually written 36 57' 31"), in the same manner as a line is expressed by so many yards, with odd feet and inches. Of course, as we can thus represent angles by numbers, we may also represent them by letters, and may have angles A,B,C; where the angle A (for instance) means that the angle contains as many degrees and parts of a degree as A contains unite and parts of a unit. * In the same manner as we may measure lines either by feet or yards, or miles, so we might take, as the unit of angular measure, any other part of the right angle than the T^jth ; and in fact at the end of last century, when the decimal notation was intro duced into France, it was proposed by certain French mathematicians, to make the degree the -j^jjth part of the right angle. The proposition was at no time extensively accepted, and is now quite abandoned. 2. Definition of the Science of Trigonometry. We are thus enabled to express lines and angles by numbers ; and this is the first step towards making calculations in which lines and angles are the data. However, before these calculations can be performed, it is necessary that the relations which exist between straight lines and angles should be investigated. It is the object of the science of Trigonometry to make these investigations. The object of the science will, perhaps, be more clearly stated, if we limit the definition so as to make it correspond more closely to its derivational meaning, by saying that the science of Trigonometry has for its object the investigation of the relations * It is usual to denote angles either by Roman capital letters, A,B,C; or else by Greek small letters, a, )3, 7 6, , ty ^MUs generally the small Roman 3, 6, c, denote lines. This ig, of courf?, only a conventional arrangement. THE CIRCULAR MEASURE OF AN ANGLE. 293 which exist between the sides and angles of triangles and the algebraical expression of those relations. The immediate application of the science is to the calculation of certain parts of a triangle from certain given parts; e.g., having given the sides BA, AC, and the angle BAG of the triangle ABC, we can calculate the magnitude of the side BC. The science has, however, very many other uses besides the one from which its name is taken, viz., the measurement of triangles. 3. The Circular Measure of an Angle. The measures above given enable us to compare arithmetically one straight line with another, and one angle with another. But it is to be observed, that an angle and a line are heterogeneous magnitudes ; and therefore, if we would perform algebraical operations in which lines and angles enter, we must devise some plan of measuring angles that shall express them by means of lines, or of the ratios of lines. In fact, when we speak of an angle (of 57 suppose) it tells us what the angle is, but does not at once give us the means of comparing that angle with given lines. The measure of the angle adopted for the purpose of such calculations, is called the circular measure. It is founded on the two well-known geometrical propositions. (a) That in circles of the same radius the angle is proportional to the arc which subtends it. (A) And that for the same angle, in circles of different radii, the arc varies as the radius. B If a = the arc BC, 6 = the angle BAG, subtended by the arc BC, r = the radius AC. we may express these propositions by the two variations a <>"'(? when r is constant. a 1/5 r when is constant. /. a rO when both vary. or the angle is measured by the ratio of the arc to the radius.' If we take the unit of angle to be the angle which is subtended by an are of tho same length as the radius, then o = ?- r In this case, the angle being measured by the ratio "of two lines, it can enter a calculation in which we are dealing with lines. N.B. We can easily find the number of degrees in the angle which is the unit of circular measure. 294 PLANE TKIGOXOMETRY. Let y = the number of degrees required. Then x is subtended by an are of the length of the radius = r. Now an angle of 180 is subtended by a semicircle ; i.e. by an arc = it >-, where r= 3-14159. _af _ _ jr^ _ \ ' ' 180 vr "* 180" 3-141.50 = 57"29577. 4. The definitions of the trigonometrical lines and ratios. It is, however, generally more convenient, and for our present purposes necessary, to dotermine an angle not by an arc, or by the ratio of an arc to its radius, but by certain straight lines, or by the ratio of certain straight lines to each other. These lines, or, as they are now more commonly regarded, these ratios, arc called respectively the sine, tangent, secant, cosine, cotangent, or cosecant of the angle. We proceed to define these terms. Let AOB be an angle A. Draw OC perpendicular to OA, and with the centre and any radius OA describe an arc of a circle, meeting OC in C. Draw B, B;, perpendicular to OA, OC ; at A and C draw A.t, Cu, perpendi- cular to OA, OC. Then BH is defined to be the sine of the angle AOB, to the radius OA. A.( is defined as the tangent of AOB. O/ secant of AOB. AM . . versed sine of AOB. Hence, also Bm (or 0) is the sine of BOG 1 . CM . . tangent of BOC. Ott . . secant of BOC. Now BOC is 90" - A. And Bz, Cu, OM, are defined as being the cosine, cotangent, cosecant respectively of AOB. Tlenco B = sine A B//J or 0? =: cosine A At = tan. A r, . / to the radius OA. C ~ cotnn. A Ot =. sec. A Ou =. cosec. A DEFINITIONS OF TRIGONOMETRICAL LINES AND RATIOS. 295 To tliis method there is the obvious objection that the sines,~&c., of a given angle have different values, according as they are referred to different radii, accordingly instead of denning the sines, &c., as lines, it is, as above stated, now more iisual to define them as ratios ; by which means all consideration of the radius to which the sines are referred is avoided. According to this method the definitions are given as follow : CAB any right angled triangle having the right angle at C. Then Hence The sine of A is f"l AB The tangent of A is . AC The secant of A is AC AC The sine of B is "AB The tangent of B is A - B C The secant of B is AB BC But because A = 90 B The sine of B is the cosine A. The tangent of B, cotangent of A. The secant of A is the cosecant of A. Hence Sine A Sec. A = BC AB "AC AC AB Cosine A = Cotan. A == Cose. A =: AC AB AC BC BC AB N.B. The angle which with another makes up 90 is called the complement of that angle. Hence B is the complement of A ; and the cosine, cotangent, and co- secant of an angle are evidently the sine, tangent, and secant of its complement. It is plain (Euc. VI. 4) that the values of these ratios depend solely on the angle, and are quite independent of the magnitude of the sides of the triangles. If, then, we can by any means calculate the value of these ratios, which correspond to any angle, these values can be arranged in a table ; and it is plain that, having such tables, if we have given any one of the ratios denned above, we know the angle ; and vice versa, if we have the angle given, we know the ratio. Such tables have been calculated on principles to be hereafter explained ; for our present purpose it is sufficient for tis distinctly to understand, that if we have given the numerical value of any one of the ratios, we know the angle that corresponds to it, and vice versa. 296 PLANE TRIGONOMETRY. 5. On the relations between the Trigonometric Ratios of the same Atnjlc. Let A B C be a right-angled triangle, A any given angle. C the right angle. Then A C- + B C 3 >= A B 2 (Eucl. 47 .) AC 3 BC~ .-. cos. 2 A + sin. 2 A = 1 .... (1.) BC Again tan. AzrjQ AC cotan - A== BC- ;. tan. A, cotan. B A =: 1 (2.) Again, B C tan.A = ^ = AB AB /. tan. A = -.-. . . f.*jV cos. A AB I Again, secant A = -- = -- AB .-. sec. A = r (4). cos. A In the same manner it may be easily proved that , cos. A /r . Cotan. A = . (5). sin. A 1 Cosec. A = - - (6). sm. A Sec. A /_ . = tan. A (7). Cosec. A It is of very great importance that the student be familiar with the relations we have just established. He will therefore do well to perform the following exercises : Show that (1). sin. A = v/ 1 cos.-A. f)\ fan \ * ->~T 1* (Z), tan. -i cos.-A (3). tan. A cosec. A = sec. A. (4). tan. A 4- cotan. A= -. r -r- x ' sin. A. cos. A i (5). cosec. A sin. A = cos. A cotan. A. (6). 1 -f co3. A _ 1 sin. -A ~~ 1 cos. A' TRIGONOMETRICAL RATIOS OF ANGLES. 297 1 + cotan. "A. w> sin. -A' (8). Express each of the trigonometrical ratios of an angle A in terms of the sine of A . There are two or three angles, the numerical values of the trigonometrial ratios of which can he easily determined. These angles are 45', 60, and 30. 6. To find the Trigonometrical Ratios of an Angle o/45. A B C, a right angled triangle. C the right angle. If A = 45, then A = B, and A C = B C. Now, A C 2 + B C 2 = A B 2 . (EucL I. 47). .-. 2 AC 2 = AB 3 , c or 2. B C 2 - A B-. BC 1 sin. 45 = AB = V2 . BC tan. 45 =77-, = 1. AB sec. 45" = -^ = V 2, and cosin. 45" = sin. (90 4i>) = sin. 45. .'. cos. 45 = 1/2 Similarly, cot. 45 = 1. cosec. 45 3 r= \/ 2. 7. To find the Trigonometrical Ratios of an Angle of 60'. ABC, an equilateral triangle. The angle A B C is one of 60". Draw A D per- *. *-* Wj -li UUCUUMJCKIM lilJ.UiJ.glC. _L HU UliJ^il Ji JJ V 1O U11U \Jt. \J\J AJLG.YJ -iX U L pcndicular to B C. Now B D = 1 B C =. AB and AD 3 = AB 3 BD 2 =: \. AB'. j A D = !_. A B. 2 . __<, AD sm. 60 == -- =: AB tan. 60 = _ = -f BD i sec. 60 = *H = 2. AB cos. 60 = . = A AB BD 1 = 1/3. AB 2 cosec. 60 = -T-^FT = ~~r~. 298 PLANE TRIGONOMETRY. Since 30 = 90" 60 D we shall have sin. 30'"= cos. 60' = . 1 2 And similarly tan. 30 = -7-5 sec. 30 = r y 6 v A V 3 cos. 30' = -. Cotan. 30 = y 3. COMB 60* = |. a 8. Generalization and Extension of the Principles and Definitions previously laid down. The definitions above given hold good for angles that are less than ninety degrees ; the definition, both of an angle and of the ratios which determine it, admit of and require extension ; the nature of- which extension and the principle on which it is made wo will now proceed to explain. 5). The use of tin Negative Sign to denote position. Let A B be a line, the length of which is a. Let ' c B C be a line, the length of which is b. Then it is plain that A C is a b. This distance, A Of is arrived at by measuring a distance (a) to the right from A, and then measuring another distance (b) to the left from B, the -{- a and the b being measured in opposite directions. It appears then that when a stands for a line measured from a given point in one direction, a will stand for a line of the same length measured in the opposite direc- tion. In other words, the magnitude of the line is determined by the number of units in a, while the direction is determined by its sign. It is generally understood that + a signifies a line measured to the right of a given point, as AB, and therefore that . a signifies a line ; , measure to the left of the fixed point, as AB 1 . 10. Extension of the Definition of an Angle. We now proceed to extend the definition of an angle. An angle, as defined by Euclid,/, c. as the inclination of one line to another, must be less than two right angles. But if we regard an angle as the space swept out by a right line revolving in one plane, about a fixed point in a given straight line, wo clearly remove the limit imposed by Euclid's definition on the magnitude of the angle. P Thus if A be the fixed point in the fixed line AB, AF the movcablc line, let the argle BAP, according to Euclid's definition, be A. Now it is plain that in one revolution A P passes through an angle equal to four right angles, or 360. A Moreover, A P will always come to its present position after one, two, or any number of revolutions ; and therefore, according to our extended definition, BAP may be either A or 360" + A, or 2 X 360 + A, or, generally, 360 n + A, where n is any integer. ANGLES AND RATIOS FURTHER DEFINED. 299 11. Negative Angles. In the same manner as we hare shown that + a and a mean equal lines measured in a contrary direction, so + A and A will mean angles measured in contrary directions. Thus, AB and AP have the same meaning as before. Let AP' be so placed that F A B = P A B ; then if PAB=A,P'AB = -A. It is plain that if AP' comes into the position A P, it must revolve through an angular space of 360' + A, as denoted by the portion of a dotted circle in figure, or through a certain number of total revolutions besides 360 -j- A. Hence BAP may also be represented by 360 + A 360 X 2 + A, or generally by 360 X n + A, where n is any positive integer. Hence we conclude that if A be any geometrical angle, its most general trigonome- trical form will be. n x 360 + A, where n is any positive or negative integer whatever. 12. Extension of Definitions of Trigonometrical Ratios. "We now proceed to consider the trigonometrical ratios of angles greater than a right angle. "We shall, in the first instance, confine our attention to the sines and cosines of angles, less than four right angles. "We have already explained that PJT AP AN Sin. B A P = Cos. BAP = AP Now as AN is measured along- AB to the right, AN is positive. And if we reckon lines measured upward, from A towards C positive, it is plain that PN, being measured parallel to that direction, is positive. The signs of the sine and cosine of an angle less than ninety dagrees are then, by this way of reckoning the signs of the measurements, positive, as they should be. Now if we consider an angle BAP,, it ie clear that P, N' stands in the same relation to BAP, that PN does to BAP. Hence we define P W Sin. B A P, = AP, Co, BAP, = 300 PLANE TRIGONOMETRY. It is plain that P, X' is positive, and A X' is negative. Hence The sine of an angle 7 90 3 Z. 180 is positive, and the cosine of an angle 7 90 ^/ 180' is negative. In like manner, if B A P, be the angle subtended by the circumference B P, P 3 , Now P 2 X' is negative, and A N' is negative. Hence the sine of an angle 7180" ^270 is negative, and the cosine of an angle 7 180Z.270 :i is negative. In like manner if P., A B signify the angle subtended by the circumference B Pj P 2 P 3 , i.e. 7270 but Z. 360'. Sin. B A P, == ?1? AP 3 Cos. B A P 3 = ^L And P 2 X is negative, and AX is positive. Hence, sine of an angle 7^70' ,/360' is negative : and the cosine of an angle 7270 Z.360 is positive. These four angles which we have considered arc said to be in the first, second, third, and fourth qua- drants respectively. . By means of the above, if we have given the signs both of sine and cosine of an angle, we can tell in what quadrant it must lie. Thus if sine 6 = -\- m and cos. , 6 must lie in the second quadrant ; i. e. ID ust be greater than 90 and less than 180. (13.) To express the Trigonometrical Ratios of any angle in terms of those of an angle less than 90. Again, the trigonometrical ratios of any angle can be expressed by means of the ratios of an angle less than 90'. For if, in the same figure, BAP, B'A P,, B'AP 2 , B AP 3 , are equal to one another, and therefore the lines P X, P, N', P 2 X', P., X, are equal in' magnitude ; as also are A X and A X. If then we take account both of sign and magnitude, Sin. BAP, = -I?- = ?-*L=: sin. BAP. A Pj A P .-. sin. (180 A) = sin. A. P, X' - P X Sin. B AP 2 = - A '-jr- = - A p- = - sin. B A P. .-. sin. (180' + A) = - sin. A. Sin. BAP, = ^ = -, r r = - sin. BAP. A JT -j .&. -L Sin. (360 - A) = - sin. A. In the same manner it is easy to show that Cos. (180 A) = - cos. A. Cos. (180 + A) = - cos. A. Cos. (360 A) = cos. A. METHOD OF EXPRESSING TRIGONOMETRICAL RATIOS. 301 For example, we have seen that cos. GO 3 = . Cos. 120 3 = cos. (180" 60 3 ) = cos. 60 = - J. Similarly Sin. 315 = sin. (360" 45) = - sin. 45 = J, If consider the case of the tangent of an augle, ZA being an angle, /.90". Tan. (180- - A) = Jl (/> -j- i 4- < c/ + i 4- ! o -f <. i c/> -4- _ i * -\- c/> 1 o + i; r, Cosine Tan-gent Cotangent Secant Cosecant CORRESPONDING ANGLES, &C. 303 The student will do well to verify carefully all the results given in this table ; ho will also observe that the trigonometrical ratios illustrate the principle that if a function of a variable changes its sign, it must pass through the values of either zero (.0) or infinity ( oo). As the values of the ratios arc continuous, the ratios increase gradually to their greatest value, and then decrease to their least. Thus, to take the case of the sine of an angle, which we call 0. Sine 6 increases from 0, when = 0' up to 1, when 9 =. 90. It then decreases to 0, "when fl=180 3 ; after which it still further decreases till it equals 1, when 9 = 270, and finally increases up to 0, when = 360". 15. To determine all the angles which have the same sine, or cosine, %c. There is another class of questions presented to us by this extension of our definition of an angle, viz., having given a trigonometrical ratio of an angle, to find all the angles corresponding to it. For example, tan. 6 p. Now, if we did not reckon any angles but those less than 180, as is the case in geometry, we could only have one value of corresponding to a given value of tan. 0. Suppose this value = a. Then, if we take the trigonometrical or generalized conception of an angle, wo shall have another = 180 -f- a. And since no trigonometrical ratio changes either its value or its sign when its angle is increased by any multiple of 360, it is plain that in addition to a we shall have a series of values, 360+ a, 2 X 360 3 + a, 3 X 360 + a + ____ > 36 D + '> and in addition to the value 180 + a, we shall have a series of values, 360 + 180 -J- a, 2 X 360 + 180 + a, 3 X 360' + 180 +', 40.r , .,35', /Co', fc THE DELATIONS BETWEEN THE TKICOXOMKTIUCAI. LTXCTIONS OF DIFFEUKNT -VNGI The formulas we have already proved hold good of the ratios of the same angle ; we now proceed to investigate the formulas which express the relates between t, or more different angles. There is a very great variety of formulas of tins kmd a they admit of an almost infinite number of combinations and modifications. however, all derived mediately or immediately from the following fo Sin. (A + B) = sin. A, cos. B + sin. B, cos. A ..... (8). Sin. (A B) = sin. A, cos. B sin. B, cos. A ..... (9). Cos (A. + B) = cos. A, cos. B sin. A, sin. B ..... (10). Cos. (A - B) = cos. A, cos. B + sin. A, sin. B ..... (U)i These four formulas can be easily rcmcmbcred-and it is of great importance (hat they should be remembered_by observing that the sine of the sum of two angle, .he sum of the product of the sine of the first angle and the cosine of the second, ay the product of the sine of the second angle and the cosine of the first ; while the sine of the difference of the angles is the former product minus the latter. sum of two angles is the product of the cosines of those ang es minus the pro of their sines ; and the cosine of the difference of two angles is the sum of the pro of the cosines and of the product of the sinos. METHOD OF PROVING DIFFERENT FORMULAS. 305 (16). To prow the Formula Sin. (A -f- B) = sin. A cos. B + sin. B cos. A. Let A B be the angle A, B C the angle B ; .-. A C is the angle A + B. In C take any point P, and from P let fall P X, P M, perpendiculars on A and B, and from M let fall M Q and M B, perpendiculars on P X and B. Then ~ Q P OP Q _X , P^ MB , PQ OP Ol 3 OP OP' MB OM OM OT" MB (Tx PM 4- ?_& PM U P And since PMO is a right angle, QPM = QMO MOB, P Q _ B PM Xo\v = sin. A = cos. B sin. (A = = cos. P M O.M sin. A cos. A 0? = B + sin. B cos. A. (17.) To prove the Formula Cos. (A For as before, Cos. (A -f- B) = B) = cos. A cos. B sin. . A sin. B. And Xow _ QM OB OM OX _ OP OB OM OM OP = cos. A OP QM PM OB OP QM OP PM OP = - * MB . . PM T>. = OM = Sm ' A OP - 8in>B ' sin. B. /. cos. (A + B) =. cos. A cos. B sin. A 18. To prove the Formula Sin. (A B) = sin. A cos. B sin. B cos. A. Let A B = A. B C = B. Take in C any point P from P ; draw P M and P X perpendicular to B and A. Draw M B and M Q perpendicular to A and P X produced. Then A C = A B. sin (\ B) = ?- X = ^~ QP _MB_QP_MBO_M_Q L P PM. OF OP OM'OI' PM" OP MB OM . OM OP = C S ' MATHEMATICAL SCIENCES. No. X. 30 '5 PLANE TRIGONOMETRY. Again Z. MPQ = QMB = BO A = A. PM PQ QR . MP = M = C S - A OP = sine B. .'. sin. (A B) = sin. A cos. B sin. B cos. A. 19. To prove the Formula Cos. (A B) cos. A cos. B + sin. A sin. B. MQ cos. (A-B) = , OR Ut OM = C S ' "OP ON _ OR OP ~ OP ^ OP OM MQ . MP " "i. A and OR OM MQ MP OM' "OP "*" MP' "OP S ' B> MP = - OP /. cos. (A B) = cos. A cos. B -\- sin. A sin. B. 20. To extend the above proofs in special casts. The proofs above given arc clearly limited to the cases in which A, B, and A -f- B or A B, are each less than 90. They admit of extension to any case whatever. Thus If A is 7 180o Z. 270', B 7 90' /_ 180=, and A B 7 90 ^ 180'. To show that Sin. (A B) = sin. A cos. B sin. B cos. A. 1 n this case the figure will be the following : AOB is the angle A, measured as indicated by the dotted circle. BOG is the angle B. .-. COA is the angle A B. From any point P in 00 draw PN, PM per- pendicular to OA and OB produced, and from M / draw MQ, MR perpendicular PN and OA. Then / Sin (\-B}- " MR " "PO PQ PO PQ PM PM' "PO MR POM MO' "PO"' cos.. MPQ = cos. AOM = cos. (A 180^) = co.?. (180 C A) = - cos. A. PM = sin. POM = sin. (180" B) = sin. B. MR Similarly ^ = sin. MO A = sin. A OM pjj = cos. POM = cos. B ; .'. Sin. (A B) = sin. A cos. B sin. B cos. A. PRINCIPLES OF PROOF. 307 In the same case, cos. (A-B)r=-^p = - NR - OR QM OR OL' ~*~OP : MQ OP MQ MP OJl MP ' OP+OM OM P' Xow =rp- ; = sin. MPQ = sin. MOA == sin. A. MP OP OR OM OM OP = sin. POM = sin. POB = sin. B. = cos. MOA = cos. A. = cos. POM = - cos. B. .'. cos. (A B) =2 cos. A cos. B -f- sine A sine B. Again, to snow that sin. (A -f B) = sin. A cos. B + sin. B cos. A. When A + B 7 270' Z 3GO D . A 7 90^ 180 B7180=./27 B 7 90= 21. Taeprinc'plt on which the proof may lie considered as established generally '. The above examples will be sufficient to satisfy the reader t^a 1 : the four formulas 308 PLANE TRIGONOMETRY. above given hold good for all angles whatever. It is -worth while to observe, however, that, independently of these examples, this follows from the circumstance that the extension given above to the definitions of angular magnitude, and of the trigonometric ratios, are made in strict accordance with the extension given to the meaning of the negative sign in algebra. Thus, a b primarily signifies that the number b is to be subtracted from the numher a ; if therefore b be greater than a, b is impossible, unless we generalise the definition of the negative sign. If we do this so as to mider n b susceptible of meaning for all values of a and b, then whatever theorem we prove to be true of a b, and its combinations with the restriction, will be equally true of a - b, and its combinations without the restriction. In like manner, if we prove a trigonometrical formula to hold good for all geome- trical angles, those will equally hold good of the angles when defined according to the trigonometrical conception of an angle explained above. The principle which we have to guide us in all these generalizations is called " The Principle of the Permanence of Equivalent Forms," and is that which lies at the root of all extensions of merely Arithmetical Algebra, as explained in the treatise on Logarithms and Scries. The reader who wishes to see a full account of the application of this Principle to Trigonometry, will do well to consult Dr. Peacock's Algebra, vol. ii., p. 144, &c. 2nd edition. 22. Relation between the Four Fundamental Formulas. It is to be observed that the last three of the four formulas given above can be derived from the first of them. Thus, sin. (A + B) = sin. A cos. B -f- sin. B cos. A. ' For B write B. Now sin. ( B) , sin. B. And cos. ( B) = cos. B. /. sin. (A B) = sin. A cos. B sin. B cos. A. Again cos. (A + B) = sin. (90 - A - B) = sin. ( 90 A) cos. B sin. B cos. (90 A) = eos. A cos. B sin. A sin. B. Again (Cos. (A - B) = sin. (9lJ~^A + B) sin. (90" A) cos. B -f cos. (90 - A) = cos. A cos. B -\- sin. A sin. B. 23. Formulas derived from the Fundamental ones. From these four the following formulas of frequent occurrence" can easily be derived : Sin. A cos. B + sin. B cos. A = sin. (A -}- B) Sin. A cos. B sin. B cos. A = sin. (A B). .-. adding 2 sin. A cos. B = sin. (A -\- B) -\- sin. (A B) ; and subtracting 2 cos. A sin. B = sin. (A -{- B) sin. (A B). Similarly, 2 sin. A sine B = cos. (A B) cos. (A -f- B) (12) 2 cos. A cos. B = cos. (A B) + cos. (A + B) (13) VARIOUS FORMULAS AND EXPRESSIONS. 309 The same formulas are of frequent occurrence in a different form. Evidently + _,0-4> 2 ~2~ ^ _ 1+ <*> _ f 2 2 4- e 4, . e A e .-. m. S = sm. -* cos. -~ -\- sm. - -i cos. . 4- <*> *] . = fl # sin. ^ = sm. ^ cos. - sm. a ^ cos. - *2 2 , . 5J .-. sin. e 4- sin. = 2 sin. fl - "JT *** cos. - Z_? . 2 2 sin 6 - - sin. <^ = 2 cos. ^~_J> :- sm . ft ; . . . (1,5.) 2 2 . Similarly, Cos. + cos. 4. = 2 cos. l_i_* cos. ~ * . . .". (10.) Cos. ri> - cos. 9 2 sin. -- i_ ^ cos. g ~ ^ . . (17.) 21. Formula for the Tangent of the sum of two Angles. Again, we can easily derive from the formulas for the sines and cosines of A and A B, expressions for the tangents of A + B and A B. Thus, , . . . T . _ sin.' (A + B) _ sin. A cos. B -\- sin. B cos. A cos. (A + B) cos. A cos. B sin. A sin. B Divide both numerator of this fraction by cos. A cos. B. , . . _. sin. A cos. B sin. B cos. A /. tan. (A + B) - + - r cos. A cos. B cos. B cos. A cos. A cos. B sin. A sin. B cos. A cos. B cos. A cos. B sine A i_ sine B __ cos. A cos. B J]_ tan. A + tan. B j _ sme_A_sine_B 1 tan. A tan. P7. . . (18.) cos. A cos. B Similarly, Tan. (A - B) - tan. A - tan. B 9 . 1 - tan. A tan. B 25. Expressions in which the sum of three angles occur. We can easily derive from the above expressions for the sines, cosines ... of the sum of three or more angles. Thus, Sin. (A + B + C) = sin. (A+ B) cos. C + cos. (A + B) sin. C = (sin. A cos. B 4- sin. B cos. A) cos. C + (cos. A cos. B sin. A sin. B,) sin. C = sin. A cos. B cos. C + sin. B cos. C cos. A + sin. C cos. A cos. B sin. A sin. B sin. C. In the same manner, Cos. (A 4- B 4- C) = cos. A cos. B cos. C cos. A sin. B sin. C cos. B sin. C sin. A cos. C sin. A sin. B. 310 PLANE TRIGONOMETRY. And hence, . A , T> i p\ _ tan - A + tan. B + tan. C tan. A tan. B tan. C J lI~tan7B tan. C ta^TClan. A toOTtanTl 26. Certain other Formulas. Again, since Sin. e + sin.

cos. ~ = tan. -* cotan. -=^ ..... (20). ^ ^ In like manner, Cos. d> cos. 6 + 4> 6 Cos. + cos. e - --T ..... - There are many similar combinations of the trigonometric ratios besides those above given. These arc of very frcqiient occurrence, and the student who has thoroughly mastered the above will be at no loss in investigating other combinations that may occur in his subsequent reading. I 27. The Sines, Cosines .... of Multiples of given Angles. We have already seen that Sin. (A + B) = sin. A, cos. B + sin. B, cos.'A. This being true of all values of A and B is true when A = B, and .'. when A + B = 2A. .-. Sin. 2A = 2 sin. A, cos. A ____ (22). Similarly, since Cos. (A + B) = cos. A, cos. B sin. A, sin. B. .-. Cos. 2A = cos-A sin. 2 A ..... (23). Now, 1 cos 2 A + sin. 2 A. Add this equation to (23), and we obtain Cos. 2A = 2 cos 2 A 1. And Cos. 2A = 1 2 sin. 2 A. Similarly, Tan.2A= 2tan 'i ..... (24). 1 tan. 2 A Again, Sin. 3A = sin. (2A + A). = sin. 2A, cos. A + cos. 2A, sin. A. = 2 sin A, cos. A, cos. A + (Cos-A sui.-A)- sin. A. -\- 3 sin. A, cos 2 A sin. 3 A. = 3 sin. A 4 sin.'A ..... (25). Similarly, Cos. 3 A = 3 Cos. A + 4 cos 3 A ..... (26), AMBIGUITIES ARISING FROM USE OF FORMULAS. 3Jld l-3tan~A I 28. Determination of Sine, SjC., of an Angle in terms of the Sine, fjC., of the Sub-mul- ples of that Angle. From these expressions we may derive others expressing the sines, &c., of an angle in terms of the sines, &c. of the submultiples of that angle. Thus, writing - for A, we have From (22) Sin. = 2 sin. - cos. -. L L From (23) Cos. = cos 3 -- sin. 2 - = l-2rin..| = 2 cos 3 - 1. 2 tan. | From (24) Tan. . And writing for - we have o From (25) Sin. ^ = 3 sin. ^ 4 s.'a 3 = From (26) Cos. ^ 3 cos. | + 4 cos 3 |' From (27) Tan.

= tllJ 1 . = cotan. 75 Sec. 15' = 2 ^ 2 - . = cosec. 75' V/3+1 Coscc. 16 = V 3 ~ 1 . = sec. 75 2^/2 The student will observe that this investigation of the ratios of an angle of 15, together with those previously investigated, gives the ratios of the series of angles 15% 30, 45, 60-, 75', 90. 316 PLANE TRIGONOMETRY. 31. To investigate the Trigonometrical Ratios of an Angle of 18 3 , 3G 3 , 54, 72. Since 54 =90" -36, if we -write for 18, we have 30 = 90" - 20, and therefore Cos. 30 = sin. 20. /. 4 Cos'O 3 cos. 6 = 2 sin. 0, cos. 0. .-. 4 Cos-0 3 = 2 sin. 0. .-. 1 4 sin. ? 0= 2 sin. 0. 4 Sin. 2 -|- 2 sin. 00=1. /. [4 Sin.-0 + 2 sin. + .- = |. This is a case of ambiguity similar to those above explained, and if we only had the equation 4 Sin. -0-|-2 sin. = 1, we should have the two values of sin. just given, viz., - and -- ^ - . But as we not only have the equation, but also know that = 18, this enables us to choose the only admissible value. Sin. 18 ^ . 4 For the other value of sin. being negative cannot be the sine of 18\ 1 _ 3 1 t/ 5 _ 5 -ft/ 5 i. ~i n _l- 9 7, cos. 18^ = llcnce, cos-ISs =1 sin. 2 18, o 4 Hence, sin. 36' = sin. 2 X 18 = --'- -~ 8 and cos. 36 =~ 4 and sin. 36 = cos. 54, and sin. 18 = cos. 72\ Hence we evidently can obtain the trig, ratios of the angles 18, 36, 54, 72. Again Sin. 3 = (18 15.) = sin. 18" 1 cos. 15' cos. 18 sin. 15". -- 2^2 _ (i' 15 - 1) (V 3 + 1) - (y/ 3 - 1 (y/lO + 2 V~5 8-1/2. Hence we may clearly obtain numerical values for the trigonometrical ratios of the , scries of degrees, 3, 6, 9, 12, 15, &c. TRIGONOMETRICAL RATIOS OF ANGLES. 317 N.B. In reading the preceding pages the student will have observed that in many instances, when a method of reasoning has been applied to one case, it has been merely indicated that the same method is applicable to a similar case. In all these instances he will do well to write out at full length the reasoning in these similar cases. By this means he will ensure a thorough comprehension of this part of the subject, and become familiar with the various combinations that trigonometric ratios can form. In regard to this very subject Dr. Peacock observes : " It should be the first lesson of a student, in every branch of science, not to form his own estimate of the importance of ele- mentary views and propositions, which arc very freqiicntly repulsive or uninteresting, and such as cannot be thoroughly mastered and remembered without a great sacrifice of time and labour." To assist in obtaining this familiarity he may perform the following exercises : (1). i. + cos. 2 A = 2 sin. (GO' - A) sin. (6 + $ = 90' show that (a) Tan. # tan. ^ -f- tan. >|/ tan. + tan. tan. < = 1. (J) Tan. -f- tan. + tan. ty = tan. -\- tan. tan. if + sec. sec.

+ and cos. (0 + $ -j- /o \ jj tn tan, (a x) __ n tan, x cos-* cos 2 (o x) Then, L^l2l) = ?. sin. 2# ;H and tan. (a 2.r) = ; tan. o. + m (9.) If tan. A -\- 3 cos. A = 4, show that A has two series of values, one of which is 45, 225', 40.5^, 585', \/ . . . (i. e. tan. A = 1). (10.) If cos. A + cos. (n 2) A = cos. A, show that A must have the values 30' 330 3 390 J?90^ 750^' n I'M 1 ' n 1 1 ' n 1' ' (11.) If sin. (x + a) + cos. (x -\- a) = sine (x a) + cos. (x o), thcn x must be 45 3 , 225', 405 ..... ( 12 -) If sin. A + sin. (0 A) + sin. (20 -f- A) sin. (0 + A) + sin. (20 - A). 318 PLANE TRIGONOMETRY. (13.) If 2 sin. (e ) = 1, and sin. (0 ) r= cos. (0 + <;>), then we shall hare = 45, =. 15. S,/, m t _ 1 (14.) If Tan. 6 = tan. J and cos. - . / o Then w 2 1 _ ,cos.? sin.* (K 2 3~ " \-os.i + sin. iV- (15.) If sin. 2 2 sin. 2 = . Show that sin. = + -- j^ , and show that the values of 6 are given by the series. 18= 162 378 628' ....:.' 54' 126' 416 486' ...... 198 342 3 558' 702' ...... 234 306' 594' 666' ...... (16.) Show that the scries of angles in No. 9 can be expressed by the formula, m. 180" +45". i *i f i In No. 10, by the formula, . 1 In No. 11, by the formula, m. 180' + 45.' In No. 15, by the formulas, m. 18^ -I- 18\ m. 180> + 54'. 31. On Inverse Trigonometrical Ratios Explanatory . The following notation, which is part of a general system of notation originally proposed by Sir J. Hcrschcl, is very generally adopted, and is very convenient : If tan. e p. Then 6 = tan. -1 p. i.e. tan. ~ l p. means the angle whose tangent is p. In like manner sin. l p means the angle whose sine is p, and cos. l p the angle whoso cosine is p. The system of notation originally proposed was the following : If sin. 6 /> and 0' an angle, the arc subtending which, divided by radius, is equal to p ; then, sin. 6 is the sin. of p, and therefore sin. 0, sin. (sin. 0), and Sir J. Herschel proposed to write Sin. (sin. 0} = sin. 2 6), reserving the notation (sin. A) 2 for the squares of the sin. of 6. Upon this principle, Sin. ( sin. ( sin. ( . . . . sin. ) ) ) :=: sin" 0, and evidently Sin" (sin m 0) = sin"*" 1 or the notation follows the law of indices, and the interpretation that sin ~~ l p must obtain, is that it is the angle whose sine is p. In like manner log ( log ( log. a) is written Iog 3 # and log~ ' a, signifies the number whose logarithm is a. Of this system the only part that has obtained any extensive currency is that given above in the case of the inverse trigonometric ratios. 32. Formulas connecting inverse Trigonometric ratios. There are some formulas in which these inverse ratios occur that are worthy of notice. FORMULAS CONNECTING INVERSE TRIGONOMETRICAL RATIOS. 319 (a.) Tan - 1 + tan - ! = tan - 1 ~t- . 1 urn For if tan. 9 =: i and tan. < == tan. 5 + tan. _ OT -f- a. Now = tan ~ l m = tan ~~ l n. .-. -f = tan. - l m -\- tan. - l Hence, .'. tan- 1 ** X tan-'nrztan-i ^-- W ---- (28). 1 mn 2 tan. l m = tan. * , -- 1 m 1 . and Tan. - l m tan. ! = tan. l - 1 ;w. Again, Sin. - ! w 4- sin. - J = sin. - 1 { ??z \/ I ri~ -\- n \/ 1 TO- j . For let sin. ~ l m = 0, sin. ~ l n = . /. jw = sin. = sin. '- = cos. y' 1 w s = cos. <^. Xow sin. (6 + tf>) = sin. cos. < + cos. 6 sin. + = sin. ! { m V 1 3 + w |/ ! m ~ } Or, Sin. -i m + sin. -1 = sin. -, { m ^TI^j ( . n y i_ w ^> ---- (29). Similarly, Cos. - 1 m cos. - l n = cos. _ i { mn y 1 Cos. ! m -f- sin. -1 z? = sin. _i mw + \/ 1 w 2 y^l^l 33. Examples. Show tnat Tan. -JJ + tan. -i = 45' -+- For tan. -Vl-f tan. - = tan. - 1 2 3 . , , o , 1.1 ^ . _ . O - 1 It is to be observed tliat when we say tan. _ i = 45 we mean that this is one value. All the values of tan. -i 1. are, of course, given by the formula ml8Q 4- 55. Show that (1). Sin. -i _ + tan. - i - = 45. V 5 (2). Sin. -i M== tan. - (3.) Cos. - 1 m + tan. - ! m = tan. - , - 9 1 m 2 ~ m in y 1 in- 320 PLANE TRIGONOMETRY. (4.) 4 tan. -1 V _ tan. -'J 1 119 (o.) 4 tan. -' __ _ tan. = 45 ') 239 (6.) tan. ~ l in -\- tan. l n=. sin. l 1 - (7). If 6 =. tan - ' - - = tan - l - ; then sin. (0 + <|>) = sin. 60= cos. 36 V 3 y 15 OX THE USE OF SUBSIDIARY ANGLES. 34. Explanatory. In making trigonometrical calculations it is nearly always necessary to conduct them by means of logarithms. For the purpose of preparing a formula for logarithmic calculation, it is often necessary or convenient to express the sum or difference of two or more magnitudes by means of a product : this can generally he performed by introducing the sine, tangent, or some other ratio of an angle chosen for that purpose, which is called a subsidiary angle. One or two examples will explain the means employed for this purpose. 35. Examples. (a) Thus, let * 2 = or sin. 6 ~/ 1, which is impossible. (i) The following ease is one that frequently occurs x a sin. A -\- b cos. A. Assume z=: tan. a f thi'n x a (sin. A 4- cos. A) 1 a a (sin. A -f- tan. 6 cos. A) pin. A cos. -<- sin. cos. A = a- cos. 8n\ CA + 6 a cos. 6 THE USE OF SUBSIDIARY ANGLES. 321 (c) Again, if we have ) Assume = tan. 9 a I - X K" 1 -i- tan. 2 Q cos. 2 9 sin. - cos. - 6 -f- sin. 2 6 cos. 9 sin. - & = cos. 2 #. N.B. It -will frequently happen in calculations that we have previously used certain logarithms, and when this is the case the calculation is very materially shortened. Thus, in the above example, suppose we already know log. a and log. b. Then L. tan e = log. b log. a + 10. which immediately enables us to find L. coe. 2 6 ; and therefore x, by only using the tables twice. (d) Sin. A = cos. B cos. C, cos. a + sin. B sin. C. cos. C cos. a Assume --- : ^ - = tan. We obtain sin. C. sin. (rf> + B.) \rr / cos. <[> sin. A = If = l + 1 where f Assume = sin. c 9. Then p- a x = p. sin. 2 - M 1 -(- A/ 1 c-' Assume c. sin. 0. Then *=tan.'|. ) Such an example as the following frequently occurs in Astronomy : If x = MI cos. 9 + n cos. (9 + a). Express x in the form A. cos. (9 + B). x = m cos. 6 + n cos. a cos. 9 n sin. a sin. 9 = (m + n cos. a) cos. 9 n sin. a sin. . / cos. cos. 9 sin. where tan. . cos. (0 + 4>). n sin. a m + n cos. a It will be observed that the expression for tan. $, is not expressed in products and quotients only ; to effect this we must introduce another angle, <'. Thus : n Assume tan = cos. a m Then m + n cos. a = m (1 -f- tan. ' + sin. jQ cos. <>' m (sin. $' cos. 45 + cos. ' sin. 45) %/"2 cos. ' m sin. (ft' + 45) A/ 2 cos. $' n sin. o cos. ' and m sin. (' + 45) V 2 Wi V2. sin. (*' + 45) cos. (6 + - Then a; is in the form required. ; - . n. sin. a cos. < cos.

A/ 2ab. .-. 2 Z 1. a + 6 We may therefore assume ' \ cos. - - = cos. 6 a + b /. C 2 = (a + 6) * (1 cos. Z 0) = (a + b)- sin. _ c = (a + b) sin. .... (37) Or we may proceed aa follows, C C Since 1 = cos. 2 + sm - s -K" C C And cos. C = cos. 2 -- sin. 2 .-. c- = (a* 326 PLANE TRIGONOMETRY. c r - (a 2 + 2 ab + b 2 ) sin. 2 + (a 2 2 ab + b"-) cos. 2 , r* r* (a + 6) 2 sin. 2 + (a 6) 2 cos. 2 . Q * O * i a & C Assume tan. = r- cotan. -. a + 6 2 Q .'. c s = (a + Z>) 2 sin.*-. (1 + tan* 0) 40 Q ( + 6) sin. -jj- or c = /OQI \ cos. i (ol.) Also the calculation of tan. = r. cotan. can be simplified, if we 2 a + b '2 iiave already log. a and log. b by introducing a subsidiary angle. Thus, b b Assume tan. 6 = . a b 1 tan. tan. 45 tan. 6 ' ' a + b 1 + tan. 1 + tan. 45 tan. for tan. 45 = 1- a ~ b = tan. (45 0) a + o and tan. A ~ B = tan. (45 0) cotan . (39) !a " There are several Theorems which can be deduced by means of the relations proved above. The following are a few. (41.) To find the area of a triangle in terms of its sides. In figure (10) we evidently have Area triangle = -J . CN. AB. Now CN. = AC sin. A = b sin. A. be . .-. area = sin. A a _ 6f 2_ Vs. (s a) (< 6) ( c) ( 35 )- ~ 2 be = Vs. (s a) (a 6) (s c) So N.B. If a = Z> =: c. or triangle equilateral, then s = .'. area = 7 If a b. or triangle isosceles THEOREMS. 327 c a - c / a- c 2 .= 2 V t (42.) To find the radius of Inscribed Circle on terms of sides. ABC the triangle. the centre of inscribed circle Join OA, OB, OC. Now area ABC = area BOG + area COA + area AOB. Let r required radius. Then, area BOG = rb area COA = area AOB = 2* a + b + c area. /. r = Vs (s a) (s b) \s c). Fig. IS. The circles which touch one side of a triangle, and the two other sides produced, are sometimes called the suscnbed circles. Let r a r b r c be the radii of those circles which touch the sides a b c respectively. Then if 1 be the centre of the circle which touches the side BC, join 0, A, OjB,- OjC. We clearly have - Area B0 i; C + area C0 lf A + area AO,, C= area ABC. r a a r a b r c -r + ~~ + -- Similarly r a (s a)=^s.(s a) (s b) (s c) \/ s. (s a) (s b) (s c) (s a) ~ ^/ s. (s a) (s b) (s c) <-&) s. (s a) (s b) (s PLANE TRIGONOMETRY. (43.) To find the Area of the Circumscribed Circle. ABC the triangle. Circumscribe a circle about it, the centre of which is 0. Join BO and produce it to meet the circumference in D. Join DC. Then (Euclid III. 21) BDC = BAC, and BCD is a right angle. (Euclid III. 31-) Now BD sin. BDC = BC. .*. If R = radius of circumscribed circle, 2 R sin. A = a. .'. -T V. ( o) (s b) (*<) = a abc R = Fig. 19. s. (s a) (n b) (-1 c) (44.) To find tlie Area of a quadrilateral inscribed in a Circle in terms of the sides. ABCD a quadrilateral inscriptible in a circle, let the four sides AB, BC, CD, DA, be respectively abed. Join AC. Now, if ABC = e ADC = 180 6 (Euclid III. 22.) Hence, AC 2 = a'-' -)- o 2 2ab cos. 6 and AC 3 = c' 2 + d- 2cd cos. (180 6) .'. 2 + t* 2ab cos. 6 c" + d- + 2cd cos. .'. cos. = 2 + 6 2 f triangle ADC (ab + cd)" AD ' PG . sin. ADC =^ sin. (180- 6) 2 2 dc . . - sm. 6 Area of triangle ABC = sin. BC. THEORKMS. 329 .-. The whole area == ao - 2 (ab + dc). Hence area of quadrilateral ) ( J) (s c) (s (45.) To find the area of a polygon of n sides inscribed in a given circle. If AB is one side of the polygon, the centre of the circle, let r be the radius, and n the number of sides. Then area n of polygon n x (area triangle OAB). Now angle A OB = n Draw Op perpendicular to AB, 180 Then A Op = A AOB .-. AOp = . n Then area of triangle = = Ap x Op. . 180 180 Ap = r sin. . Op = r cos. . n n , . , . 180 180 . Area triangle = r- sm. cos. . n n = A - sin. 360 If a is one side of the polygon, a 180 Then A Fig. 21. 180 2 sin. 71 area of triangle 180 180 A 2 18 4 sin.- 5 a 2 180 = cotan. . 4 n n 360 Hence area of polygon = -^ r- sin. wben radius is given, And area of polygon = na" 180 - cotan. \vhen side is given. 4 n The area of circumscribed polygon can in like manner be proved to equal 180 n r - tan. n 330 PLANE TRIGONOMETRY. ON TRIGONOMETRICAL SERIES AND TABLES. (46) General Explanations. The preceding pages give the theory of plane trigonometry. To complete thin branch of mathematical science it remains to explain the mode of deducing numeri- cal results from the above theory, and to perform the actual calculations of the parts of triangles from which the science derives its name. We have already seen that the trigonometrical ratios of certain angles are known ; for example, we know that sin. 45 = T- sin. GO = and some others, and knowing these values we \/z 2 can determine log. sin. 45", log. sin. 60, and so on. Now instead of knowing only the trigonometrical functions of certain angles, we want to know the trigonometrical functions of every angle from up to 90. And as the calculations are for the most part carried on by means of logarithms, we require to know also the logarithms of these functions. These values have been calculated and arranged in tables in a similar manner to the tables of logarithms as before explained; we purpose in the following pages to explain the principles on which this calculation has been effected. In our article on series and logarithms, we found it necessary to investigate certain algebraical series before proceeding to discuss the nature of logarithms ; in like manner we shall find it now necessary to investigate certain trigonometrical series before explaining the construction of trigonometrical tables. In the follow- ing article on trigonometrical series we shall always keep this object strictly in view, a circumstance that will account for the absence of certain series that are often given in treatises of trigonometry. There is a very large number of such series; we treat the series here simply as the means of arriving at the numerical results, and on this principle make the following selection from that large number. The student will do well to observe, in studying any branch of mathematics, in which algebraical formulas occur, that unless he understands the method of obtain- ing arithmetical results from his formulas, he has not as yet mastered the subject. The first article is an example of a limiting value. We would recommend the student to consider it very carefully. The proposition to be proved is that the limiting value of = 1, where, of course 8 is in circular measure. The u following will sufficiently explain the meaning of the statement. When 6 becomes very small, - becomes very nearly equal to 1, and the smaller 8 becomes the 6 raore nearly becomes equal to 1 ; but so long as 8 has any value, however D small, can never actually equal 1. Then the value which limits the value of o - is 1, while the value which limits the values of 6 is 0, and we assume that if V actually equals 0, actually equals 1. The assumption is in point of fact an TRIGONOMETRICAL SERIES AND TABLES. 331 axiom. We cannot, however, discuss the questions here that this statement gives rise to. There are several methods by which the proposition can be proved. The following is a modification of Newton's sixth Lemma. (47.) To show that when = ive must have Let AP ,P be the arc of a circle, the centre of which is 0. AT a tangent, AP a chord. Produce AP to p, arid ATto t, drawPT and pt parallel to AO, draw po parallel to PO. Then angle Apo = APO = PAO. Since AO = PO, and therefore Ao = op, with centre o and radius op describe a circular arc, Ap. Then the angle AOP being equal to Aop, we have arc AP _ arc Ap ~AO Ao ' since each measures the equal angles. Again, draw PN and pn parallel to At. AT PN" Then ^= = sin. AOP. At pn - = = sin. Aop. Ao po ' AT At A O Ao arc. A P _ AT arc. Ap At Pig. 22. Now, suppose P to move along the arc to P,, and suppose pt to remain fixed, produce the chord AP i; to meet pt in p, make the chord AP t to meet pt in p lf make the angle AP Z = Ap 1 o. Then as before arc APj arc Ap x ATj : At so that in all cases the equation (a) holds good. Now when P moves up to A, p moves up to t, and when P coincides with A, p coincides with t, and then p and t coinciding, the changing arc Ap coincides with At, and therefore in the extreme or limiting case arc Ap ~At~ =L and hence on the limiting case, when AP vanishes, arc A P AT = L Now let AOP = 6, where is in circular measure, then arc AP and AO AT _ AO" ~ FIST PO" 332 PLANE TRIGONOMETRY. Now arc AP arc AP PO sin. 6 AT AT PO Hence, iu the limit when 6 equals zero, Q. E. D. Cor. Under all circumstances tan. 6 sin. 6 1 e e x cos. tf ,, . sin. 6 JN ow when = x 6 cos. Hence, in the limiting case, when 0=0, tan. _ (48). DC Moivre's Theorem. (a) To prove that (cos. + V 1 sin. 6) x (cos.

) = cos. (6 + $} + V 1 sin. (0 + $). For by actual multiplication (cos. 6 + V 1 sin. 6) (cos. + V 1 sin. ) = cos. 6 cos. + v 1 sin. 6 cos. + v 1 cos. 6 sin. sin. 6 sin.

+ v 1 H sin. 6 cos. < + cos. 6 sin. ) (cos. ^ + v 1 sin. ^) = j cos. (0 + <(>)+ V 1 sin. (6 + ) J (cos. ^ + V 1 sin. ^) cos. (0 + $ + $) + A/ 1 sin. (0 + <(> + if). And generally if we had n angles 6\ . . 6-2 . . . O n , we should have {cos.ej + A / 1 siu.0] | "[cos. 6-2 + V^lsin. 6-2 I'-'Icos.e.,, + "/ lsin.0 m = cos. (Oi + 2 + 6 n ) + VUTi s i a . (0, + = (cos. + V 1 sin. 6) (cos. + V 1 sin. 6) .... (cos. + V 1 sin. 6) = (cos. e + V 1 sin 0)" . And cos. (0i + 0; + . . . + 0) + A 1 sin. (0j +;+.. + n ) TRIGONOMETRICAL SERIES AND TABLES. 333 = cos. n + V 1 sin. n 6. Hence (cos. 6 + */ 1 sin. 0)" = cos. n 6 + V 1 sin. n 6. (c.) To prove the same theorem when n is a negative whole number. For by multiplication, (cos. n 9 + V I sin. n 6), (cos. n 6 V 1 sin. n 8) = cos. 2 n 6 + sin. 2 TO 6 = 1. (cos. + A/ 1 sin. 0) B now 1 = - / . - (cos 8 + v 1 sin. 6) * = (cos. e + V 1 sin. 0), n (cos. + V 1 sin. 0) - = (cos. n + A/ 1 sin. 0), (cos. + V 1 sin. 0) - .'. (COS.H + A/ i sin. n 0), (cos. + V 1 sin. 0) - " = (cos. n + A/ 1 sin. n 0), (cos. nO V 1 sin. n 0) ; /. (cos. + V 1 sin. 0) ~ " cos n V 1 sin. n 0) = cos. ( n 0) + A/ 1 sin. ( n 0). Since cos. ( n 0) = cos. n sin. ( n 0) = sin. n. 0. Hence the theorem is true when n is negative. (d.) To prove the same theorem when n is fractional. Let n = - (p , - 7> 0V cos. + v 1 sin. ) = cos. > = (cos. + 1 sin. 0) P- n , _ p 6 7 ~^^ .'. cos. ' + V_ i s in. = A/ (cos. + V 1 sin. 6)*> = (cos. + A/ 1 sin. 0) 9 and hence, in all cases (cos. + A/ 1 sin. 0) = cos. n + V 1 sin. n (41.) Whether n be positive or negative, integral or fraction ; which theorem is called De Moivre's theorem. (49.) To express sin. m and cos. in in terms of powers of sin. 9 and cos. 0. Since (cos. m + A/^1 sin. m 0) = (cos + A/ 1 sin. 0) m . m i m. m 1. m 2. m 3 = cos. m jTg COSt m ~" 6 siu< " & + ~ 1.2.3.4 cos m ~ 4 sin. 4 &c. r m . m 1 i. 2 -i + V 1 J. m. cos. 1 "- 1 sin. ; ^ cos. m - 3 sin. 3 + ^ 1 . .4 . O Hence equating possible and impossible quantities, cos. m B = cos." 1 mm 1 _ . . .,. . cos. m ~" sin. - 1.2 m . m 1 . TO 2. m 3 V o o ^ c s- m ~ 4 sin. 4 &c (42.) j, * a o v 334 PLANE TRIGONOMETRY. and m . m 1 . m 2 Sin. m 8 = m cos. "- 1 6 sin. =r ^ 5 cos. w ~ 3 6 sin. 3 + ... (43)" (50.) To express sin. , in terms of. Let m 9 , and .-. - Hence in the expression of the last article for cos. m 6, we have (\ I* -i\ / -A" 1 " 2 2 j^ \ m """^ - 2 /* Bi / sin. J- m I 4 This formula is true for every value of m, however large m may be ; but if we suppose m to become very large, then in the limiting or extreme case when m is infinity wi is zero, and hence in this limiting case * We assume in this article that if o + V 1- Z> = A + J 1 B. where a. 6. A. B. are real. Then a A and B = b. To prove this, suppose A to be unequal to a and let A = a + x. and B = b + y. .'. a + b Ji = a + x + (b + y) Jl .-. o = x + y V 1 .-. - x = y V-l .-. x* = y* .-. z 2 + y = o. Now x"- and ^ must each be positive. .'. x - o and y = o .*. A = a. and B = 6. TRIGONOMETRICAL SERIES AND TABLES. 335 o> sin. 1. Also 1 - = 1. ! -=!. fee. Also ( cos. ) 1. \ m I And therefore . *=!-. ^ + -172^374 - &c....(44.) By reasoning in precisely the same way from the formula for sin. Jit 6 we shall obtain COE. If be so small that we can omit (f> 3 , we shall clearly have sin, =

I. 3 5 J =- tan. - or if, = tan. -1 .r. Then i -^ a ,5 tan. a;=a; + _ &c. . . . (51.) 3 5 If x be positive, this value of tan. ~" 1 x is, of course, that value which is less than -', Hence, if we wish to express all the values of tan. x in terms otV, we ought to use the formula -1 r y3 tf> I tan. x = m IT + { x 4-^ &c. > I 3 5 J Where m is any positive or negative integer. TRIGONOMETRICAL SERIES AND TABLES. 337 (53.) To calculate the Value of TT. i. e. the ratio which the Arc of a Semicircle bears to its Radius. Let a be the arc to radius r subtending an angle of 45". Then since the semicircle subtends an angle of 180,^= - Hence j is the circular measure of an angle of 45. /. tan. . =. 1 4 and 7 = tan. 1 Hence, from equation (51). * ? = tm .- 1 = ] -i + .l-l + ... and from this we might obtain the numerical value of TT. But it is much more expeditiously determined in the following manner, 2 -il -i T ~ 1 5 2 tan. - = tan. - = tan. 5 - l I* ~ _ -il -15 -16 -120 4 tan. ? = 2tan. _ = tan. ^ = tan. 1 144 120 Tan. " _ tan.'' 1 = tan. " H5 - 1 _ -_!_ 120 ' 239 * 119 -' 120 -1 1 ... tan. -' 1 - tan. - tan. ^ -il -il = 4tan. ^-tan. -il -il /. l = 4 tan. - 5 - tan. 1 - 1. I + I !-&e.} 5 3 5 3 5 5 | J = 4 . 5 3 _L _ i._l +1 J_-&c. 239 3 239 3 5 239 5 The expression it will be observed consists of the difference between two series. Their values can be calculated separately, and the value of T. be found from thence in the following manner : (1.) We obtain the value of . . + - 77 &c. 5 o 5* 5 o' to the 12th place of decimals in the following manner. 1 MATHEMATICAL SCIENCES. No. XI. 338 PLANE TRIGONOMETRY. - = '00032 5" i- = -000012 8 -i = -000000 512 -L = -000000 02048 5" 4r, = -000000 000819 513 -ip = -000000 000032 5 la r, = -000000 000001. 5" From these we can easily obtain, first, the negative, and then the positive part of the expression ; and subtracting we obtain the whole value required. _ - = . 002666 666666 _ . - = . 000001 828571 gjj = . 000000 001861 JL J_ = . 000000 000002 15 5 15 002668 497100 Again = 2. o JL . _L = . 000064 5 5 5 ~ = . 000000 056888 9 5 a JL . 1- = . 000000 000063. 13 5 13 . 200064 056951 . 002668 497100 197395 559851 . 789582 239404 Which is the value of decimals 4 / . + &c. ]. to 12 places. (_ 5 3 > J r i 11 (2.) To calculate the value of i _-^r 5-. 239 s T * -L -004184 100418 239 -L. = -000000 073249 TRIGONOMETRICAL SERIES AND TABLES. 339 1 . -000000 000001 i39 __L_ = -004184 100418 239 1 1 _ -000000 024416 T 239 ~~ -004184 076002 Whichisthe value of-! . + r to 12 places of decimals. 1 239 3- 2893 J If we substract this from the result last found we shall obtain . and .'. multi- 4 plying that difference by 4, we shall obtain the value of ir. required. Thus, 789582 239404 004184 076002 785398 163402 4 3141592 653608 The calculation is carried to 12 places of decimals, but is not trustworthy beyond the first 10. Hence (to 10 places of decimals) = 3-1415926536 If we had wanted to obtain the value .of ir to a larger number of places of decimals, we should have had to carry the calculations throughout to a corre- spondingly greater extent. Thus, we can show that ir = 3-141592653589794 "We can very conveniently apply de Moivre's formula to the solution of Binomial Equations, i.e., equations of the form x n + a = (54.) To explain what we mean by the roots of a Binomial Equation. If we take the case x n + a 0, and if a be &, then the equation becomes 0" + 6" = 0, and if x = by. x n = J y n , and then the equation becomes y n + 1 = where the 1 may be either positive or negative. It might seem at first sight that this has only one root, viz., unity, but to conclude so would an error, as a little con- sideration will make quite plain ; for take the case x* 1 = Then x is either -f 1 or 1 ; i.e., has two values. And if we take the case y? I = Then because x* 1 = (x 1) (x 2 + x + 1) we shall have (x 1) (a? + x + 1) = whence x will have values corresponding to x1 = x- + x + 1 = From the former of them we find that x = 1, and from the second that 340 PLANE TRIGONOMETRY. x- + x + 4 = I 1 4- V 3 =- And hence a; lias the three values. _ 1 + V"H"3 1 A/IT and 2 which are three roots of the binomial equation yS [ _ And hence we manifestly have that the roots of y? a 3 = are __ 1 i A/ Q .,__,_ 1 , ^ Q ff CL Clll tl - Ci. ~~2~~ 2 In like manner x n + 1 = has n different roots, which indeed follows from the general principle, that every equation of the n th degree has n different roots, real or imaginary. (55.) To find tJte roots of the Equation, X* + 1 = 0. Since, cos. (2 p + 1) it + A/ - i sin. 2p + 1) -a = 1 whatever be the value of p. provided it be an integer. And since * = 1 we must have x n cos. (2p + 1) it + *J 1 sin. (2|> + 1) IT. a- = (cos. 2j + 1 4- ^ 1 sin. 2 p + I ir)" 2 p + 1 / . '2p + I tt. or x cos. IT + v 1 sm. - 27541 2 p + 1 Hence cos. ' . v + V 1 sin. v. is a root of the binomial equation whatever integral value we may give to p. We shall prove that there are n different values of this formula, corresponding to the different values of p, viz. : 0, 1, 2, 3 n 1, and that there are no more than n values. (a). Let> and q be two values, each < n ; and if possible let 2 p + 1 . 2p + 1 2g + 1 . 2 ;> + 1 cos. 1 ** A* ~ A *" i ' * / 1 -. IT + V 1 sin. . * = cos. - IT + V 1 sm. ic, 94 Hence equating possible and impossible terms, 2 71 + 1 2g + 1 cos. . if = cos. - T. 2p + 1 27 + 1 sin. * = sm. ^ . TT. Hence the angles must differ by some multiple of 2 , say by / x 2 IT. TRIGONOMETRICAL SERIES AND TABLES. 341 2p + 1 2 q + I '. -n= * = . 2 i *. i n - 1 - .'. jp q = Tc.n. a multiple of n, and tlierefore either p or q must be greater than n. Hence all the values of x which correspond to the values of p < n, viz., 0, 1, 2, 3, .... n 1, are different from each other. (&) Again, if we give p a value r greater than n, then the corresponding value of x will be the same as one of those which is given by some value of p, that is less than n, as for instance q. For suppose p = Jen + q. where i is a whole number, and q is less than n. 2 p + 1 . - . 2 p -f 1 Then, cos. - IT + V 1 sin. - it = n n 21cn + 2q + 1 - . 2 Jen + 2q + 1 cos. -- v + v 1 sin. - TT = n n - f sin. (2 * 27 + 1 , - 2 9 + 1 . - - TT + V 1 sin. - IT. cos Hence by giving to p. successively the values of 0.1.2.3 ..... n 1. in the formula 2 p + 1 - 2 p + 1 a: = cos. - it + v 1 sin. - ir. w n we obtain all the roots of the Equation x + 1 = 0. Thus. To find the roots of the Equation x* + 1 = 0. the roots are given by the formula 2? + 1 - - . 2p + 1 x = cos. - g - TT -r v 1 sin. - g - TT. and are therefore respectively TT - 7T - A/3~ + A/ 1 cos. ^-+ v 1 sin. ^--= cos. 30 + v 1 sin. 30 = - ^ - 6 3 7T 3 7T cos. -S" + v 1 sin. = cos. 90 + V 1 sm. 90 = v 1 o o 5 v 5 if . v 3 + v 1 cos. -7-+ A/ 1 sin. --= cos. 150 + V 1 sin. 150 = ~ O ^ cos. -g- + \/ 1 sin. -Q--= cos. 210+ A/ 1 sin. 210 = cos. T + V 1 sin. -3-.= cos. 270+ A/ 1 sin. 270 = V 1 o o 11 IT , 11 IT VS" A/^l cos. Z-+V 1 sin. 5--= cos. 330+ V 1 sin. 330 = ^ bo z Similarly if we have to find the roots of the Equation a 1 = 0. Since cos. 2 p IT + A/ 1 sin. 2 ^ ir = 1 342 PLANE TRIGONOMETRY. wo shall have all the roots given the formula 2 P v , / ; 2 of = cos. [- V 1 sin. (56.) To resolve 3? n 1 into factors.* If oo <*i a* . . . . O-J B _ i are the roots of the Equation a; 2 " 1 0. Then x* 1 = ( x a,,) (x ai) (x 02) ---- (x oj _ i). Now all the roots of the Equation are included in the expression 2 P V / - : . n p cos. - + v 1 sin. 2 1 - 2n 2 n If then we suppose p < n. and a p be the corresponding root n IT - p f a,, = cos. + V 1 sin. ^ Now observe that there will be another root a? n p . which will equal 2 n p ir , - 2 7i p cos. -- s + V l sin. - - 71 yt cos. (2 -^ Hence ir / - . pit aon-p = cos. V 1 sm. - 71 71 where p < n, and so we have a pair of factors x a p and x a 2n - p , which are / r> IT , p ir\ I pit i ; . p * \ equal to I x cos. - v 1 sin. i x cos. + V -l sin. I \ n n ) \ n n / A} ^p = x* 2x cos. - -f- 1, by multiplication. n Where p may take any value, 1-2 3 n I. It will be observed that x a n x + 1. ' aUd * _ 1 = fgft _ 1) ( X * 2XC03.-+1\ \ f (x* 2x cos. + 1 J (x* 2x cos. "~ + 1 J . (52) Corol. Hence, a*. - a-- = (*-a*) (x' - 2 ax co= + o ) . ^ ' a / ' * If ai, 2, *3 . an are the roots of the equation, x* 1 = Then a: n i = (xai ) (x a, 2 ) (x 3 ) . . . . (x a n ) for all values of x. For, if we actually divide am 1 by x i, we shall obtain a remainder ai 1 ; now, since i is a root of the equation z 1=0, we have a, i*-l=0, i. e. x* 1 is divisible by x <*i, or x i, is a factor of z 1. Similarly a:-a2. a;-s .... a; n are each factors of x I, and since x* 1 cannot have more than n factors, it is plain that *-!=(*- !)(*-<* 2) .... (*-n). TRIGONOMETRICAL SERIES AND TABLES. 343 + x 2 a x cos. V For since [ x 2 a x cos. - - + a'' \ . . . (53) V / 22/cos. ~ + U ---- fs/ 2 n 1 IT \ _ cos + 1 J n ) x 2 * I x"" ,\ /a.- aJ IT \ /a? a w IT , \ -- 1=1 1)1 2 - cos. - + 1 1 I 2 - cos. - + 1 I ' ' a-" \ a- J \ a- a n j \ a- re y Now as there arc n factors, if we clear the equation of fractions, multiplying the right-hand side by a 2 " will be the same thing as multiplying each factor by a 2 . Hence : a-:* a-*= to?- a 2 J [a 2 2ax cos. +a 2 ] [x- 2ax cos. - +c \ \ )\ n J \ n J By a precisely similar process we shall be able to prove that = [ x- 2 x cos. ;r- + l) [x' 2x cos. ;r~ + l) ...... [ x- Sa: V 2 J \ 2n J \ 1 - cos - And htnce / . ir A / , 3 IT A / x -" -f a 2n = I x 2 ax cos. ^ \rtrt I x 2 2 ax cos. + a- 1 fa; 2 . V / \ w /\ 2 1 IT A 2 a a; cos. + a- J (oo). (57.) To prot-e that 2 2 - J sin.- ^ sin. 2 -^ sin. 2 ^ . . . . sin. 2 ^T" ~ 2 n ' \Vc have by actual division - = a - a; - ---- x . *J7 ~~~ J. X 1 " 1 Hence, when a; = 1. The limiting value of - will equal 1 + 1 + 1 + *C ~~* J. + 1 + 1 to 2 n terms = 2 n. Again by the last article a*- 1 / * \ f 2*\ _ = (x + 1.) ( x- 2 x cos. ~ + 1 J I a^ 2 a? cos. ~^~ + !! (n 1 ir A x -- 2 x cos. - + 1 I n ) ( 2n _ and hence when # = 1. The limiting value of 344 PLANE TRIGONOMETRY. is 2 (2 - 2 cos. L) ^2-2 cos. ^V . f 2 - 2 cos. n ~ 1 and therefore these two limiting values are equal, and 2 n = 2" (l - cos. ^-) A - cos. *lV . . . (l _ cos. """I 2 IT 2 IT 3 IT T - 1 IT = 2 x 2 sin. - 2 sin. 2 - 2 sin. 2 - . . . . 2 sin.* - - 2 sin. 2 5 TT 2 ir 3 IT n 1 IT 1 sin - 2 2^T sin;: 2^T sin -2 ST ' sin - 2 ~2ir ' < 56 > Jn the same manner we may prove that 2 = 1" n sin. 2 -. sin. 2 -. sin. 2 - .... sin. 2 (57 \ 4 n 4 n 4 n 4 n ^l = (*-2*co...M)(rf.-2*oo-^ + l).... (27t 1 *. \ z 2 2 a; cos. + 1J Let, x = 1, and then obsenring that 7"r / ff \ a,- 2 2 a cos. + 1 = 2 ^1 cos. g - J 4 n. We shall obtain the required formula. (58). To express sin. x in a series of factors. The above formula (53) for the factors of a; a is true for all values of x and a. s z Suppose x = 1 + s ; and a = I ^ & tv ~ It P If Z z" Then*' --2 a* cos. + a- = 1 + ~ + - 2 ^- / J9ir\ 2 2 = 2(l-cos. P ~) + 2^ j>ir 2 2 - /)ir = 4 sin. 2 ^ + 4. -. . cos. 2 x . 2 re 4 w- 2 n pir ( z- P v 4 sin.- ^ . I 1 + -; r cotan.- .T 2 ?i \ 4 7i- 2 n TRIGONOMETRICAL SERIES AND TABLES. Hence the factors on the right-hand side of Equation (53) will be equal to 2z TT f z" TT\ 2 IT . 4 sin." H~ I 1 + A ~- cotan.- 77 14 sm. 2 ;, x n 2 n\ 4 w- 2 n) 2n ( T \ . (w l)w/, 2 2 lir\ 1 + - ; cotan. 2 J 4 sin.- - - I 1* 7-1 cotan.- 1 4 n- 2 nj 2n \ 4n- n f 2s . . IT . . a 2-ir Sir nlir . 4 sm/ - 4 sm.- 4 sin.- - .... 4 sm/ n 2n 2 it, 2n '2n f 2 2 01 . 1-fc-T (2 2 7T \ / Z" 2 1T\ / 1 + Ttf' C0tan ' - 2^J I 1 + 4^ cotan '" ijj' s IT 2ir -ITT =; . 2 -" 1 sm. - ( 2 V \ /, 2 2 2 7T\ / 2 9t 1 IT \ [ 1 -i- - -. cotan. 2 ) ( 1 + 5 s-. cotan. - - I ... I 1 + - r cotan. - - 1 I 4 ? 2 nj \ 4 n 2 2nJ \ 4 w- '2 n / ( z- TT \ ( , z- 2 ir\ / c 2 = 2 z I 1 + r cotan. 2 - 1 I 1 + -^ cotan. 2 - ) .... (1 + - - x \ A ri 2 2 i/ \ 4 rt 2 2 / \ 4 ?i 2 cotan. - 2n Since by the last article 7T . - 71 - 1 T 2 -"- 1 sin. 2 .... sm. 2 - - = 2n. 2n 2n Again **>_a*is/l+ -M - fl \ 2w/ \ and expanding each by the Binomial Theorem. 2 n.2 n 1 2 2 2 n. 2 n 1 2 TO 2 L2 ' Tro 2 " + ~ L2^3 (z 2 . 2 TO 1 2 2 2n. 2n 1 1 . + ?T-S- 2 n 1.2 4 TO 2 1.2.3 op 8n? + ' ] ' 1 2 TO 2 n + __! 2 n 1.2.3 Now (a) and (b) are equal under all circumstances, and therefore their limiting 12 p values when 2 n becomes very large are equal, i.e, when - - .... are each 2 ft 2 91 A 'il equal to zero. -2 n jg T 2 2 J Now since cotan.-^ ' - = 7 . 4w 3 2?i 4.'- TJTT tan. 2 -7T- 346 PLANE TRIGONOMETRY. pjr 2rt tan' ?L tan. 1 2 L 2n But in the limit when 2re is very large ^5 __ 2 And the limit of ~ when 9 o is unity. Hence the limiting value of = 1. f f)Tf tan. *- z" J91T 2^ And therefore the limiting value of -T;, cotan.- - = -^-5- 4.n 2 2n p-r Hence th? limiting value of the expression (a) is 2, { 1 + !1 j {l + i'l f I *il I ir- J L 47T2 J I 971-2 J Again since (6) can be written I 1.2.3 1.2.3.4.5 The limiting value of (&) ia 2s / 1 1.2.3 1.2.3.4.5 Hence, 2 2 2 4 + T2~3 + 1-2-3-4.5 " or writing c 2 for z'-, we have 1 ^4rr, + 1.2-3 1-2-3-4-5 Multiply both sides by 2, and we obtain g 3 2 s / _\ (. __ &_\ ~ T2T + 1.2.3.4.5~~ Z \ ' 7T-J ^ '" 2%-J 2 3 v.-hich is the expression required. In the same manner, by taking x* + a-" and its factors from formula (.55), we can obtain cos. x in factors. Thus, . 2.2n 1 2ft 2 2 ^ = 2 1 2ro 2 2?i 3 2 4 ) 1.2.3.4 " 16n 4 ' J TRIGONOMETRICAL SERIES AND TABLES. 347 And as before the limiting value of 2n + 2 "i*2(l + 3 + 1^74 +....) () Again, since s ~ cos. 1p + 1 TT z 2 , 2 j? + 1 - cotan< - The factors of as 2 " + a 2 " become _ a- STT 5ir . '2m 1 2 2 " sin. 2 sin. 2 sin. 2 - . . . sin. 2 u _. cotan. 2 T-) 1 + 733" cotan -" 1^ ' ' 1 ,!> A 8 a For tan. 6 = 2 tan- 2 .'. tan. 6 > 2 tan. - SB Hence by same reasoning as before, tan. > 2 n tan. tan. . .-. tan. 6 e > e e Now, when n is infinitely large -- TRIGONOMETRICAL SERIES AND TABLES. 349 tan. 2 and r in tins case = 1. or tau. 6 > 9 (2.) To show that sin. 6 > . 4 6 For sin. = 2 sin. -77-. cos. -;p : e = 2 tan. -TC-COS.'- -77 * = 2 tan. (l sin. 2 2 \ 2 tan. _ f /sin. _ 2 tan. f sin. . 1 Now ^ bein S > 1 and | - V being < 1. 1 _ . must be < 1 . 4 4 / fl 2 \ /. sin. e must be > 6 I 1 J .-. sin. > (60.) To calculate the value of sin. 1' and cos. 1'. Let 6 be an angle of 1' measured by the circular measure. Then, _. ~* + 180 ~~ 180 x 60' .'. Taking the value of v previously given, e = -00029088820. e 3 = -000000000006 4 Hence, if we only take in the first ten decimal places, sin. 1' = 0002908882. And since cos. 1' = -v/1 sin. 2 1' = 1 ^ sin. 2 1 sin. 4 1' 350 PLANE TRIGOXOMETKY. = . 99999,99577. COR. Hence, it is plain that if our approximation do not extend beyond ten places of decimals, Circular measure of angle of 1' = sin. 1'. Similarly Circular measure of angle of 1" = sin. 1". . Circular measure of angle of n" = n sin. 1", provided n < 60. (61.) To calculate the Sines of 2' 3' 4' - Since sin. (A + B) + sin. (A B) =2 cos. B, sin. A, we have sin. (n + 1)' -t- sin. (n 1)' = 2 cos. 1' sin. n'. Now cos. 1' = 1 0000000423 = 1 k. .. sin. (n + 1)' )- sin. (n 1)' = 2 sin. n' 2 k sin. n' .'. sin. (n + 1)' sin. n' = sin. n' sin. (n 1)' 2 k sin. n. Tliis formula is very convenient for calculating the sines and cosines of successive angles. Thus, Sin. 2' sin. 1' = sin. 1' 2 k sin. 1'. Sin. 3' sin. 2' = sin. 2' sin. 1' 2 t sin. 2' Sin. 4' sin. 3' = sin. 3' sin. 2' 2 k sin. 3'. a;id so on. It will be observed that the first member of the right-hand side of each equation is given by the former equation. So that the only term requiring multiplication is 2 k sin. 1', 2 k sin. 2', 2 k sin. 3' in each equation. This multiplica- tion can be greatly facilitated by forming a table in which 2 k is multiplied by each digit, thus : 846 1692 2538 3384 4230 5076 5932 6778 7614 where the seven zeros in front of the significant digits of 2 k are suppressed. By means of this table, which resembles that of the proportional parts in the table of logarithms, the multiplication can be performed by means of addition only. It will also be observed that in the case of the sines of the first few minutes the products of k will have more than ten zeros, and therefore can be omitted, and that under all circumstances they will have at the least seven, so that the multiplication is soon performed. Thus, suppose Sin. p' = -3759264827 2 k. sin. p' = -0000000253,8 59,3 4,2 7 0000000318 By this means we can successively obtain the sines of 1' 2' 3' 4' and so on for every minute up to 45. METHOD OF CHECKING PRECEDING CALCULATIONS. 301 (G2.) To obtain the Cosines of the Angks 2' 3' 4', &c. Since ct>s. (A + B) + cos. (A B) = 2 cos. A cos. B. .*. Cos. (n+ 1)' + cos. (n 1)' = 2 cos. 1 cos. n'. Hence as before Cos. (n + 1)' cos. n' = cos. ' cos. n 1' 2k cos. n'. or, since the cosine continually decreases, Cos. n' cos. n + 1' =cos. n 1' -f cos. n' -t- 2k cos. n'. Hence Cos. 1' cos. 2' = 1 cos. 1' + 2k cos. 1'. Cos. 2' cos. 3' = cos. 1' cos. 2' + 2k cos. 2'. Cos. 3' cos. 4' = cos. 2' cos. 3' + 2k cos. 3'. The method of calculation is precisely similar to that of the sines. It will "be further observed that since sin. (90 A) = cos. A, that we need not to continue the calculations of the sines and cosines of the angles beyond 45. For example, if we know sin. 23 15' and cosin. 23 15', these are respectively cosin. 66 45' and sin. 66 45'. (63). Simplification of calculation in case of certain Angles. Again, the calculation of sines and cosines of angles greater than 30 can be very much simplified, for Sin. (30 + 8) + sin. (306) *= 2 sin. 30 cos. 6 = cos. 1 Since sin. 30 = & .'. sin. (30 + 0) = cos. 9 sin. (30 0) ; now if 30 + 6 is less than 60, then 0, and 30 6 are each less thun 30. Hence by our previous calculations we know both cos. 6 and sin. (30 0), and therefore obtain sin. (30 + 9) by subtraction. Thus sin. (41. IS') = cos. (11. 15 7 ) sin. (18. 45'). It is plain that by this formula we can calculate the sines of angles from 30 to 60. And since cos. (60 6) = sin. (30 + 6) this calculation of the sines from 30 to 60 gives the sines from 30 to 45 and the cosines from 45 to 30 ; which is what we want to complete the tables from to 45. Thus if we calculate by the preceding formula, sin. (51. 33'), this is the same thing as cos. (38. 27'). (64.) Method of checking the calculation. It will be observed, that according to the method above given, the sine of a given number of degrees or minutes is inferred from the sine of the number preceding. Hence, if an error is made at any one point, say in sin. 3 15', it will be propagated into the sines of every succeeding angle ; to arrest the progress of any such error, as well as to verify the correctness of the calculations, at different points of their progress, when no such errors exist, it is usual to interpose the values of any such terms as can be calculated by independent methods. Thus we have already seen that the sines, &c. of the angles 15 . 30 . 45 ... can be readily expressed, as well as those of 18 . 36 . 54 ... And that from hence we can obtain the sines, &c. of an 352 PLANE TRIGONOMETRY. angle of 3. These and others will act as stops in the series, and also serve to verily the accuracy of the calculation up to the point where they are inserted. (35.) The calculation of Tangents and Cotangents. By the methods now explained we can calculate the natural sines and cosines of all angles from to 90 for every minute of a degree. The natural tangents of angles between and 45 can be obtained by simple division, since tan. 6 = cos. C From 45 up to 90 we can obtain the tangent by the formula, Tan. (45 + 6) = 2 tan. 2 + tan. (45 6.) To prove this formula. Tan. (45 + 0)-tan. (45- 0) = _ cos. (45+ 6) cos. (45 0) sin. (45+ 0) cos. (45 9) sin. (45 6) cos. (45 + 0) cos. (45+ 0) cos. (45 6) sin. 2 cos. (45 0) cos. (90 45 2 sin. 2 6 2 cos. (45 6f sin. (46 6) 2 sin. 26 2 sin. 2 8 cos. (90 2 6) cos. 2 = 2 tan. 2 6 :. tan (45 + 0) = 2 tan. 20 + tan. (45 0) Since 6 is less than 45. 2 6 is less than 45 + 0. and hence, whatever be the value of 0, we shall have already calculated tan. 2 6. before we need tan. 2 9. for the determination of tan. (45 + 6), Thus tan. 81 = 2 tan. 72 + tan. 9 where before we calculate tan. 81 we shall already have calculated tan. 72. It is plain that if we know the tangents of angles from to 90, we also know the cotangents of the angles from 90 to 0. (66.) Formulas of verification. There are many formulas by which the accuracy of the tables, when calculated, can be tested. The following are some of them : (1.) Cos. 9 = sin. (30 + 6) -f sin. (30 6). (2.) Sin. 9 + sin. (72 + 6) sin. (72 0) = sin. (36 -f 0) sin. (36 0). (3.) Cos. (90 0) + cos. (18 0) cos. (18 -f 0) = cos. (54 0) - cos. (54 + 0). The student will readily verify these formulas if lie remembers the numerical values found in previous articles for the sines, &c., of 18. (67.) Another method of calculating Tables of natural Sines and Cosines. Besides the method already given for calculating the natural sines and cosines of angles, there is another more convenient than that. The following is an account of this second method. TABLES OF NATURAL SINES. We have ~. a^ , , m -IT Now suppose x to be an angle = . . Then n 2 m IT m IT /m\ /irV s 1 />"V f^V Sm ' n ' 2 = ' 2 + (n) (z) X 120 + (nj UJ _ J ^V f-} 3 1 ^V /nV 1 \nj {>) 6 + V?J \2/ 5040 + 362380 Now we have already seen that - = 1-570796326794897. Hence if we reduce the in /m\ 3 coefficients of I I .... to numbers, we have n \nj Sin. -.90 = - x 1.570796326794897 /'-Y* 0.645964097506246 ".in \ n / Y X 0.079692626246167 -Y x 0.004681754135319 X 0.000160441184787 -x 0.000003598843235 n/ \.nJ + /m\ 13 x 0.000000056921729 f^Vx 0.000000000668804 \n/ \ns + (Y'x 0.0000000000060G7 f-V 9 x 0.000000000000044 Vy \ n / + &c ..... A similar formula can be calculated for cos. . 90. It will be observed that n n is always a fraction less than -, since we only require the sines and cosines of angles less than 45. Hence these series converge very rapidly, and from them we can easily calculate the sines and cosines for each degree from 1 up to 45. When these are known the sines and cosines of the angles for intervals of 1', or if necessary 10" or of 1", can be found by the "method of interpolations." The method of interpolations involves mathematics of a higher order than is admissible in an elementary course. The advanced reader will find an account of the application of this method to the calculation of Tables of natural sines and cosines in Airy's " Treatise on Trigonometry," in the " Encyclopaedia Metropolitana.'' "We now proceed to give an account of the method of constructing Tables of the logarithms of sines, cosines, &c. of angles. N. B. The shies and cosines of angles are never greater than 1. Consequently their logarithms are negative. The numbers printed in the tables are always the logarithms of the sines, &c. with 10 added to them, thus if L denote the tabular logarithm, and %. the ordinary logarithm L. sin. = log. sin. + 10. This notation will be observed throughout the following pages. The tables may be constructed by calculating the logarithms of the natural sines and cosines. And then as we have already found the natural sines and cosines of angles for intervals of 1' from to 4 5, this table will give us the log. sines, and cosines for intervals of 1' from to 45. MATHEMATICAL SCIENCES. No. XII. 354 PLANE TRIGONOMETRY. They are, however, more generally calculated by an independent process of which the following is an account : (68.) To obtain an expression for log. sin. . 90 in a form adapted for calculation. Tb We have already seen that hence m w _ m TT /, __^\ /-, m? \ / m? \ / 5in ' V 2" " n' 2' V n- } \ f*tf } V~" -1 o ~\ (cotan. 0.5 - ) - - (5 cotan. ) . . . .1 \ 2 \ -/ J M. ( S cotan. 6 . S 2 ) omitting 5 3 . V 2 sui.- d J = M. 8 cotan. 6 ( 1 } . \ sm. '26 ) Now if 8 is an angle of n" then 8 the circular measure of this angle, equals n. siii. 1". .-. L. sin. (8 + n") L. sin. 8 M TO sin. 1" cotan. 6 ( 1 . ' ' ) (60.) \ sm. A I I * In this and the following articles are several instances of approximation which must bo care- fully attended to. We say if 2 be so small that we may omit i 3 and all the higher powers of S, : then cos. S = 1 and sin. 2 = S. For this, see cor. to article (50). We then obtain log. sin. (0 + 2) log. sin. 6 = M -< (cotau. 6 . 3 - sA - (cotan. 6 t ^ \ 2 which is of course = M I cotau. Be S 2 cotan. 2 6 + - cotan. fl S 3 + &c. J \ 2 / which = M ! cotan. 6 S ^-o* (1 + cotau.2 fl) l | if we omit 5 3 , S 4 , &c., and hence the result in the text. So again in another article. If we omit fl j , and all higher power of 9, we have P sin. . 0* sm. 6 = .'. - by the Binomial Theorem LOG. SIXES OF SMALL ANGLES. In like manner we may prove that L. tan. (6 + n") L. tan. 6 = ' j - { 1 n sin. 1" tan. 9 } cos. ~0 \ / Now suppose that our tables are calculated for angles that differ by so small an angle 5", that we can neglect the second term of Equation (60) so that L. sin. (0 + 5 ") L. sin. 8 = M 5 sin. 1 " cotan. 6. Also, if re < 5, a fortiori, L. sin. (e + n") L. sin. 6 = M n sin. 1" cotan. 6. Now, for any value of 6 the Tables give us L. sin. (0 + 5") L. sin. call this A A = M 5 sin. 1" cotan. 6 n .'. L. sin. (0 + n") L. sin. Q = A. 5 i. e. when two angles are nearly equal, difference between log. sines is proportional j to the difference of angles. Now -r- is the difference corresponding to 1". This can easily be multiplied by n. I Also L. sin. is given by the Tables, and hence we can find L. sin. (0 + n") and conversely having given L. sin. (0 + n"), where only L. sin. is given in tables, we can easily find the n seconds. If the numerical values of formula (60) are taken for different angles, it will be found that in order to make the difference between two consecutive L. sines, given in the Tables proportional to the number of seconds, we must have 5 = 1" from up to 1. 30'., and 5 = 10" from 1. 30' up to 5, and 5 == 60" from 5 upward : the log. sines being calculated to 7 places of decimals. The practical mode of employing the Tables will be readily understood from the following examples : Thus (1). Find L. sin. (15 11' 16"- 5). By tables L sin. 15 11' = 0-4181495 Do. diflf. 1" = 7755 x 10" = 7755 C" = 4653 5 = 387 L. sin. (15 11' 16". 5) = 9-4182765 (2). Find L. sin. (3 19' 37"- 4). By tables , L sin. (3 19' 30") = 8'7634252 Do. diff. 1" = 362-3 x 7" = 2534 1 4 = 1449 L. sin. (3 19' 37"- 4) = 8-6736931 (3). Find 6. If L. sin. 9-7645827 L. sin. 9-7645827 L. sin. (35 33') 97644849 978 Diff. 1" = 29-45 < 30 893-5 84^5 2 58.9 358 PLANE TRIGONOMETRY. 25-6 8 23-56 .-. = 35. 33'. 32". 8. We proceed in precisely the same manner with the logarithms of the other trigo- nometric functions, with this exception, that what is true of the L. sines and L. tangents of small angles, is true of the L. cosines and L. cotangents of angles which are nearly equal to 90. For, since sin. = cos (90 0) and tail. = cotan. (90 0), then if be small, 90 is nearly equal to 90. (71.) Another way of treating the L. sines of small angles. If the tables give the log. sines, &c., for intervals of 1' throughout, and not the refined tables for small angles before spoken of, we can find the accurate value of a small angle whose L. sine is given, and vice versa, by the following process. If 6 is small, so that we can omit 4 s , &c. we have (Art. 50) sin. = 6 fi' & cos. = 1 Jn. 0* /, 0*\s 5 Hence - =1 = [ 1 5- )~ cos - e u u \ u ' Bin ~0 G \~ 2 Now suppose to contain n" /. = n siii. 1" Kin. i - = cos. J n sin. 1" .-. log. sin. log. n. log. sin. 1" = log. cos. u L. sin. 6 log. n L. sin. 1" = L. cos. o o .-. L. sin. = log. n -\ -L. cos. + L. sin. 1" 3 o & L. sin. 1" = 4-6855749. .-. L. sin. = log. + ~p L. cos. + 1-3522416 and log. n = L. sin. ^--L. cos. 1.3522416. o It is to be observed that when is very small, cos. changes very slowly. Hence in the term L. cos. we may use the number of minutes in 0, omitting the odd seconds, and then, the former formula gives us L. sin. 0, when or n" is given, and the latter gives us the n" when L. sin. is given. For instance, Find 0, having given that L. sin. = 7.2777613. On looking into Tables we find that is between 6' and 7', hence in this case cos. = cos. 6' L. sin. = 7.2777613 J L. cos. 6' = 6.6666668 10. - + 1.3522416 = 8.6477584 10. 2.5921865 .5921768 LOG. SINES OF SMALL ANGLES. 359 97 90 7. 391.0086 = n. .'. = 6', 31" .0086 If we find n" by more refined tables L. sin. 6 7-2777613 L. sin. 6'. 31" = -2777514 99,00 75,21 1 23,790 22,563 3 .-. 6=6'. 31".013. /. by this method n = 391-013 which does not differ materially from the result we obtained by the less refined table and the formula. Find L. sin. 6 when 6 = 7'. 31". 37. Here n 451.37 L. cos. 6 L. cos. 7' L. sin.6 = log. n = 2-6545327 + \ L. cos. 7' = 3-3333330 3 + 1-3522416 = 1-3522416 7-3401073 The more refined tables give us at once L. sin. 7' 31" = 7-3397511 Diff. 1" = 9619 x *''*'.= 2885,7 7 = 673,33 7-3401070 In like manner if we have so small an angle that 6 4 6 s ... may be omitted, and if 6 be an angle of TI seconds, -* 'n. 6 _ 6_ l ~2 tan. 6 _ 1 _ 6 3 \ 2 / 2 .*. log. tan. log. n sin. 1 " = - . log. cos. 6 o 2 L. tan. 6 = log. n + L. sin 1 " + - . (10 L. cos. &). o L. tan. 6 = log. TI + 11-3522416f L. cos. 0. O This formula can be used in the same manner as the one for L. sin. 0. 3GO PLANE TRIGONOMETRY. (72.) Ddumbres method for L. sines of small anglts. There is auother method of treating the L. sines of small angles, of which the following is an account. Sin. 6 = 6 - *L + _*L _ & c . sin. e ( /e- l \\ ... log. _ = log. 1 1 - (-- _ _) I . Onuttmg 6> J /* ^\ l / 02 _ e ' V I \9 120 / + a U 1207 * . ' From this a table for log. - can easily be calculated. Now sup- pose 9 to contain n" Then n sin. 1" = sin. n" sin. 0. sin. 1" sin. 5. sin. 1" L. sin. w = log. n + L. a and since L. - - is given by the Tables, if we know n we can at V once find L. sin. n". And conversely if we have given L. sin. 0, we can find n; for sin. 6 = 6 n sin. 1" nearly. .'. log. n = L. sin. L. sin. 1" nearly, whence we know n approximately. This will enable us to obtain the value of log. - from the Tables with sufficient accuracy ; which V being known, we have log. n = L. sin. e log 'which gives us n V accurately. sin. 8 It can easily be shown, that in calculating tables for log. -^ from the formula, , sin. 8- ,, / 0- 0' \ log. - = Ml + - . 9 \ 6 18U / that for tables of seven places of decimals up to 5, 6 4 can be omitted. For when 6 is an angle of 5, 8 = ^JL 180 Let us now look back and consider what we have done in the previous pages. (1.) We have investigated the various general relations existing between the trigono- metrical functions. (2.) We have investigated the general relations existing between the sides and angles of triangles. (3.) We have fully discussed the mode of constructing trigonometrical tables, i.e. the means of obtaining numerical results from our trigonometrical formulas. Wo now proceed to the actual numerical calculation of the sides and angles of NUMERICAL SOLUTION OF TRIANGLES. 361 triangles, which we shall find will invariably consist in reducing formulas to numbers. For instance, Given sin. (m + x) sin. (m x) = cos. (m ri) cos. (m + n) Find x when m 12 ' 13' and n 7- 37'. Since sin. (m + x) sin. (m x) = 2 sin. x cos. m and cos. (m n) cos. (m + n) = 2 sin. j sin. n, .'. 2 sin. x cos. m = 2 sin. m. sin. . .'. sin. x = tan. MI. sin. n. .'. log. sin. x = log. tan. m. + log. sin. ?i. .-. log. sin. x + 10 = log. tan. m + 10 + log. sin. n + 10 10 and /. L. sin. x =. L- tan. m. + L- sin. n 10. L. tan. m = 9-3354823 L. sin. n = 9-1223624 10 L- sin. 1 33'- 40' 8-4578447 8-4578369 diff. 1"= 732,8 x -1 Aus. x = l-38'- 40" 78 73-28- THE NUMERICAL SOLUTION OF RIGHT-ANGLED TRIANGLES. The parts of a right-angled triangle are the three sides, two angles and the right angle, if any two of the former five, one of the two being a side, are given, we can calculate the remaining three. The following are the methods employed in the various cases, which are these. (1). Given the base and perpendicular. * (2). Given the hypothenuse and another side. (3). Given the base or perpendicular and an angle. (4). Given the hypothenuse and an angle. '^ (1). Given the base and perpendicular i.e. given a. b. find A. B. c. tan. A = - gives A. b B = 90 A. Gives B. Cos. A = -. cos. A Gives c. These can be put into forms adapted for logarithmic calculation, as follows : log. tan. A = log. a log. b. /. 10 + log. tan. A = log. a + 10 log. b. or, L. tan. A = log. a + ar : comp : log. b. (a). Similarly. Log. c = log. b. log. cos. A. = log. b. + 10 (10 + log. cos. A) log. b. + 10 L. cos. A. 362 PLANE TRIGONOMETRY. .". log. c = log. b. + ar : comp. : L. cos. A. (a), and (b). are the formulas actually used in calculation. Thus Given a = 7564 5 yds. b = 3987 ' 4 yds. Find. A. B. and c. To find A. from (a) we have. Log. tan. A = log. a + ar : comp : log. b. = log. 7564-5 = 3-8787802 + ar : comp : log. 3937-4 6-3993102 (b). L. tan. 62 3 12' 10-2780904 10-2779915 = 51-05 ) 989-00 ( 19 5105 A = 62 12' 19" 90 47850 45945 and B = 27 47' 41" To fiud c we have. Log. c = log. b + ar : comp : L cos. A log. 6 = 3-6006898 ar : comp : L cos. 62 12' 19" = ar : comp : 9-6686623 = -3313377 3-9320275 9320271 85512 08 diff. = .'. c = 8551-208 (2). Given the hypothenuse and another side, e.g. given a c to fiud A B and b. a We have. sm. A = . c .-. log. sin. A = log. a log. c. L. sin. A = log. a + ar : comp : log. c which gives A and .-. B which equals 9U A. b Again cos. A = . b c cos. A. log. b log. c + L. cos. A 10. Ex : Given a = 724-5. c = 1005-4. We shall have A = 46 6' 17". b = 607-036. B = 43 53' 43" NUMERICAL SOLUTION OF TRIANGLES. 363 We have Again Again (3) Given the Base or Perpendicular and an Angle. e. g. given a. A., find B. 6, c. B = 90 A. sin. A log. c = log. a log. sin. A log. c = log. a + ar : comp : L. sin. A. b (a) cos. A log. 6 = log. c + L. cos. A 10. (b) In using formula (b). It must be remembered that in the calculation of formula (a) we have already found log. c. Example. Given a = 7643-5 A 37. 13'. We shall find B = 52. 42' c = 12613-4 b = 10033.53 (4) Given the Hypothenuse and an Angle. e. g. given c. A, find B. b. a. We have B = 90 A. b = c. cos. A. log. b = log. c + L. cos. A 10. (a) Again a c sin. A. log. a = log. c + L. sin. A 10. (b) Example. Given c = 7234-5 A 33. 19'. We have B = 56. 41'. b = 6045-493. a = 3973.665. The cases of oblique triangles are in like manner reducible to four, viz. (1). Given three sides to find the angles. (2). Given two sides and the included angle. (3). Given two sides and the angle opposite to them. (4). Given one side and two angles. (1). Given three sides to find the angles, t. e., given a. b. c. to find A. B. C. We have 2 s = a + b + c. Fig. 24. .'. 2 log. tan. = log. (s b) + log. (s c) log. s log. ( o) Si J^ :. 2 log. tan. - + 20 = log. (s b) + log. (s c) + 10 log. 8 + 10 log. (* a .*. 2L.tan. = log. (s b) + log. (s c) + ar. : comp. : log. s + ar : comp log. (* a) 364 PLANE TRIGONOMETRY. This formula gives us . We shall have to find B from the formula M 2 L. tan. = log. (s c) + log. (s a) + or : comp. : log. s (- or : comp : log.(-J) It will be observed that all the logarithms needed for finding B and C have been already used in finding A, whence, when A is known, B and C are found with very little trouble. A Instead of using the formula which gives us tan. -, we may use those which give i A A sin. or cos. -- w It Example given, a = 3756-3 b = 5923-7 c = 4652, we shall find that A = 39- 18' 28" B = 49- C' 50" C = 51- 40' 42'' .-. A + B + C = 180 -0-0 which circumstance is a test of the accuracy of the calculation. (2.) Given two sides and the included Anyk. e. g. given a b C, find A B c. We have A B a b , A + B tan. - - = - - tan. - a + b -2 . . L tan. - = log. (a 1) -f ar : comp: log. (a + b). + L tan. A --? - 10- (a) A -t- T* r* It must be observed that - = 90 and is therefore known. Wherefore (a) gives us = o, a known angle. Then, 2 A B - = a 2 A = 90 + a _ C and B = 90 a -. whence A and B are known. To find c we have .". log. c = log. a + L sin. C + ar : comp : L sin. A 10, which will give us c. We may employ a subsidiary angle 6 in the following manner (see Art. 40). NUMERICAL SOLUTION OF TRIANGLES. 365 a - fi C Assume tan. = 7 . cotan. a + o 2 <% c = ( + J ) sin --2- cos. iOg. tan. = log. (a b) + ar : comp : log. (a + 6) + log. cotan. 10 (c) log. c = log. (a + b) + log. sin. + ar : comp : log. cos. 10 (d) By using (c) and (d) we can obtain c without first finding A and B. If we solve by the former methods we require six logarithms, viz. : those of A + B (a b) (a + b) tan. - a. sin. C. sin. A. C C If by the latter we require five logarithms (a b) (a + b) cotan. . sin. . and cos. 0; the latter method possesses a slight advantage over the former. The in formula (c) is evidently the same as - in formula (a). If 6 > a the formulas become B A b a A + B tau. - . tan. 2 b + a 2 b a C and tan. 7 . cotan -pr. o -f a 2 Example : Given, a = 562. b = 320. C = 128 4' We shall find A = 33 34' 40" B= 18 21' 20" c = 800-008. If we take the second method we shall find log. tan. = 9-1258960 .-. log. cos. = 9-9961568 And, as before we shall find, c = 800-008. (3.) Given two sides, and an Angle opposite to one of them. e. g. given B. b. c. find A. C. a. We have sin. C = -. sin. B. .-. log. sin. C = log. c + log. sin. B + ar : comp : log. 6 10 (a) which gives C. Then A = 180 (B + C) which gives A. Then , sin. A a = o -. ~ Sin. B or, log. a = log. b + log. sin. A + ar : comp : log. sin. B 10 (6) which gives a. N.B. This is sometimes called the ambiguous case of a triangle ; for we have 366 PLANE TRIGONOMETRY. sin. C given us, whence we know G a, where o ^ 90. Now, sin. a = sin. (180 a) .'. C may also equal 180 a. Hence this equation may give us two triangles, one which has C = a, the other having C = 180 a. In many cases, however, this ambiguity does not exist. (1). B = or > 90. Then C= 180 a is inadmissible, since C would then be > 90, and B + C > 180, which is impossible. (2.) IfB < 90. Then (a). If c sin. B = b then sin. C = 1 or C = 90', and the other value of C =180 90 = 90, or in this case there is no ambiguity. (JB). If c < 6. Then C < B, hence if C = o. a < B, and 180 a + B > 180. or in this case the angle 180 a is inadmissible, and there is no ambiguity. (7). But if c > b. Then both c = a, and c = 180 a are admissible, and there are two triangles determinable from the given values, b. c. B. This can be illustrated geometrically as follows : AB = c ABc = B. with centre A and radius 6 describe a circle. Then, (a). If b = c sin. B, the circle will touch Be in c, and the triangle will be ABc. (j8\ If c < 6 the circle will cut Be, in two points C 2 C,, on dif- ferent sides B, and the triangle is ABC,. (7). If c > 6 the circle cuts Be in two points c, c 2 on the same side of B, and the triangle may be either ABc 2 or ABc, ; which is the reason of this case being called ambiguous. Example : Given B = 45 b = 305 c = 219*5 Then C = 30. 36'. 22" A = 104. 23' 38" a = 416. 8344. (4). Given, one side and two angles e. g. given, a B C, to find A. c, b. We have A = 180 (B + C). which gives A. Now, b _ sin. B c _ sin. C a sin. A- a sin. A' .'. log. b = log. a + log. sin. B + ar : comp : log. sin. A 10. and log. c = log. a 4- log. sin. C + ar, comp : log. sin. A 10. Whence we can immediately calculate b, &c. Example : Given B = 49. c = 29. 19' and a = 95.4, we shall find A = 101. 41' b = 73-52255 c = 47-69952 Several practical applications of the science of Trigonometry -will be found in the following short treatise on Mensuration. HEIGHT OF A TOWER. 367 MENSURATION. THE object of the following short treatise is to show the practical application of the truths investigated in previous treatises of geometry and trigonometry, to solve a variety of questions about measurements such as may frequently occur in practical life. For instance, the determination of the area of a field, the distance between two inaccessible points, the solid content of a pyramid, and others. We shall endeavour in each instance to refer these questions to the principles on which they rest. By doing so, the attentive reader will be enabled to apply the same principles to cases that may occur practically, but which our limits forbid us to treat of. We shall take in order a variety of questions concerning the determination (1) of Heights and Distances, (2) of Areas of Surfaces, (3) of Contents of Solids. (1.) ON HEIGHTS AND DISTANCES. In the following articles we are supposed to have the means of measuring the angle subtended at the eye of the observer by the line joining two points. The instrument by which this can be done is called a sextant. For measuring the angle subtended by two objects on a horizontal plane, and for determining the vertical elevation of one point above another, an instrument called a Theodolite can be used. Fig. I. 1. To determine the height of a Tower standing on a Horizontal Plane, the base of which is accessible. AP the tower. B any convenient place for the observer. At B measure the angle ABP ; also measure the line AB. Then, if AP =: x, x = a tan. B. .-. log. x = log. a + L tan. B 10. which gives x. N.B. If P is the top of a steeple, half the breadth of the tower must be added to the measurement, which is made to the outside of the tower. It will be observed, moreover, that B is the place of the observer's eye, and .*. A P, or x, is the height of P, above the horizontal plane passing through the observer's eye. Hence, if h is the height of the eye above the ground, x + h is the height of the summit of the tower above the ground. (2.) To determine the height of a Tower, Hie base of uUch is inaccessible. Let ABX be the horizontal plane ; PN the height r. required. At B measure angle PBN-; move backward to another point, A, taking care that A and B are both on the same vertical plane passing through P, and measure AB, and the angle PAN. Let PBN=:/3. PAB a. AB = a. BP = x. PN = y. Now angle APB = (,3 a) 368 MENSURATION. and " sin. (ft - a) x = y sin. B. sin. a- sin- B .-. x = a -.73 r- sin. (B a) . .*. log. x == log. a + L. sin. a -f L. sin. B + ar. comp. L sin. (B a) '2Q. (3.) To determine the heiyht of (he Tmcer in the last article when the nature of the ground does not admit of the observer moving backward. Let PN be the vertical height required, place pickets at two points, A and B, APN and BPN being two different vertical planes. Measure AB = a. Let PN = x. Measure angle PAN rr: o. PAB = A PBA = B. x AP sin. a. APB = 180 (A 4- B) sin. B Then Now AP = sin. (A + B) sin. a- sin. B Fig. 3. sin. (A + B)' .-. log. x = log. a -(- L. sin. a + L sin. B + ar. comp. L sin. (A + B) 20. In the above case such an angle as PAN being the vertical elevation of P above the horizon is measured by a theodolite. While the angle PAB, which is not in a horizontal plane, being the angle subtended by the distance between two objects, is measured by a sextant. If the object is a long way off we may measure the angle subtended at the eye by the line joining that object and its image in still water, by a sextant; half that angle will be the vertical elevation of the object. Example: "We observe that the altitude of a hill is 3 3 15' on proceeding lj{ miles towards it. Its altitude is 15 '67'. Find the height of that hill, neglecting the sphericity of the earth. Here (fig. 2) AB = If miles. PAB = 3 15'. PEN = 15 37'. /.PEN PAB = 12 22'. Then by the formula proved in Article (2) p,,,. _ sin. PEN sin. PAB ' siu. (PEN -PAB) .-. log. PN = log. a + L. sin. PEN + L. sin. PAB + ar. comp. L. sin. (PEN PAB) 20. log. 1-75 -243 0380 L. sin. 15 37' 9-430 0750 L. sin. 3 15' 8-753 5278 Ar. comp. L. sin. 12 22' -669 2473 20 THE DISTANCE BETWEEN TWO POINTS. 19764 G 197646 of a mile or ^ of a mile very nearly. l-2y58881 2958748 133 132 (Answer.) (4.) To find the distance "between two points, one of which is accessible and ilie other inaccessible. Let P be the inaccessible point. A the accessible. Drive a picket in at A, and drive in another at any con- venient point, B. Measure the angles PAB and PEA, also measure the line AB. Then by the fourth case of the solution of triangles sin. APB AP = AB. = AB. sin. ABP sin. (PAB + ABP) sin. ABP .-. calling AB = a. PAB = A. ABP = B. we have log. AP = log. a + L. sin. (A + B) + ar. comp. L. sin. B 10. Fig. 4. (5). To fold the distance between two points neither of which is accessible. Let A, B, .be the two points ; place on the ground two pickets C, and D, such that the dis- tance between C and D can be measured, and that from each of C and D, the two inaccessible poiuts and the other picket may be visible. Measure CD = p ACB = C BCD = C' ADB = D ADC D' Then in triangle ACD we know one side and two angles, and therefore can calculate AC. Fig. 5. In triangle BCD we know one side and two angles, and therefore can cal- culate BC. And finally in triangle ACB we have already calculated AC, BC. and have measured the angle C, and therefore can determine AB. Of course from the first two triangles we can also determine AD and BD, and then in triangle ABD we know two sides, and the included angle, and hence can determine AB, and can use these two calculations for checks upon each other. The calculation is performed as follows : Call the sides and angles of ABC a.l.c. A.B.C. and observe that CAD = 180 (C + C' + DO and CBD = 180 (C' + D + D'). Then from triangle ACD siu. CDA /. log. AC = log. CD + L. sin. CAD + ar. comp. log. sin. CDA 10. MATHEMATICAL SCIENCES. No. XII. 370 MENSURATION. log. 6 = log. p + L. sin. (C + C' + D') + ar : comp : log. sin. D' 10 (1) Similarly log. a = log. p + L. sin. (C' + D + D') + ar : comp : log. sin. C' 10 (2) From triangle ACB \ve have = 90 5 2 At A B 5 A + B tan. - = - tan. - - 2 a + b 2 b a 6 1 tan. assume tan. = - -- =. -- = tan. (45 ) a a + 6 1 + tan.

= 43 24' 9.9757318 4218)159600(37.8 12654 = 43 24' 37".8 33060 45 29526 45 = 1 35' 22".2 35340 (4) L. tan. A ~ B = L. tan. (45 ) + L. tan. A + B 10. THE AREA OF A RECTANGLE. 371 = L. tan. 1 35' 22"-2 = 8'4432492 L. tan. 28 22' 30" 97326024 10 A B 51' 30" 2 = 51' 32" 8-1758516 8-1755658 1408)2858(2 2816 42 = 28 22' 30" A == 29 14' 2" B = 27 30' 58" (5). log. c = log. a + L. sin. C + ar. comp. L sin. A 10. = log. a = 2-6843172 L. sin. 56 45' 9-9223549 ar. comp. L. sin. 29 14' 2" -3112458 10 82778 6 827.786 .*. c = 827.786 yds. 2-9179179 9179149 30 32 (Answer.) II. THE MENSURATION OF AREAS. (1.) To find the Area of a Rectangle. Let ABCD abed be two rectangles. Then (Euclid. VI. 23.) the areas of these rectangles are to each other in the ratio com- pounded of the ratios of the sides, i.e., in the ratio compounded of the two, ab : AB and be : EC, and if we suppose these lines to be repre- sented by numbers, this compounded ratio is ab x Ic : AB x BC. Hence rectangle ac : rectangle BC : : ab x be : AB x BC. Now suppose ab = 1, and cb = 1. Then the area ac is the unit of area, i.e., is a square inch, or a square foot, or a square yard, according as ab is an inch, a foot, or a yard. In this case Rectangle BC = AB x BC. Hence, if a and b be the sides of a rectangle, its area is ab, i.e., it contains as many units of area as the product of the number of units of length in one side, by the number of units of length in the other. Cor. If o is the side of a square, its area is a 2 . Fig. 6. 372 MENSURATION. (2). To find the Area of a Triangle. Let ABC be the triangle ; from A draw AN per- pendicular to BC. Then, since the area of ABC is half that of a rectangle, whose base is BC and BC x AN height AN, the area of triangle = a Cor. (1). If we have given AB. BC and angle B. BC x AB sin. B ac sin. B D Area triangle = = Since AN = AB sin. B. (2). If we have given BC and the angles of triangle, c sin. C Then, since = a sin. A Area triangle := a- sin. B sin. C 2 sin. A (3). If we have all the sides given (by Trigon : Art : 41) Area triangle = A/ * (s ) (s b) (s c) where 2s = a + b + c. (3). To find the A rea of a Parallelogram. Ifa.l be the sides of the parallelogram, and A the angle contained by those sides, then by the last article the area of half the parallelogram ab sin. A ~ 2 .'. Area of parallelogram = ab sin. A, Or, if p is the perpendicular distance between two parallel sides, each of which a, then Area of parallelogram = ap. (4.) To find the Area of a Trapezoid. Let ABCD be the trapezoid of which the side AB is parallel to the side CD. Join BD, draw DM perpendicular to AB and BN perpendicular to DC, or DC produced, and let a = AB. b = DC and p = DM or BN. Then Area triangle ABD = op Area triangle BCD = 4 bp .'. Area ABCD = p (ft + b) Or the area of a trapezoid = 4 sum of parallel sides + perpendicular distance between them. Fig. 3. AREAS OF TRAPEZIUM AND POLYGON. 373 (5.) To find the Area of Trapezium. Lei ABCD (Big. 9), be the trapezium ; join AC, from B and D let fall perpendiculars B m, D n, upon AC. The area of triangle ABC = A Bw. AC and area of triangle ADC = y DM. AC. /. Area of trapezium = \ AC^Bw + Dn.J If instead of having the perpendiculars B m and D given we have the sides AB. BC. CD. DA. and a diagonal, we must find the area of each triangle separately by the formula Area triangle = V s. (i a) (s b) (s c) Or, again, if we have the sides and one angle, as A, given, we can proceed as follows. Let AB. = a. BC = 6. CD c. DA = d. Join BD (Fig. 10), then we can calculate BD from triangle ABD by second case of oblique-angled triangles, and having calculated BD we can determine the areas of the triangles as before. (6). To determine the Area of an irregular Polygon. Suppose A B C D E F to be an irregular polygon, its . area can be determined by effecting the following measurements ; join AD, the longest distance across the figure from B C S F, draw perpendiculars to AD., viz. Em, Cn, Ep, Fq, respectively; measure Am, mn, nD, Cn, Bi, Aq, qp, pD, ~&p, Fq. Then area of polygon = ABwi + BmC + o CwD + DpE + Ep/F + FqA.. = ^ Am x B?n. + - mn x (Bm + Cn) + Cn y. nD + Dp + pE + ^ pq (Ep + Vq) + ^A.q x qF. (7.) To determine the Area of a regular Polygon. Various formulas have been already given for this in the "Treatise on Trigonometry" (Art. 45). It is there shown that if n be the number of sides of the polygon, and a the length of one of the sides, Fig. 10. Area = Hence in case of pentagon, r>a- 180'- cotan. . Area = - cotan. 3Q"' 4 In case of hexagon, Area = And so on in other casec. 3 a 2 cotan. 30 ~ Fig. 11. 5 (A/5 + 1) a 2 4 vTo'^^2"v r 5 A/3 a 2 374 MENSURATION. (S.) To determine the Area of a Circle, If n is the number of sides of a regular polygon inscribed in a circle whose radius is r, then, as we have already seen (Trig : Art : 45), nr 2 . 2ir Area polygon = sin. * 2* Now if we increase the number of sides of the polygon, it becomes more and more nearly equal to the circle, and in the limiting case when the number of sides 2 T is infinitely great, it becomes the circle; * but when n oo .*. = Oand (Trig: 2* Art : 47) in this case sin. = 1 27T .' Area of circle = r- ir. Cor. Hence area of quadrant = -j = the length of arc on which quadrant stands. U .'. Area of quadrant = - x arc on which quadrant stands. (9). To find Area of a Sector of a Circle. Let AOB bo any sector of a circle, OA =r r. AOB 9 (in circular measure), 7T Now if AOB were a quadrant, AB would equal M and the area of the quadrant = j- A' But in equal circles sectors arc to each other as the | arcs on which they stand. TTT Area sector : area quadrant : : r 9 : Area sector : : : 6 : 4 2 .*. Area sector = Fisr. 12. Compare witli the statement in the text what is said on page 160. AREAS OF PORTIONS OF CIRCLES. 375. =. r x (arc on which sector stands). (10). To find the Area- of a Segment of a Circle. Let ACB be a sector of a circle. Join A B. " We are required to find the area of the segment ABC. Draw ONC perpendicular to AB. Let AOB r= 9. OB r. 9 6 .'. ON = r cos. AN = r sin. A 2 9 01 /. area ABO r- sin. -- cos. = - Now area ACBO = -i~0. A Fig. 13 .'. area of segment =: -^ (9 sin. B). Cor. From A draw A n perpendicular to OB. Then A n = r sin. 9, also arc ACB = r 9. Hence the area of segment = 2 (r& r sin. 6) = -, x (ACB An) (11.) To find the space between two Concentric Circles, viz. ABCEDF. Let OA = r OB = r x . ,* Then area of interior circle = irr* and area of exterior circle = ir r^" .". Area of the space between the circles On OB as a diameter, describe a circle OPB cutting the interior circle in P; join OP. PB, then OPB is a right angle, BP = OB 2 OP 3 = J-, 2 ?" also since OPB is a right angle, BP touches the circle, or the area of the space in question is Fig. 14. equal to the area of a circle whose radius is the length of the line drawn from any point in the exterior circle to touch the interior circle. The above cases of areas of plane figures are the chief of those which belong to Elementary Mathematics. The determination of areas bounded by curved lines belongs, of right, to the Integral Calculus ; the following propositions, however, are best given here, though the reasoning is not of a strictly elementary character. 376 MENSURATION. (12.) To find the Area of an Ellipse* Fig. 10. Now, by a property of the ellipse, NQ : nQ Let aBA be a semi-ellipse. OA its semi-major axis = a. OB its semi-minor axis = 6. On aA as a diameter describe a semicircle aCA. In AB take any point P and draw an ordinate PN perpendicular to OA and produce it to meet tlie circle in Q and draw another ordinate qpn parallel and near to QPN and complete tha parallelograms P, nQ. NP : : o : 5 nP : : a : b and hence, if we suppose a series of parallelograms to be described in circle and ellipse, the same proportion will hold good between each of these, and therefore Sum of parallelograms in semicircle : sum of parallelograms in semi-ellipse : : a : b. and this being true, however great the number may be, is true in the limit. Now the semi-ellipse is the limit of the parallelograms inscribed in it, and the semicircle is the limit of parallelograms inscribed in it. /. semicircle : semi-ellipso : : a : b .'. circle : ellipse : : ir a 2 : ir a b But area of circle = ir a" area of ellipse = IT ab. * Let AI'Q be an area included by two straight lines AB. AP, and the curve PQ ; divide AQ into or[ual parts Aa, ab, be, cd, dQ, and on those Hues draw the rectangles Ap, aq, Ir, a within the -.;rvcd areas, and complete the parallelograms Ap', aq', Ir', cs', dt. Then it is plain that thft (difference between the interior parallelograms (Ap, aq, Ir, cs)and the exterior parallelograms (Ap', aq', br', eg 1 , dt) will equal Ap'. Now, by making the number of parallelograms very large, Aa, and /. Ap' will become very small, and may be made less than any magnitude that may be assigned. Now the curvilinear area is clearly greater than the interior and less than the exterior parallelograms, and therefore differs from the interior parallelogram by a quantity less than Ap' ; i. e., a quantity that can be made less than any that can be assigned : and therefore the curvilinear area is the limit to which the sum of the interior parallelograms continually approaches when their number is increased. AREA OF A PARABOLA. 377 (1 3.) Let q PQ be portion of a Parabola. It is required to find its Area. Bisect Qgin V, draw the diameter PV. Through P draw r PR parallel to QV q, and draw QR and qr parallel to PV. Then the area q PQ is two thirds of qr RQ. Take p and p\ , two points in PQ, and through p, draw p { mi , pi n } parallel to PV and Q V, and through p draw pm pn also parallel to PV, QV, and produce them to meei pi i"in 51 and pi m\ in q respectively. Then Parallelogram pm x : parallelogram pj : : p n x MM, : pm x m m x . : : pn x (Pw, Pn) : Pn x (p, rij pz) Now, by a property of the parabola Pn : P t i : : .pw 2 : Pi i 2 .. PTI : PWj PM : : pn" : pi MI 2 pn" or Pn : Pw 1 Pn, : : pri 2 : (p\ i pn) (p, M, + pn) /. Pn x (pi n\ pn) : (Pni Pn) pn : : pn : pi ni + pn .-. parallelogram pnii : parallelogram pni : : pn : pi ni + pn. Now in the limit p\ n diners from pn by a quantity less than any that can be assigned. .-. In the limit pn : pi n\ + pn : : 1 : 2 /. In the limit parallelogram pmi : parallelogram pni : : 1 : 2 which proportion is true for each pair of parallelograms inscribed in PQR and PQV, and .'. is true of all. .'. In the limit sum of parallelograms in PQR : sum of parallelograms in PQV : : 1 : 2 But the area PQR is the limit of the sum of the parallelograms inscribed in it; and PQV is the limit of the sum of the parallelograms inscribed in it. .'. PQR : PQV : : 1 : 2 .-. PQVR : PQV : : 3 : 2 Now PQVR is the half of qr QR, and PQV is the half of PQg. /. PQ Pi'Ps' Pz>Pa> & c - t ke drawn from point to point of AB, these chords in the revolution will describe frustums of cones; now tho arc AR is the limit of the chords Ap, +.PI.PS + p 2 Pa -\- . . and hence the area of the portion of sphere described by AR will be the limit of the sum of the frustums of cones described by Fig. 27. perpendicular to PQ, Let PQ (Fig. 28) be one of these chords: draw PM, QN perpendicular to AO. Draw and tn perpendicular to AO. Now tn is the radius of the mean section of the frustum of cone described by PQ, and therefore area of that frustum = 2 TT tn x PQ Now tn = Ot sin- JO A. and PQ sin. PQN = JSM. or since PQN = tOA. PQ sin. tOA = NM .-. tn. PQ = Ot. NM. And area of frustum = 2 TT 0^. NM. In (fig. 27) draw p^ M X p 2 n 2 p a n s perpendicularly AO. Then, since Ap 1 p 1 p 2 p 2 p a are all equals, the perpendiculars on them are equal, and therefore the sum of areas of frustum of cone 2 it Ot x (An x + ij n 2 -\- n 2 n a -J- ) = 27T Ot x AN. if we only consider the portion of sphere described by AR. This is true, however great the number of chords, and is therefore true in the limit ; but in the limit Ot =: radius of sphere .-. Area of portion of sphere, whose height is AN 27T. OA. AN. If we take the whole sphere. AN 20A .'. Area of sphere = 4ir. (OA) 2< Cor. In the sphere 2ir x OA = circumference of a great circl*. .-. Area of portion of a sphere = height of portion x circumference of a great circle. If we imagine a right cylinder to be described about a sphere, its curved area circumference of a great circle x diameter of sphere = area of sphere. Hence, " area of sphere = area of circumscribing cylinder." INCLINATION OF THE PLANES. 383 Exain : How many square miles of sea are visible from the top of a mast 80 feet above the surface ? Let be the centre of earth, draw OA. 00., so that AC is perpen- dicular to OC. draw CN perpendicular to OA. Then, if BN = x. OB r. The area visible from A will be the area of the part of the sphere whose depth is BN i.e. will := 2ir r x. Now, let AB = p. Then, by similar triangles r + p : r : : r : r x. Fig. 29. r + p x r P r + p' and visible area = . r +p 2trrp very nearly. Now, IT = 314159. r 3958 miles 5280 66 .*. visible area = 2 x 3-14159 x 3958 x ~ DO = 378 nearly. III. THE MENSURATION OF SOLIDS. (1.) If two solid Angles are eacJt, contained "by three Plane Angles, that are equal each to each, then the inclination of the Planes will lie equal, each to each. Let 0, o, be two solid angles contained by the plane angles AOB, BOG, COA, and O aob, Joe, coa, respectively, Then the plane AOB is inclined to plane COA at the same angle that aob is inclined to aoc. For, take OA = oa and let the plane CAB be 384 MENSURATION. perpendicular to OA, and cab perpendicular to oa. Then angles CAO, BAO, coo, bao, are right angles ; and CAB, cab, are tlie inclination of the planes in question. Now in triangles AOB, aob. we have OA = oa and angles BOA, OAB = angles boa, oab, each to each, /. OB = ob and AB = ab (Euclid, 1. 26) ; similarly iu triangles COA, coa, we have OC = oc and AC = etc. Then in triangles BOG, boc, we have the sides BO, OC = sides bo, oc, each to each, the included angle BOC = boc. .'. (Euclid, I. 4) the base CB = base cb. Hence, finally, in triangle ABC, abc, we have the sides BA, AC = the sides ba, ac each to each, and the base BC base be. .-. (Euclid, I. 8) CAB = ca&. Q.E.D. Cor. 1. Hence (fig. 30), if the solid angle be superimposed on the solid angle o, BO that AO coincides with oa, and the plane AOB with plane aob, then, because angle AOB = angle aob the line OB coincides with ob, and because inclination of plane COA to AOB equals that of coa to cob the plane COA will coincide with coa, and hence OC with oc. Cor. 2. If ABCDEF and abcdef are two prisms, the edges of which are equal each to each, and the plane angles at B equal those at b, the prisms are equal in all respects. Fig. 31. For, if the triangle ABC be applied to ale so that AB coincides with ab, and BC with be, it ia plain by the last corollary that BD coincides with bd. And hence the prisms will coincide throughout. COR. 3. If ABCD abed be a parallelepiped * (I.e. a prism on a parallelogram for a base) it' can be divided into two equal prisms by a plane ACca passing through the diagonals of its bases ; for it is obvious that the edges of these prisms are equal, each to each, and also that the plane angles containing the solid angles at D and B are equal. COR. 4. Again (in fig. 31) if we suppose the base DEF of the one prism to be in all respects equal to that of the other def, and if the angles at D are equal to the angles at d, each to each, then it is plain that if DB is > db the prism BDFE is > prism bdfe, and if DB < db, the prism BDFE < prism bdfe. Also if BD is double of bd, the prism BDFE is double of the prisin bdfe, and generally if BD is any multiple of bd, then BDFE is the same multiple otbdfe. Cor. 5. If we suppose (fig. 31) BD produced to K so that DK is any multiple of DB, the prism KDFE is the same multiple of BDFE ; and if we suppose db produced to k so that Ted is any multiple of bd, then kdfe is the same multiple of bdfe. But if KD > M, KDFE > Mfe; if equal, equal; if less, less /. (Euclid, 5. p. 136). BDFE : bdfe : : BD : bd. Cor. 6. The results proved in Cors : 4 and 5 to be true of prisms, are manifestly true of paral"'. * We shall use throughout the abbreviation paralPP d for parallelepiped. EQUALITY OF PARALLELEPIPEDS. 385 (2.) Parallelepipeds on the same Base and of the same Altitude arc equal to one another. If the ParalPP* are on the same base and of the same altitude, it is manifestly that the ends opposite the common base are in the same plane. (a) Suppose two of the edges of the ends opposite to the base of each figure to be in the same straight line. Let DBEG, DBey, be the parallelepipeds on tho same base BD, and of the same alti- tudes, and suppose that EH, eh, are in the same straight line, and also FG, fy, in the same straight line. Then it is obvious that Ee = H7t, HC = ED, and angle hUG = angle eED. .-. the base eED = base liRG, also HG = FE, and the plane angles that form the solid angles at H and E ara equal each to each, .'. the prism ACHG = the prism eDEF. Hence if we suppose the former prism taken from the whole figure hCDEF, and the latter to be taken from the same figure, the remainders will be equal .'. DBEG = DBegr. (6) Suppose that no two of the edges of the side opposite to the base are in the same straight line. Let DBEG, DEeg, be the paralPP" 1 on the same base BD, and being of the same altitude, their ends EG, eg, are in the same plane. Produce EH, FG, fe, yh, to meet, they will evi- dently form a parallelogram, let this parallelogram be KLMN, then LK, MN, and FE, GH = AD, EC, each to each, and KN, LM = Eli, fg = DC, AB, each to each, and angle MLK = angle Fig. 31. GFE = angle BAD. .-. LKMN is equal to ABCD. Join AL, BM, CN, DK. Then DB MK is a parallelepiped and by the first part of this proposition, BD MK is equal to each of EDGE, and BD#e. .'. EDGE = BD^e. Q. E. D. MATHEMATICAL SCIENCES. No. XIII. 386 MENSURATION. (3.) Solid Parallelepipeds on equal Bases and of the same Altitudes, are equal to each other. (a). Suppose the edges to be perpendicular to the bases. To avoid a complicated figure we will only letter the T bases, and will call the edges of figure that are perpendicular to the bases by the letters at the angles of the base. Thus the edge B means the edge perpen- dicular to the base at the point Ei.e., Eb. Let AC, DF be the bases where we suppose the solids to be so placed as to have a common edge D, and the sides AD, DE in the same straight line, produce CD, and FE to meet in H H XI F = 35 H, through G draw LGK parallel CD and produce BC to meet KG in L. since DH = GK, and HE is evidently = KF and the angle DHE = GKF .-. the base DHE = base GKF ; and since edges H and K are perpendicular to base, the solid angle at H is contained by plane angles respectively equal to those containing the angle K ; and hence the prisms whose bases are DHE and GKF are equal. Add the solid on base DEKG to both, .-. paralPP 4 on DK paralPP d DF. Xow paralPP d BD : paralPP d DL : : AD : DG : : BD : DL and paralPP d DL : paralPP d DK : : CD : DH : : DL : DK .-. (Ex equali). paralPP d BD : paralP? d DK : : BD : DK But BD = DK, since (Eucl. I. 35) DK = DF /. paralPP d BD = paralPP d DK = paralPP d DF. (6). If the edges are not perpendicular to the bases, call the paralPP d P and Q. If on P's base a paralPP d p. is described, whose edges are perpendicular to base and whose altitude is same as P, then p = P by last Proposition, and if q be in like manner described on same base as Q, but having its edges perpendicular to the base, then (last Proposition) q = Q, and by the former part of this Proposition, 3 = p, .'. Q = P. Q.E.D. COR. 1. Hence if there be two paralPP ds of the same altitude, but the base of the one is double of the base of the other, the former is double of the latter, and generally if the base of the one is any multiple of the base of the other, the former paralPP d is the same multiple of the latter; and hence, by reasoning similar to that employed in Cor. 5. Prop. 1. of the present article, it appears that paralPP ds of equal altitudes are as their bases. And again, if they have equal bases, they are to one another as their altitudes. CCK. 2. Let there be two paralPP ds P, P', the base and height of one of which are A and h, and one of the other A' and h', where Ah A.'h' are in numbers. And suppose S to be a third paral ppd on the base A' and the height h. VOLUME OF PRISM. 387 Then P : S : : A : A S : P' : : h : h' .-. P : P : : Ah : A'k' COB. 3. If we suppose P' to be a cube which has one of its edges equal to unity, then A' = 1. and /.' = 1. .-. P : P' : : AA : 1. hence if we consider P' to be the unit of solid measure (a cubic inch, or foot, for instance) then P = AA. Hence the volume of a paralPP d is found by multiplying the area of one face by the distance between that face and the opposite one. COR. 4. We have seen that a prism on a triangular base is half of a paralPP d of the same altitude and on a base which is double of the triangle. Hence if A is the area, of the triangle, and h the height of the prism, the volume of this paralw^ = 2A x h, and .". the volume of the prism = Ax h ; or the volume of a prism on a triangular base is found by multiplying the area of the base by the altitude. COB. 5. It is plain that a prism pn a polygonal base can be divided by planes passing through one edge of the prism into a number of prisms on triangular bases, each having the same altitude as the original prism. Let A, A 2 A 3 . . be the areas of these triangles, and h the common altitude, the volumes of all these prisms = (A, + A 2 + A 3 + ... ) h. But the polygon is equal to all the triangles ; hence if A is the area of the polygon, A = A, + A 2 + A 3 + .'. The volume of the prism on a polygonal base = AA, or is equal to the base multiplied by the altitude. Cor. 6. It is plain that Cor. 5 is true of a regular polygon of any number of sides, and therefore is true in limit ; now when the number of sides of the polygon is increased, its limit is the circumscribing circle, and the limit of the corresponding prism is the circumscribing cylinder. Hence the volume of a cylinder = AA, or base x height. N.B. When we speak of the volume of a solid, a prism for instance, being equal to the base multiplied by the height, it is of course understood that all the measurements are referred to the same unit. Thus, if we were asked what is the volume of a prism whose base is 2 square feet, and height 18 inches, the answer is, 18 not 2 x 18, but 2 x = 3, and the 3 is in CUBIT FEET. The cubit foot and \ the square foot being the unit of content and of area corresponding to the linear unit one foot. (4.) If P,ABC is any pyramid on a triangular base, and if through the middle point (D) of one of the sides (AP) we draw planes (DGH and DEF) parallel to one of the faces (PBC) and to the base (ABC) of the pyramid, then if we suppose a plane to be drawn through DE and DH, cutting off a prism EFD.HKC, the figure EDGHCB is double of the prism EDF, HKC. Since the plane EDF is parallel to ABC, ED is parallel to AB. Now AP is bisected in D. .-. PB Fig V 36. is bisected in E. Similarly AC is bisected in H. Also since plane DEF is parallel 388 MENSURATION. to ABC. .*. ED is parallel to KH, and /. KH is parallel to AB. Draw CN perpen- dicular to AB, meeting KH in M, then since CH : HA : : CM : MN (Euclid VI. 2), and CH = HA /. CM = MN. Hence, if from N and C perpendiculars are dra\vn on the plane, DEKH, they are equal. Hence paral ppds on the base DEKH, with these perpendiculars respectively for altitudes, are equal. Now the prism BEKHDG is half the former, and the prism KCHDFE is half the latter. Hence the prisms are equal, and they are together double of one of them. But the two prisms make up the figure EDGHCB, which is, therefore, double of the prism EDFHKC. (5.) Pyramids on triangular Eases and of equal Altitudes are to one anotlier as t/teir Bases. Let P, ABC, p, abc be the two pyramids, and let them be divided by planes as in the last proposition. Fig. 37. Now efd : bca in the duplicate ratio of fd : ca ; but/cZ : co : : 1 : 2 .*. efd : oca : : 1 : 4. Similarly EFD : EGA : : 1 : 4 .'. efd :bca:: EFD : BCA or efd : EFD : : lea : BCA. But the prism efdkch : prism EFDKCH : : efd : EFD : : abc : ABC. .'. (by last proposition) The double prism edghcb : EDGHCB : : abc : ABC. Now if we suppose pd and PD, and also da and DA figures will be formed p,dfe, and P,DFE of the same kind as in p,abc and P,ABC, and also in d,agk and D.AGH, and these figures will have to each other the ratios efd : EFD and agh : AGH respectively ; which ratios are each equal to the ratio abc : ABC. And the same will be true, however often we continue to bisect the bisections of the sides, and hence All the double prisms in p,abc : all those in P,ABC : : abc : ABC. And the number of successive bisections being as great as we please, this is true in the limit. Now the pyramids are the limits of the sum of these double prisms. Hence p,abc : P,ABC : : abc : ABC. Q. E. D. Cor : Hence pyramids on equal bases and of the same altit ude are equal. VOLUME OF PYRAMID. 389 (6.) Every Prism on a triangular Base can le divided into three equal Pyramids. Let ABE, DFC be the prism ; draw a plane through DE and E, cutting off the pyramid E, DFC, and through ED and B draw a plane dividing the remainder into two pyramids E, ABD and E, BCD. Now since these two latter have their vertexes coin- cident at E, and since their bases ABD, BDE are manifestly equal being halves of the parallelogram AC, the two pyramids E, ABD and E, BDE are equal. Now the pyramid E, ABD is clearly the same as the pyramid D, ABE. But the base ABE = base DEF, and the perpendicular from E on DFE, is equal to that from D on ABE, since these planes are parallel ; hence the pyramid D, ABE m pyramid E, DFE, and therefore the three pyramids are equal. COE. 1. Hence if a pyramid and a prism have equal triangu- lar bases and are of equal altitudes, the pyramid is one third part of the prism. Con. 2. Let A :=: area of base, h height of pyramid. Then the volume of the prism on base A and of height h = Alt. And the volume of pyramid = 3 COR. 3. If we have a pyramid on a polygonal base of which the area is A, it can be divided into triangles whose areas we will suppose to be A t A 2 A 3 . ... so that A = A, -f A 2 + A 3 + ... Now if h be the height of the pyramid, then its volume being clearly the sum of the pyramids whose bases are A t A,, A 3 .... and height h, will equal - h + A + A = -AA. COR. 4. This is true, however great the number of sides there are to the polygon, and hence is true in the limit. Now if we suppose the polygon to be regular, its limit is the circumscribing circle, and the limit of the pyramid is the circumscribing AA, cone .". volume of cone where A is the area of the circular base. o (7.) To determine the Volume of the Frustum of a right Prism, on a triangular Base. Let ABC, DEF be the frustum, where ABC is per- pendicular to the edges. Let A =. area of ABC, and let &j 7t 2 Ji 3 be the edges AF, BE, CD respectively, and V the required volume. Join FC and suppose plane to pass through FCE, cutting off the pyramid C, FED, and another through EC A catting off the pyramids E, ABC, E, ACF, these three make up the volume V. Join FB, DB, and DA. Now volume of E, ABC = A ^ ; volume of E, FAC = volume of o B, FAC; since perpendiculars from E and B on the plane ACF are clearly equal, and B, FAC is the same pyramid as F, ABC, the volume of which is A . 3 Again, since FA is parallel to DC, the triangle ACD 390 MENSURATION. is = the triangle FCD, and we have already seen that the perpendiculars from E and B on the plane ACDF are equal, .'. the pyramid E, CDF = the pyramid B, ACD, i. c. = D ABC, the volume of which is A ^. o V = A 3 ' 3 3 ' 3 Cor : If the prism be ABC FED, in which neither of the cuds is perpendicular to the edges, take ale an area whose plane is perpendicular to the edges, and let A = area of dbc. Let AE = A x BD = A 2 CF = h 3 aE = x v ID = x 2 cF = x 3 aA = y t bE = y a cC = y a Then if V = volume required V = a&c DEF + ale ABC = A *' + * + * 3 + =: * y i A, + A Fiff. 40. A 2 A 3 A 4 be the edges at (8.) To determine the Volume of a Frustum of a Right Prism on a base which is a Parallelogram. Let ABCDEFGH be the frustum in question; let A. B. C. D. respectively ; draw AC, and if a plane E pass through AC and E it divides AB . . . . H into two frustums similar to that in the last proposition. Let V be the required volume, and A, the area ABC which is = area ADC (Euclid 134). Now the volumes of the two triangular prisms are respectively hJ>2 , and A AI + * + A I A If we had divided the figure by a plane passing through BD, we should have had Similarly if we had divided at C and D, we should have had respectively 1 1 A J, 7 V = 1 A, (A 4 + A! + A 2 ) -f - A, (A 2 + A 3 -f A 4 ) 3 3 Now let S = A, + A, + A 3 + A 4 . Then adding the four values of V we have 4V = ! A 3 = 2A r S "VOLUME OF FRUSTUM OF PRISM. 391 .*. If A is the whole area ABCD. Since A = 2A, V _ A S _ A h, + h a _ + h 3 t- h t 4 4 (9). To find the Volume of a Frustum of a right Prism on a regular Pentagonal Base. Let P! Pff 3 P P 5 be the base, and let h t A 2 h 3 h t h 3 be the edges corresponding to these angles. Join P x P 3 and P 1 P 4 , and let the areas of P! P 2 P 3 and Pj P 4 P 5 , which are equal, be each A 1? and the area of P 1 P 3 P 4 be A 2 . Let V be the required volume, then this volume consists of three portions of prisms of the same kind as in article (7) and hence V = ^ (\ + ** + A s) -I' ^ <*i + A s +*,) + % & + A* + fy 66 6 Now divide the base by lines drawn through P 2 the areas will be the same as before, and hence and similarly by dividing the base by lines drawn successively through P 3 P 4 and P 5 we obtain V = i Pig. 42 Then adding these together and writin = S we have 5 V = ^L (S + S + S) + ^L (S + S I S) + - 1 (S + S r S) 3 3 '3 .-. 5V = (2A X + A,) S = A. S. If A = area of base. The student can easily prove that a similar formula is true in the case of a prism on a hexagonal base, or indeed on any base which is a regular polygon. 10. To find the Volume of a Prismoid. A DEF. A prismoid is a solid of the form represented in the accompanying figure. ABCD is a rectangle, and ABFE, DEHG, are planes perpendicular to _ the planes of the rectangle, ED and CF are planes inclined at given angles to the plane of the rect- angle, the lines EF and GH being parallel to AB and CD. Through AD and BC draw planes ADLK and Fig. 43. MENSURATION. BCNM perpendicular to the plane of the rectangle, these divide the given figure into three, viz., AEKDLG, BDN, andBMFCHN. Let AB = a AD = h AK = x DL = x i EK = b GL == b l MF = c NH = c l Then (a) the figure BDN is of the same kind as in article 8. Now the area DB = ah, if therefore Vj is its volume, Vl = ^ (X + X -r X, + Xj = ^ (X + X,) (b.) Let V 2 be the volume of the figure AEKDLG, draw planes through ED,K and GD,K. dividing the figure into three pyramids, DGL,K, AEK.D, and DEG,K. The volume of GLD,K KL x l b l Ji h , Similarly volume of AEK,D = xb. 6 Now the volume of GED.K is to that of ADE,K as GED is to ADE, or as AE to GD, or since triangles AEK, GDL are similar, these volumes are as AK to DL, or as EK to GL. .- Volume GED,K x b x or = ? . xb x ~ 6 x 6 o h li 44. (c.) Similarly, if V 3 is the volume of MBFCNH, h f x,c + xc l V 3 = e" [*& + XC + 2~~ Now if V be the required volume, V = V, + V 2 + V 3 . .- V = - [ 3 a (x + x,) -f x t (6 X + c,) + x (b + c) D I b + c b. + or re-niTauging the right-hand side of the equation, + (a: + ^) Now in this expression, 6 + c ft, + c ' x f a+ ^-~-\ is the area of ABFE- (Fig. 43.) TI ( a + ^T^) is the area of DCHG ' VOLUME OF A RAILWAY CUTTING. 393 and (x 4 x b + c I + c , which equals I is clearly four times the area of a section made by a plane parallel to ABFE and DEHG, and half way between them. Hence the solid content of a prismoid is found by the following rule : " To the by areas of the ends add four times the area of the mean section, multiply this sum ith of the height of the figure ; this product is the volume required." In practical cases it will generally happen that 6 = c, and b L = c t . This affect the enunciation of tJte rule, but simplifies the formula, which becomes V = ~ { 3 a (x + jcj + x^ (2 b i + V) + x (2 6 + 6 a ) 1 . (11.) To find the Solid Content of a Railway Cutting. In the last article the figure is very nearly that of a portion of a railway cutting, iii which ABCD is the road, AG and BH, the sloping sides of the embankment, and hence the solid content required can be found by means of the rule given in the last article. It is to be observed that the rule requires E G and FH to be straight lines, or, as an approximation, to be very nearly straight lines, or EH to be a plane, which is not true if the cutting is a long one. For this case we derive the following rule from the above formula. P2B + 1 Pig. 45. Let ABC represent a section made by a vertical plane of the hill to be cut through, AB the level of the road, and suppose sections of the cutting to be made by planes perpendicular to AB, at equal distances along that line, viz., at M t M 2 M 3 . . . M^n + i, let the terminal sections at A and B be a and b, and the sections at PjMj P 2 M 2 . . PiPzPs P2nP-2n + i, the number of sections being odd, and let the common distance between the sections be h. Now by last article the volume of portion AM P 2 D is ' 2 . (a + 4 pi + jp 2 ) or (a + 4 p\ + p- 2 ) o 7i and the volume of the portion P 2 M e M 4 P 4 is . (p z + 4 p 3 + p ), and so on, and the volume of Po M 2 BC is . (p* + 4 p% + i + 6). 394 MENSURATION. Hence, by addition, the whole volume required is _. (a + b + 4 (p x + p s + . . . + p., m + which may be expressed as a rule as follows : Between the first and last sections make an odd number of sections at equal distances along the road, the planes of the section being perpendicular to the road ; then one third part of the common distance multiplied by the sum of the first and last sections, with four times the sum of the odd sections, and twice the sum of the even sections, gives the volume between the first and last section. The student will observe that this rule is the same as that for finding the area of a figure bounded by a curve, which has been already given (p. 379), excepting that the ordinates ia the former rule are replaced by trapezoidal sections in the latter. The formula given in article 10 is called the Prismoidal formula. It will be observed that the material oil each side of a cutting being generally the same, is the reason the inclination of the planes AG and BH to BD being generally the same as stated in the last article. Mr. Macneill, to whom the prismoidal formula is due, has constructed tables founded on that formula, by which the volume of a cutting is very readily calculated. The volume of an embankment is to be found by the same formula, since, as a question of mere figure, an embankment is only an inverted cutting. If the calculation is made directly from the formula it is very tedious ; the value of the Tables above referred to is therefore very great, and is enhanced by the following circumstance : In constructing a long line of railway, the earth taken out of the cutting should be sufncient to form the embankment, otherwise land must be purchased for the mere purpose of obtaining earth ; to effect this end of making the volume of the embankments equal that of cuttings, the ascents and descents (gradients) of the line of road have to be properly chosen, and this can only be done by trial, so that the calculation may have to be performed two or three times before a right adjustment can be hit upon. It is worth adding that, as a general rule, a cutting is followed not by a long level, but by an embankment; if possible, these two are adjusted to each other to prevent the need of carrying earth from long distances. 12. To find ike Solid Content of a Military Earth-work. The form of a military earth-work will be understood from the following explanations : The form of a section of the work made by a D vertical plane perpendicular to the face of the work is such as ABCDEF from B,C,D,E draw perpendiculars to AF, viz., Em, Cn, Dp, ~Eq, then Am, mm, np, pq, q~F, are of known magnitudes, as also are B/n, Cn, Dp, ~Eq. The plan of the work will be of the accompanying kind, viz., F'f pj_ 4 g (Fig. 47) is the line corresponding to F (Fig. 46) E'e the line corresponding to E, and so on for the others ; the lengths of all these lines are known ; we will call them a,b,c,d,c.f, respectively. In fig. 46 join pE, pC, p'E, dividing the section of the work into triangles ApB, BpC, CpD, DpE, and E/>F; call these areas respectively A 1 A 2 A 3 A i A 5 , then the cont^its of VOLUME OF AN EARTHWORK. 395 the work are clearly equivalent to the frustums of five prisms (similar to that in the Corol. to article 7), which have the area of perpendicular section A 5 and edges f,e,d, section A 4 and edges e,d,d, section A 3 , and edges d,d,c, section A 2 and edges c,d,b, and section AI and edges b,d,a, it being evident that the edge through p d. Hence if the whole volume equals V, we have V - -f (f + e + d) + ^ (e + d + d) C' D' Fig. 47. -o 2 (d + d + c) . + -g 5 (c + d + b) + (b + d + a) .-. 3 V = a Ai + 6 (A. t + A 2 ) + c (A 2 + A,) + d(A 3 + A,) + c(A, + A 5 ) +/A 5 ) + d (A, + A, + A 3 + A, + A.) In which formula it will be observed that each line in the plan is multiplied by 'the triangle, or by the sum of the triangles, which have an angular point in that line, and that the line dD' is also multiplied by the whole area of the section. Hence if these products are formed and added together, the required volume is one third of the 13. To find the Volume of the Frustum of a right Cone made ly a Plane parallel to the Base. Let ABCD be the frustum of the cone PCD. Join P00 t where centres of the ends of the frustums. Let AO = r C0 l i\ 00, = k. PO = x. Then volume ABP = " g !^L volume PCD = ^ ( x + ^ 3 3 .-. Volume offi\istum = . r* h + v ^ TI ' r * ' 3 3 Now x + h : r 1 : : x : r. .'. h : x : : t\ r : r. .-. Volume of frustum = r,-h + * ( r ^~ hr 3 r,r are the which is the same as the volume of three cones whose common height is h, and which have the radii of their bases respectively r, r 1 and a mean proportional between r and r,. 398 MENSURATION. J \ Fig. 49. 14. To find the Volume of a Portion of a Sphere. Let ABO be a quadrant of a circle, draw BC, AC, tangents at A, and B, join OC, and take any point P in AB and draw QPN ^ a parallel to BO ; then if the whole figure revolve round AO, the quadrant ABO will clearly describe a hemisphere, the square AOBE will describe a cylinder, on a base whose radius is OB, and height AO, and the triangle AEO will describe a cone on a base whose radius is AC and height AO. Also APN will describe a portion of a sphere, ACQN a cylinder whose height is AN and radius AC, and ACMN a frustrum of a cone, the radii of whose ends are AC and MN, and height AN. Divide AN into any number of equal parts, of which let m m^ be one, through m, draw mk parallel to BO, meeting AB in p, and OC in Ic, and through m 1 draw a line m^ Tc^ parallel to ml;, and from p and h draw pp^ and hh 1 perpendicular to TO X ij. Join Op. Now Op* mp- + mO 2 since Omp is a right angle. But since triangle Om/t is similar OAC and AC AO /. Om = mh also km = BO = Op __ mk- = mp" + mh- .-. ir x mk- = TT x mp" + TT x mh- or (Mensuration of Areas, Art. 8) the circle, the radius of which is mk is equal to the sum of those whose radii arc mp, and mh ; also since kJu 1 = pp^ = hh^ we have it mk- x kki = IT. mp" x pp^ + IT mh- x .7^ x or the cylinder, the radius of whose base is mTc and height k k\, is equal to the cylinder, the radius of whose base is mp, and height pp v together with that the radius of whose base is mk, and height hh 1 ; but these cylinders are those which will be described by mk 1 mp 1 and mli v when the whole figure revolves round AO. And the same is true of the cylinders corresponding to any other one of the equal parts into which AN is divided, and therefore is true of all of them. Now all the cylinders corresponding to km l make up the cylinder described by AQ, and all those described by !j p, make up the series of cylinders inside the sphere, and those described by m l h make up those described about the cone. Hence, Cylinder AQ - sum of cylinders inside portion of sphere + sum of cylinders outside portion of cone. This being true, however small m m 1 may be, is true in the limit, but the portion of the sphere is the limit of the inscribed cylinders, and the frustrum of cone is the limit of the cylinders outside the cone (compare note, p. 376). Hence, Cylinder described by AQ =r portion of sphere described by APN + portion of cone described by ACMN. Now let r radius of sphere, h = height of AN, and V, = volume of portion of sphsre, NM = ON = r h. Now column of cylinder described by AQ = IJT -h. And volume of frustum of cone described by ACMN = ^- x -I ?- + (r h) r + (r h) 2 i- ir/4 r 1 /. trr-h = V, + r- + (r J>) r + (r /i) 2 } o I J VOLUME OF A SPHERE. 397 = STT r"k ir/i -j r- + (r /t) r + (r 7i) 2 [ V, = 3** Hence if V = volume of the whole sphere, in which case h = 2i: V = 47T r 3 Now the volume of cylinder circumscribing sphere in 2 x 2?-. . o .'. V = . (circumscribing cylinder.) 8 Con. 1. The above proposition may be demonstrated in the following manner. Suppose a solid having any number of plane faces to be described in the sphere, and let A! A 2 A 3 &c., be the areas of these faces, and p^ p z p 3 &c., be the perpendicular distances of these faces from the centre of the sphere : now this inscribed solid may be conceived to be made up of pyramids, the bases of which are the faces of the solids, having the centre of the sphere for their common vertex. Hence, if the volume of the solid is Y! V, = A 2 p, + Now this is true, however great the number of faces may be, and /. is true in the limit, but in the limit 2^ P 2 P 3 become equal to one another and to r the radius of sphere. Hence, Limit of A! p l + A.,, p., + A 3 p 3 + . . . = r x (limit of A t + A 2 x A 3 . . .) Now the limit of A t + A + A 3 + . . . = surface of sphere = 4ir r" (Mensuration of Areas, Art. 19). Also the limit of V t = V the volume of sphere. , , T 1 4JT7- 3 . V = - r x 4?r r* = - . 3 3 COR. 2. To find the volume of the portion of the sphere corresponding to BPNO. Let the volume be called V, and let h = ON which is = MN. Then V = cylinder BN cone ONM. Now volume of cylinder BN = *r% Volume of cone UNO = - w/t 3 3 V = 70-274 3 = irk 3 ITJ \ - - /t j I 3 / 15. To find the Volume of a Spheroid. DEF. A spheroid is a figure formed by the revolution of an ellipse about one of its axes ; if about the major axis it is called a prolate spheroid, if about its minor axis it is called an oblate spheroid. Let ABa be a semi-ellipse, OA its semi-major axis, OB its semi-minor axis. "With centre and radius OA describe a semi-circle ACa, draw QPN through any point in the ellipse parallel to OC, Fig. 50. 398 MENSURATION. draw Mwm. parallel to NPQ and complete the rectangles MP, MQ (Mensuration of Areas, Art. 12). Now if the figure revolve round Act the semi-ellipse will describe a prolate spheroid and the circle a sphere, also MP and MQ will describe cylinders with altitudes MN and having the radii of the bases NP and NQ respectively, and /. having volumes it MN x PN 2 and TT. MN x XQ 2 . Now by a property of the ellipse, QN : PN : : a : b /. QN 2 : PN 2 : : a 2 : 6 2 But Cylind. MQ : cylind. MP : : MQ 2 : MF /. cylind. MQ : cylind. MP : : a 2 : b- and the same is true of any other cylinders described hi like manner within the sphere and spheroid, and .'. is true of their sum .*. all cylinders within sphere : all within spheroid : : a 2 : 6 2 and this being true, however many cylinders there may be, i. e. however small we suppose MN to be, is true in the limit, but the sphere is the limit of its inscribed cylinders, and the spheroid the limit of its inscribed cylinder, c, , . . . 70 4ira 3 4ira5 2 . . Sphere : spheroid : : a- : o- : : : - . 3 3 But volume of sphere . o Volume of spheroid := . 3 Cor : In like manner the volume of an oblate spheroid is o 16. To find the Volume of a portion of a Spheroid, cut affly a plane Perpendicular to the Axis of Revolution. In figure 50, suppose we wish to find the volume of the portion of ths spheroid corresponding to the portion APN of the generating ellipse ; then, as in Art. 15, we shall have Portion of Spheroid : portion of sphere : : 6 2 : a* Now (Mensuration of Solids, Art. 14.) if AN = h the volume of the portion of the sphere is equal to -!LL.(3ft h.) .*. If V x is the volume required, 3 VTT li~U" in J \ = _ (3a A). 3a- Similarly if we wish to find the volume of the portion corresponding to ONPB, and if V 2 is this volume, and h = ON, / 1 7/ 2 \ V = M- I 1 . _ ) V 3 a- / as is evident from corol. 2. Art. 14. Mensuration of Solids. 17. To find the Volume of a Cask. "We may consider a cask to be either the middle portion of a spheroid, or two frustums of equal cones joined together at their bases, though it will not coincide VOLUME OF A CASK. 399 with either of these forms exactly. It is to be observed that excise officers generally consider ca.-ks to be of the first form. The measurements that are most easily made in practice are the diameters of the end, and of the middle section, and the distance between the ends ; we will call these, d, D, and Tc , respectively, and investigate the rule in each of the above cases. (a.) Suppose the cask to be a portion of a spheroid. Its volume V will be double that of V in th last article /. V = 27rA6 2 ( 1 1 \ 3 a? Also by a property of the ellipse, (See fig. 50.) PN 2 ON* OB 1 + OA 7 ' PN* ' 1 PNii I A 8 _ I 3" ~~& + 3"' a* ~~ "3* .-. 2 1 PN 2 1 h* + "3 ' ~^~ =1 ~3c^ r ... v = ^w (? + 1 P _E 3 6* Xow 2k = L 25 = D 2. PN = d Also = .2618 very nearly, hence the rule. " To twice the square of the 1 a middle diameter add the square of the end diameter, and multiply the sum by the length of the cask, this product multiplied by .2618 gives the content of the cask." (b.) Suppose the cask to hare the form of a double frustum of a cone. If v is the volume required v is clearly double of the volume ABCD, fig. 48. And hence by Article 13, Mensuration of Solids, * = ~ ( *i' + ". + >" ) Xow D 2i\ d 2r and Jc= 2h .-. v = . Tc ( D 2 + Dd + dA . 12 V / Hence the rule " To the product of the diameters add the sum of their squares, multiply this by the length of the cask, then the whole product multiplied by .2618 gives the contents of the cask." N.B. If the measurements are made in inches, the above rules give the required contents in cubic inches ; to obtain the contents in gallons we must divide by 277.274, since 277.274 cubic inches go to one gallon. COB. It is evident that v, the value given by the second rule, is less than the true contents of the cask ; it is to be observed, also, that V, the value given by the former rule, is generally greater than the true value ; so that the true value will lie somewhere between these two results. Hence we can easily estimate the amount of 400 "UENSCRATION. Accuracy in each of the above determinations. Thus, V r is clearly greater than the difference between the true result, and either of those given by rule, and v is less than the true value of the contents, hence the error committed by either way of (Y v \th J . of the whole. Now V_r = *JL f 2D* + rfO - ^ ( D- + Dd + d> ) = = !D (D-d). 12 ,.T- TI D{D - d > D(D_* jja i" D " + Dd + d " 12 If D = d + n. Then Dd = D- Dn, and d- = D ! 2D?i + n V v Dn 3D r 3D- 3D + n- 71 n* " D + 3D 3 which is ari expression for a limit of the part of the whole, by the approximate differs from the true value, c. g. Suppose the diameters to be 18 and 20 inches respectively, n 1 then n = 2 inches and = -, and there the error committed by calculating 1 "so" according to either rule cannot be so much as of the whole, or so much 1 L a. _ 10 300 ;is -^rr th or Ky th (very nearly) of the whole. If the cask in question had an interior length of two feet, then, by the first rule, its contents are 25'47 gallons, and, by the second, its contents are 24-56 gallons. So that the error committed by either way of gauging must be less than one gallon. INTRODUCTORY. 401 SPHERICAL TRIGONOMETRY. BEFORE reading the following treatise, the student will do well to reperuse the treatise of Spherical Geometry already given (p. 251, &c). He will there find the definitions enunciated and the chief properties of spherical triangles proved which are employed as the premises from which the formulas of the following treatise are deduced. It is stated in the introduction to that treatise that the chief applications of this science are found in practical astronomy and geodesy ; also it is stated on p. 256, that the side of a spherical triangle measures the angle it subtends at the centre of the sphere, and hence is spoken of as an angle ; now it is to be observed that in practical astronomy, the measurements made by the various instruments are invariably the angles subtended at the eye of the observer by arcs of the great sphere, for instance, the altitude of a star is measured directly as an angle, so that in these cases the radius of the sphere never enters into consideration ; but in the case of measurements on the earth's surface, if we have a distance measured along a great circle in miles or yards, which is to enter into our calculations, we must determine the angle these yards or miles subtend at the earth's centre ; thus if a is the length in question, r the radius of the earth, 6 the angle, then 6 = where is r in circular measure. If 6 contains n, then n = . Further it will be vr observed that in case the sides of a spherical triangle are small compared with the radius of the sphere, the triangle does not differ sensibly from a plane triangle: e.y. a triangle on the earth's surface the sides of which are each about a mile long will not differ sensibly from a plane triangle, unless the measurements are made with very refined instruments ; hence it is manifest that the plane triangle is the limit of a spherical triangle, and accordingly we shall find that the formulas for the solution of spherical triangles are quite analogous to those that have been already deduced for the solution of plane triangles (pp. 322, 325), and we shall see that the latter can be deduced from the former by considering the plane triangle as the limit of the spherical triangle. N.B. The following results already proved on p. 259 are very important. Let A, B, C, a, 6, c, be the angles and sides of any spherical triangle, and A', B', C', a' b', c', the angles and sides of the corresponding polar triangle. Then A + a'=B + 6' = C + c'^: 180 And A' + a = B' + & = C' + c = 180 We shall employ this notation for the angles and sides of a spherical triangle and of its polar triangle throughout the following treatise. MATHEMATICAL SCIENCES. No. XIII. 402 SPHERICAL TRIGONOMETRY. (1.) To show that the Sines of Ike Angles of a Spherical Triangle are proportional to the Sines of the opposite sides. Let ABC be the triangle, the centre of the sphere, join OA, OB, OC ; through A draw a plane ANT perpendicular to OB, cutting the plane AOB in AN and BOG in NP, these lines are perpendicular to OB, and the angle ANP measures the inclination of the planes, and is /. equal to the angle B of the triangle. Through A draw another plane AMP perpendicular to OC, cutting AOC in AM, COB in MP, and AMP in AP, then AM and MP are o< perpendicular to OC, the angle AMP is equal to the angle C of the triangle, and AP is per- pendicular to the plane BOA, and .*. APN and APM are each right angles. Hence AP AN' Sin. B = A n and sin. C = AM sin. C Also since c is the angle AOB and 6 the angle AOC AN , . , AM sin. b sin. c = and sin. 6 = .-. = OA OA sin. c sin. B _ sin. b sin. B _ sin. C sin. c sin. c sin. 6 sin. c The same proof holds good of the other sides and angles, sin. A sin. B sin. C AM AN Hence sm. a sin. b sm. c (1). Q.E.D. COR. 1. Suppose a, b, c, to be the lengths of the sides BC, CA, AB, then the angles denoted in formula (1) by a, b, c, are - . r being the radius of sphere. Hence b sin. 5 r Bill. sin. B _ r _ _ _ sin. C sin c _ in circular measure, r sm. r sin. _ sin.* Xow in the limiting case when r is infinite, = o and = o r r b c and /. (Plane Trig. Art. 47) = 1, and = 1. /. in the limit -. = The formula for plane triangles (p. 322). sin. O c FUNDAMENTAL FORMULAS. 403 COR. 2. If through and AP a plane be drawn cutting the surface of the sphere AP in Ap, then Ap is perpendicular to BC, and equals the sine of AOP : i.e. is equal O*\. to the sine of Ap, which we will call p. AP AN AP Now sin. B sin. c x 1- sin. Ap. AN OA AO sin. B sin c = sin. (2). (2.) To prove the Formula cos. A = cos. a cos. 6 cos. c sin. 6 .siu. c As before, let be the centre of the sphere, and ABC the triangle, join OA. OB. 00, and produce the planes AOB, BOC, COA indefinitely ; at A draw a plane API perpendicular to OA, cutting the planes AOB, BOC, COA hi Ap, pq, qA respectively, then since pA is on the plane AOB, and perpendicular to OA, and qA is on the plane COA, and 2; perpendicular to OA, pAq is the angle between the planes, and .". is equal to the angle A of the triangle, also the angle pOq is the angle subtended by BC : i.e. is the angle a. Hence (Plane Trig. Art. 37) Fig. 2. Ap 2 + Ap "OA 2 Ap. Aq cos. A = pq* = Op" 4- C2 2 2 Op. Oq cos. a. = tan - c = Aq sin. b %Z = tan. & = r OA cos. b OP 1 777 =sec. c= OA cos. c 0? 7. 1 = sec. o = - - OA cos. 6 sin. 2 c sin. 2 b coa. 2 c cos. 2 6 2 sin. c sin- b - 2 sin, c sin, b cos. c cos. 6 cos. A = + ra .- 2 cos. cos. c cos. 6 and 2 cos. a 1 cos. A = , - COS- C COS. COS/ C 1 sin. 2 c cos. 2 c cos. 2 b c cos. c cos. b Similarly cos. 2 1 cos. 2 c sin. 2 6 cos. 2 c cos. 2 6 = 1. cos. 2 b cos- 2 6 cos- 2 c 2 sin. c sin. 5 = 1. cos. A 2 cos. 2 7 WW J-*. ' ' T COS. C COS. COS. C COS. sin. c sin. 6 cos. A = cos. a cos. c cos. 6 cos. a cos. b cos. c cos. A = .... (3) Q. E. D. sin. b sin. c COR. 1. If a.b.c. represent the lengths of the sides of the triangle, and r the radius, SPHERICAL TRIGONOMETRY. Then . - . . are the angles of formula (3) in circular measure. r r 6- e 4 & Hence, remembering that cos. 6=1 _ + . . and sin. = 1 . -j -1 . . >.-!. J. ***o we have Cos. A = - aa / , tf\ /-, j2 c * \ ! ( 1 . 1 1 -- -- 1 + terms involving . . . V 2r*/ \ __ 2?* 2rV & r* 6c~~~ 1 4- ... terms involving , + ... 13 4- c- a* + terms involving + . . . 2 be -r- terms involving . + . . . r r* Xow in the limit when r is infinite, the terms involving . . will all r r* [ disappear ; hence in the limit, 7* _l_ ' 2 Cos. A = 2 be a- Zr + c- 2 be cos. A. as in the case of the plane triangle. COR. 2. In formula (3) substitute 180 a' for A. ISO A' for a. 180 B' for b, and 180 C' for c, and we shall have cos. A' + cos. B' cos. C ' Cos. a' rr . T,,- . >,. sin. B sin. C' This is true of the sides and angles of every polar triangle. Now it appears from Prop, xi., p. 259, that every triangle may be regarded as the polar triangle of some j other ; hence the above formula is perfectly general, and is true of every triangle, and we have p co?. A 4 cos. B cos. C ,,, sin. B sin. C Con. 3. Forrr.ulas similar to (3) and (!) are, of course, true of Cos. B and Cos. C, i and of Cos. b and Cos. c. I (3.) To express the Formulas of the la&t A rticle in a Form adapted for Logarithmic Calculation. cos. a cos. b co?. c Since Cos. A = .'. 1 -f cos. A = I + sin. b sin. c cos. a co?. b cos. c cos. a cos. b cos. c + sin. b sin. sin. L sin. c sin. b sin. c FUNDAMENTAL FORMULAS. 405 i . i cos. a cos. b cos. c cos. 6 cos. c + sin. c sin. c cos. a i And 1 cos. A = 1 *r> ~ ss = r = sm. o sin. c sm. o sm- c A cos. a (cos. b cos. c sin. 5 sin. c) cos. a cos. (6 + c) 2 sin. 6 sin. c sin. b sin. c A cos. (6 c) cos. a And 2 sin.- 77- = - - - . i .- 2 sm. o sin. c A 2 sin. A (a -i- 6 -i- e) sin. i (5 + c a) .. 2 cos. 2 := 2 sm. 6 sin, c A 2 sin. i (a b + c) sin. i (a + b c) And 2 sin. 2 77-= = r = 2 sm. o sm. c Xow ifa-f-6 + c = 2, then 6 + c a = 2 (s a) a b -f c = 2 (s Z>), and a + b c = 2 (3 c). 2 sin. 6 sin. c sin. 2 ^ = sip ' ( 6 ) sip - ( c ) 2 sin. 6 sin. c / tan. 2 - = sin ' (' V) 8m - ( c ) 2 sin. s. sin. (s a) And since. sin. A = 2 sin. cos __ sin> s> sin> ( $ ~ ^ sia ' ^ ~~ b) sin> ^ "" sin. 2 6 sin.- c these formulas are analogous to the formulas on p. 324 of plane trigonometry, which can be shown to be the limits of these in the same manner as in Cor. (1). Art. (1), and Cor. (1) Art. (2). It is to be observed that, since any two sides of a triangle are greater than the third, s a, s b, s c, are positive ; and since all the sides of a triangle are less than four right-angles, s is less than two right-angles, and a fortiori, s a, s b, s c, are each less than two right-angles : so that, sin. s., sin. (s u), sin. (s b), sin. (s c), are each positive, also b and c are each less than 180 ; so that sin. b and sin. c are always positive. Hence, (5), (6), (7), (8), are always positive, and .*. the A A A values of cos. s j n . tan. ^- and sin. A derived from them are always real. A a' Cor : If we consider the case of the polar triangle A = 1 80 a' .'. 90 a 180 A' b 180 B' c = 180 c'. .-. s 270 I (A' + B' + C') = 270 S' if S' = 1 (A' + B' + C') .'. s a = 90 (S' A'), s 6 = 90 (S' BO s c= 90= (S' CO- a' cos. S' cos. (S' A') Hence sm. 2 = : ^7 ^-~- A sm. W sm. C with similar expressions for cos. 2 cotan. 2 . and sin. 2 a': and since these are true 2 2 for all triangles we shall have o a cos. S cos. (S A) ,^ 406 SPHERICAL TRIGONOMETRY. a cos. (S B) cos. (S C) Similarly cos.* 3 = sm . B sin C < 10 > cos. S cos. (S A) 2 cos. (S B) cos. (6 (J) 4. r and sin. 2 a ^-., ,, . ^ 1 cos. S cos. (S A) cos. (S B) cos. (S C) [ (12) sin.~ j-> fciio." \*> [_ " j where 2 S = A + B + C. It is to be observed that A + B + C must be less than six right-angles, and greater than two right-angles ; /. S > 90 < 270. .". cos. S is always negative. Also S A being equal to - ^B + C A) = - /(1 80 (&' + c'a') j must be < 90, since V + c' > a' .'. cos. (S A) is always positive, and similarly cos. (S B) and cos. (S C) are always positive, and .'. (9) (10) (11) (12) though in appearance a a negative, are really always positive, and .. give us real values for sin. -5 > cos - -, 2 a tan. , and sin. a. & b c Similar formulas, of course, exist for sin. - sin. ^ &c. (4.) To prove the Formulas, ab a b cos. sm. Tan A *" cotan. - and Tan. A ~ B = cotan. J? 2 a+b 2 2 a+b cos. sin. Q. , A sin. (s b) sin (s c) . B sin. Since tan.* = - - tan. 8 = . . sin. s. sin. (* a) ' 2 sin. s. sin. (s 6) 2 C sin. (s a) sin. (s b) 2 ~ sin. s. sin. (s c) , B 2 C sin. (s c) sin. (s a) sin. (s a) sin. (s b) sin.* ( a) '2 2 sin. s sin. (s b) sin. s. sin. (* c) siu.*s B C sin. (s a) . tan. tan. = - sm. s C , A sin. (s b) A , B sin. (a c) Similarly, tan. - tan. _ = ^ and tan. y tan. - = _J_^ tan. B tan. + tan. tan. A "(>-) ". (*-) Button. A -J tan. 5 = 2 2 sin., sin., 2 2 . A , B 1 tan. tan. 2 2 j / \ 1 / ' 2 sin. - 1 2s a b \ cos. -Ib a /. tan. A + E tan. 5 = sin, (a a) + sin, (s b) = / 2^ 2 2 sin. s sm. (s c) n . 1 1 / 2 sin. _ c cos. - I 23 i NAPIER'S ANALOGIES. 407 sin. _ c cos. _ (a, b\ 2 2 V } sin. _ c cos. _ .'. tan. Similarly A r, 2 sin. - (a Acos. - (2s a b\ tan. A ~ B tan. = sin, (s 5) an. (-o) > 2V 2 2 sin. s + sin. (s c) . 1 / \ 1 2 sin. _ / s c \ cos. _ c. 2 V / 2 . p sin. -(a I) .'. tan. ^ = 1- - cotan. | ( 14 ) sin. 2 (a + i) These formulas are clearly analogous to (36) on p. 325 Plane Trigonometry. Cor : If we take the polar triangle, since ' + V A B If of a + b A' + B' 2 = 180 _ _ a _ ~ ~~' ~~ a l> B' A' , C c' =__, and - = 90'--. We have from (13) and (14) B' A' B' A' a> + l> COS '-2 , C ' I' -a' Sin '^ C ' - tan. -g- > , tan. , and tan. = - tan- sm. Hence, remembering that the formulas for the polar triangle are perfectly general, we have A B cos. - . A B sm (5.) To /irow <7ie Formula Cotan. A sin. C = Cotan. a sin. 6 Cos. C cos. 5. This formula is used in certain propositions : e.g. it is employed in the astronomical problem of finding the aberration in declination. 408 SPHERICAL TRIGONOMETRY. From formula (3) we have Cos. A sin. 6 sin. c = cos. a cos. b cos. c. And from formula (3) we have Cos. c = cos. a cos. b + sin. a sin. 6 cos. C. .'. Cos. A sin. 6 sin. c = cos. a cos. a cos. 2 b sin. a sin. b cos. b cos C. .*. Cos. A sin. b sin. c = cos. a sin. 2 b sin. a sin. 6 cos. b cos. C. .'. Cos. a sin. c = cos. a sin. b sin. a cos. b cos. C. Also from formula (1) we have sin. C sin. a Sin. c = sin. A sin. C sin. a . , Cos. A cos. a sin. 6 sin. a cos. 6 cos. C. sin. A Cotau. A sin. C sin. a = cos. a sin. b sin. a cos. b cos. C. Cotan. A sin. C = cotan. a sin. & cos. C cos. b. Q.E.D. THE SOLUTION OF RIGHT-ANGLED SPHERICAL TKIAXGLES. There are as many as six different cases of right-angled spherical triangles, as will appear from the following considerations : Let A B C be the triangle, having a right angle at C. Then using the ordinary notation, all possible cases are the following : (1.) Given the base and perpendicular, i.e., given a and b. (2.) Given the hypothenuse and another side, i.e , given cand a, or c and b. (3.) Given the base or perpendicular, and an adjacent angle, i.e., given a and B, or b and A. ' (4.) Given the base or perpendicular and an opposite angle, i.e., given a and A, or b and B. (5.) Given the hypothenuse and an angle, i.e., given c and A, or c and B. (6.) Given the two angles, i.e., A and B. If these cases be compared with those on p. 361, for plane triangles, it will be seen that the third case of plane triangles diverges into t\ve cases, viz., the third and fourth of spherical triangles, while the sixth case is peculiar to spherical triangles. Both of these differences are due to the circumstance, that in the spherical triangle, A + B + C is not known, whereas in the plane triangle A + B + C = 180 G. To investigate the Formulas on which the Solution of Right-angled Spherical Triangles depends. , , , sin. c sin. b sin. a From the general formula sin. C sin. B sin. A Since C = 90 and .-. sin. C = 1, we have sin. b sin. a sin. c - sin. B sin. A SOLUTION OF RIGHT-ANGLED TRIANGLES. 409 .*. sin. b = sin. c sin. B (17) Similarly sin. a = sin. c sin. A (18) A ,-, cos. c cos. a cos. b , .-. .. , Again, since cos. C - and cos. C = 0, we have sin. a sin. b :. cos. c = cos. a cos. b (19) cos. C + cos. A cos. B Again, since cos. C = -. T -. ^ and cos. C = sm. A sm. B /. cos. c = cotau. A cotan. B (20) Again, since cos. a = cos " A + cos ' B cos> C and cos. C = and sin. C = 1. sin. B sm. C /. cos. A = cos. a sin. B (21) Similarly cos. B cos. b sin. A (22) . . . sin. b Again, since cos. A = cos a sin. B and sin. B ~ sm. c oo A sm> ^ i cos. c , ,> A sin. b COR. c . . cos. A = cos. a and cos. a = by (2) .-. cos. A = sm. c cos. b bin. c cos. b .-. cos. A = tan. 6 cotan c (23) Similarly cos. B = tan. a cotau. c (24) Again from (1) sin. a = sin. c sin. A = sm ' ^ . sin. A from (2) and this = S1D ' & sin. B cos. b cos. B sin. A cos. b sin. B cos. B But by (4) cos. B = cos. b sin. A. /. sin. a = tan. b cotan. B (25) Similarly sin. b = tan. a cotan. A (26) (7.) Napier's Rule for the Solution of Riglit-Anyled Spherical Triangles. The formulas given in the last article can be included in a single rule, which is very easily enunciated and remembered. It is generally called Napier's Rule, having been invented by Napier, who, as we have already stated, was the inventor of logarithms. Leaving out C, which is 90, there are three sides and two angles in the triangle, viz., a. b. c. A. B. we will call the base, the perpendicular, the complements of the hypothenuse and the angles circular parts ; if we fix on any of these and call it the middle part, then of the remaining four two will be adjacent, and the other two opposite : then it will be found that all the formulas of the last article are included in the following rule. " The sine of the middle part equals the product of the tangents of the adjacent parts, and also equals the product of the cosines of the opposite parts ; " Or, Sin. mid. = tan. ad. = cos. op. Thus, if 90 A is the middle part, then 90 c and 6 are the adjacent, and 90 B and a are the opposite part. The rule gives us sin. (90 A) = tan. (90 c) tan. b = cos. (90 B) cos. a Or Cos. A = tan. b, cotan. c = cos. a, sin. B (a) 410 SPHERICAL TRIGONOMETRY. Similarly if 90 B is the middle part, then 90 c and a are the adjacent, and 90 A and b the opposite parts. The rule gives us Cos. B = tan. a, cotan. c = cos. b, sin. A (b) If a is the middle part then 90 B and 6 are the adjacent, and 90 c and 90 A the opposite part, then the rule gives us Sin. a = tan. b cotan. B = sin. c sin. A (c) If b is the middle part then 90 A and a are the adjacent parts, and 90 c and 90 A the opposite parts, then the rule gives us Sin. b = tan. a cotan. A = sin. c sin. B (d) Finally, if 90 c is the middle part, then 90 A and 90 B are the adjacent, and a and b the opposite parts, then the rule gives us Cos. c == cotan. A cotan. B = cos. a cos. b (e) If the five formulas of the present article be compared with the 10 of article (6) they will be found identical. Hence Napier's Rule, as was stated, comprises all the formulas of Art. 6. It is a question whether as a matter of practice Napier's llule is really more convenient than the disconnected formulas of Article (6). (8.) To explain the Method of Solution in the cases of Riyht-Anr/led Spherical Triangles. All the formulas of Art. 6, to which we refer in the present article, are expressed as products, and consequently are adapted for logarithmic calculations. The cases, as we have already seen, are the following : (1) Given a and 6, find c, A and B Cos. c is given by (19), then cos. A is given by (23), and cos. B by (24) (2) Given c and a, find b, A and B Cos. b is given by (19), then cos. A is given by (23), and cos. B by (24) (3) Given a and b, find b, c, and A Cos. A is given by (21), then cos. b is given by (22), and tan. c by (23) (4) Given a and A, find b, c, and B Sin. B is given by (21), then cos. b is given by (22), and cos. c by (19) (5) Given c and A, find a, b, and B Sin. b is given by (17), then cos. a is given by (19), and cos. B by (22) (6) Given A and B, find a, b, and c Cos. c is given by (20), cos. a by (21,) and cos. b by (22) It will be observed that each of the above determinations is clearly unambiguous, except the determination of B in (4), and 6 in (5), for these are the only two determinations made by means of sines ; for which reason if B' and b' are the values less than 90 which satisfy (4) and (5), then ISO B', and 180 b' also satisfy (4) and (o), and hence it would seem that in the former case there would in general be two values of b and two of c, corresponding to B' and 180 B' respectively; and in the latter case that there would be two values of a given by (3), and therefore two values of B given by (6). If more closely considered, however, it will appear that there is really no ambiguity in case (5). We will consider the cases separately. OBLIQUE-ANGLED TRIANGLES. 411 Iii case (4) we have given A and a. Now AB and AC being produced meet at A' where ABA' and AC A' are each of arc 180 (See B Spherical Geometry, pp. 255, 256) and the angle at A' is equal to the angle at A. Hence the angle A and the side a belong equally to triangle ABC and to A'BC. And if we take the value of B less than 90 to be ABC. Then CBA' is 180 B', the second value indicated by the solution. In case (5) from (10) it appears that tan. a = sin. b tan A. Now sin. b is always positive, and hence the sign of tan. a must be the same as that of tan. A. Hence if A > 90 a must be > 90 and if a > 90 A must be < 90, and A is given, hence only one of the two values of a is admissible. This also follows from geometrical considerations. Let C be the right- angle, then CA and CB when produced meet in C', then since we have given AB (c) and BAG (A) we determine b, i.e. cA from the equation or 180 6 i.e. AC', but AC' belongs to a triangle on which the angle BAG' is not A but 180 A, and .'. the value AC' is inadmissible. THE SOLUTION OF OBLIQUE-ANGLED TRIANGLES. (9.) To enumerate the Cases of Oblique-Angled Triangles. There are six cases of oblique-angled triangles, viz., (1.) Given three sides, e.g. a. 6. c (2.) Given two sides and the included angle, e.g. a. b. and C (3.) Given two sides, an angle opposite to one of them, e.g. a. J. A (4.) Given one side and the two adjacent angles, e.g. A. B. c (5.) Given one side, the opposite angle and another angle, e.g. A. C. c (6.) Given the three angles, e.g. A. B. C As in the case of right-angled triangles, these six cases are analogous to the four cases of plane oblique-angled triangles (p. 363.) Exit the fourth case of a plane triangle diverges on to the fourth and fifth of the spherical triangle, owing to the circum- stance that A + B + C is not known in the case of the spherical triangle, whereas in the plane triangle A + B + C = 180. For the same reason case (6) is peculiar to the spherical triangle. (10.) To solve the First Case of Oblique-Angled Triangles. AYe can obtain A from either of the formulas (5) (6) or (7), and then can obtain A B and C from similar formulas. Of these formulas (7) which gives tan. -77 is the most convenient if we wish to find both of the other angles. Compare the analogous case of Plane Triangles, p. 363. 412 SPHERICAL TRIGONOMETRY. (11.) To solve the Second Case of Oblique-Angled Triangles. Iti tins case we will suppose that we have given a b and C. Then from formulas j^ _|. g A _ B (13) and (14) \ve can determine and - and hence A and B; and then, z 2 knowing A and B, we can determine c from formula (1). If, however, we wish to determine c directly, i.e., independently of A and B, we can effect our object by introducing a subsidiary angle in a manner analogous to the corresponding case of plane triangles. (See pp. 325, 365.) Thus, from formula (3) we have cos. c cos. a cos. b Cos. C. = em. a sin. o Cos. c = cos. a cos. 6 + sin. a sin. b cos. C .-. 1 Cos. c =1 cos. a cos. b sin. a sin. 6 cos. C = 1 cos. a cos. 6 + sin. a sin. b sin. a sin. 6 (1 + cos. C) = 1 (cos- a cos. b sin. a sin. b) sin. a sin. b (1 + cos. C) = 1 cos. (a + b) sin. a sin. b (1 + cos. C) A A Xow 1 cos. A = 2 sin.- and 1 + cos. A = 2 cos. 2 c a + b C sin. 2 sin.- - -- sin. a sin. b cos. 2 ft a 2t C Assume sin. 2 6 = sin. a sin. o cos. 2 - c a + b sin. 2 = sin.- -- sin. 2 6 -j -t a + b \ / . a + b n \ sin- - + sm. 6 \ / . a + b \ / . = I sin - -- sin. 6 } I si . I /a + b \ 1 -8*1. -* cos.- 1 /a + 6 cos.- 1 fa + b \ 1 /a + b x 28in . 1 /a - (- This latter method is very much easier than the former : for by this we only j require five logarithms, whereas by that we require eight, for the determination off. (12). To solve the Third Case of Ollique-Angled Triangles. In this case we will suppose that we have given a, b, A. Then we obtain sin. B by formula (1) ; and knowing a, b, and A, B, we can determine C, by formula (13) ; aiid finally we can determine e by formula (1). OBLIQUE-ANGLED TRIANGLES. 413 For let ABC be a triangle, It will be observed that in this case, since the results depend on our determining B from a given value of sin. B, they will be ambiguous, as in the analogous case of plane triangles (p. 366) ; for if B' is the value of B < 90, which we derive from formula (1) : then 180 B' also satisfies formula (1). This amount of ambiguity depends on the data, having the angle BAG = A AC = & and BC = c. Produce AB and AC to meet in A', draw CB' = CB. Then the given data belongs as much to the triangle ACB. as to ACB'. Moreover it is plain that CBB' = CB'B, and hence if CB'A = B', we shall have CBA = 180 B', the same conclusion that we derived from the formula. (13.) To solve the Fourth Cass of Oblique- Angled Spherical Triangles. a 4-5 In this we suppose that we have given A. B. and C. Then - is given from formula (15), and ^ from formula (16); hence we obtain a and 6, and then we obtain C. by formula (1). If we wish to obtain C without the previous calculation of a and b, we must introduce a subsidiary angle 8 and proceed as in article (11). From formula (4) we have Cos. C = cos. A cos. B + sin. A sin. B cos. c. c Assume sin. 2 =z sin. A sin. And we shall obtain r C . /A+B Cos. 2 = sin. I . >) - ( A -f B - ) 2 V 2 (14.) To solve the Fifth Case of Oblique- Angled Triangles. In this case we will suppose that A. C. and c. are given. from formula (1). Now, formula (13) gives us A 4- C COS " I ( a ~ We shall obtain sin. a Tan.: K- * cotan. TT Whence we obtain B, and a similar modification of formula (15) will give us & ; or, having B, we may obtain b from formula (1). In this case a is determined from its sine, and therefore has two values, viz. a' and 180 a', and if both these values are admissible, the case is ambiguous. In the triangle ABC let AB = c BAG = A and EGA = C, and suppose BC = a', drav BC' = a' ; then if a' is greater than c, it is plain that A falls between C and C'; in this case produce C'B and C'C to meet in C". Then the angle BCC' = BC'C = BC"C. Hence the data belong equally to the triangle BAG. and BAG" ; and the case is ambiguous, provided Fig. 7. 414 SPHERICAL TRIGONOMETRY. c < a ; also it will be observed that BC" = 180 BC' = 180 BC = 180 a', as previously appeared from the calculation. If BC is less than BA, then C' would fall between B and A, and the above construction would be no longer possible. Hence if c > a, the case is not ambiguous. (15.) To solve the Sixth Case of Oblique-Angled Triangles. In this case we have given A,B,C. We shall find a from formula (11), and b and c from similar formulas. This case, however, never occurs in any of the practical applications of spherical trigonometry. On the Solution of Quadrantal Triangles. A quadrantal triangle (Spherical Geometry, def. xiv. p. 260) has one side of 90, and .'. the corresponding polar triangle is a right-angled triangle. From this considera- tion it would be easy to modify Napier's Rule to suit the case of the quadrantal triangle. In practice, however, it is better to treat them as oblique triangles on doing so it will be found in practice that the circumstance of one side being equal to 90 will introduce important simplifications. ON THE FORMULAS PECULIAR TO GEODETICAL OPERATIONS. We have already stated in general terms that the science of spherical trigonometry finds one of its applications in Geodesy. It is to be observed that this application possesses some peculiarities in consequence of the sides of the triangles employed in a survey, on even the largest scale, being small compared with the radius of the earth, and consequently sma.ll when estimated in degrees or minutes ; whereas, in astronomy, there is no limitation imposed on the magnitudes of the sides of the triangles employed in that science ; our object in the present article is to explain concisely the results of this limitation, and to deduce certain formulas depending on it. (16.) To state the Object of a Trigonometrical Survey of a Country. The object of the survey is, (1) to fix accurately the relative positions of certain chief points in the country, so as to lay them down on a map ; and (2), having fixed these chief points, then by means of subsidiary opera- tions to lay down in detail all the minor features of the country, its roads, rivers, towns, hamlets, &c. The accompanying figure will be sufficient to illustrate this matter for our present purpose, which is from an actual survey. A, is a place called Ruckinge, B, High Nook, C, Allington, D, Lydd, E, Fairlight Down, and F, Tenterden. The line AB, is measured very accurately, and is called the base line; and then the angles CAB, ABC, are measured; from these data, AC, and CB, can be calculated ; then CB being known, the angles DCB, and CBD, can be MEASUREMENT OF A BASE LINE. 415 measured, and thus CD, and DB, be determined ; and this operation continued for any number of triangles whatever. It is usual in the larger triangles to measure all the three angles of any triangle, and not merely the two at the base ; this is done with a view of keeping a check upon the various errors to which all observations are liable. When the triangulating has been continued for some distance, it is necessary to compare the calculated length of a line that has been fixed upon, and then measure it ; the coincidence of the two results is a verification of all previous measurements and calculations ; hence such a line is called a base of verification. It is usual to choose stations that are from ten to twenty miles apart ; also ifc is usual to choose for a base line, a line of about four or five miles long. In late French surveys only two bases of verification have been used. The accuracy attainable in practice will be appreciated when the fact is stated that, in some English bases of verification, of four or five miles long, the computed and measured lengths have differed only by one or two inches. The operations of a trigonometrical survey are then two, (1) the measure- ment of base lines, (2) the measurements of angles. "VVe will proceed to consider each of these. (17.) The Measurement of a Base Line. A space of open ground which is nearly level must be chosen, the line to be measured being indicated by stations and stages erected, if necessary, to secure the horizontality of the base ; the measure may be made by rods of glass, or steel, proper corrections being applied for temperature ; a more convenient contrivance for securing accuracy in the measures has been devised of late for the Irish Survey, it is of the following kind : AB, CD, are rods of platina and iron riveted together at the middle point P, and are exactly the same length for a given temperature : A.p, ~Bq, are marks affixed to the ends of p. each. Now the metals have different expansions for the same temperature ; suppose then that AB, for a change in temperature, becomes ab, while CD becomes cd, then Ap will assume the position ap, and let Aa = Sa Cc = 86, then if Cp is taken so long that -^ = j| then p will not be changed by the change of temperature; since within very large limits the expansion of a metal is proportional to the increase of temperature, and /. rr is constant; the same arrangement being made at the end, ED, the distanc e pq, will not be affected by change of temperature. ..... Now (Plane Trig. Art. 50, cor.) cos. 0=1 __. if we omit 0" ... Zi .'. AB 2 = a"- + 2 ab + I- abO- = (a + b)"- 1 (a + 6) 2 J AB = (a -i- b) jl "', MO}"= ( + &){l 9 7~^| if we omit 0'... AB = + & r-^ j- 2 (a + 6) *) .. = *000004848 'x 60 x 60 06 00000000001175 a + b The correction to be applied to the sum of a and 6, to obtain the true distance, AB. (19.) To measure the Angles of the Triangles of a Surrey. Any two of the three angular points of one of these triangles is rarely in a horizontal plane passing through the third ; the angle required is, of course, such a horizontal angle. Now the angles are measured either by a theodolite, or by a repeating circle in the case of the former instrument, the vertical elevation of each object is observed, and the horizontal angle between them that is to say, if ABC Fig. (D) are the stations, ANM the horizontal plane through A, BN and CM perpendiculars from B and C on AMX, then by the theodolite we observe the angles BAN, CAM, and MAN ; the last is the horizontal angle required. The theodolite is the instrument that has been used in the English surveys. But if a repeating circle is employed, the angle BAG is the one observed ; and it is necessary to deduce from this the horizontal angle MAN. The repeating circle is the instrument used in the French surveys. CORRECTION FOR REDUCING AN ANGLE. 417 (20). To determine the Correction for reducing an Angle to the Horizon. Let AB C M N be the same as in last article; with centre A and any radius describe a sphere, which meets the lines AB AC, AN, AM in p, q, n, m, respectively, pn and mq, if joined by great circles, clearly meet in Z vertically over A, since the circles must be per- pendicular to the horizontal plane. Then mti, or the angle mZn (Spherical Geometry, prop. v. cor. 4) is the angle required, and for its de- Fi , n , termination we have given pq = A, pn = h, qm = h'. Suppose mZn = A + 5A, then our object is to determine 8A. Now in triangle pZq we have by formula 3, cos. pZq Now pi = 90- .*. cos. (A A, and qZ = 90 h' . JA \ cos - -A- 8m - ^ sin. h cos. pq cos. pZ cos. qZ sin. pZ cos. qZ cos, h cos. h' Now we will suppose SA so small that we can omit 5A 2 . . . and A and h' so small that we can omit every power and product higher than A 2 , hh', and A' 2 . .-. (Cor. Plane Trig. art. 50.) sin. 8A = 8A, cos. 5A = 1, sin. h = h, sin. h' = h', cos. h = 1 , and cos. h' = 1 L 2 2 Now cos. (A + 8A) = cos. A cos. 8A sin. A sin. 5A. Hence cos. A 5A sin. A = _ cos ' cos. A hh' . ,,, A 2 + A* . = ' COS. A ft/I -r _ COS. A 5A sin. A = hh' cos. A Now let h + Jif = p t and h h' = q. ,'. 2A = p + q, and 2A' = p q .-, 4AA' =piqt y and 2 (A* + A' 2 ) =^2 + 5= .-. 4 5A sin. A = p* q* (p* + f) CO s. A = .p 8 (1 cos. A) - Now i c a b cos. cos. - cos. _ Cos. C = a> b sin. - sin, But if we suppose that we retain all terms up to the fourth order, i. e. / a 4 \ / 6 4 \ a 2 b 2 (r ) (r)*&- ^ 1 1 1 1 / a 2 + 6 2 \ -^- ^- I 1 -4- I a 6 /a a 3 \ /b b 3 \ ab f a" + b 2 \ ab V &"* I r ^'r 7-^ r--^ J3 l - -$*- ~* ^ Sin. and COS. COS. - COS. - 1 r r r 2 >2 24H \ 2r~ 24 r' _j i g2+i 2r 2 ' 24r*' L 2r 2 24 r 1 If c 2 24r 4 Taking in every term involving 7- 420 SPHERICAL TRIGONOMETRY. i f , .'. Cos. C = r- I a : 2a& L 4 c 4 + 6u- bs 2ab Now 2a 2 6 2 + 2& 2 c 2 + 2c=a 2 c 4 6 4 a 4 = 26 2 c 2 + 2c2a 2 c 4 (& a 2 ) 2 = 4c6* 2c* (& 2 a-) c 4 (6 2 a 2 ) 2 = 4c 2 6= (c 2 + b" a 2 ) 2 = (2c6 + c 2 + 6 2 a 2 ) (2 c6 c 2 6 2 + a 2 ; = ! (6 + c) 2 a 2 i /a 2 (6 c) 2 1 = (6 + c + a) (b + c a) (a + 6 -c) (a b + c) = 16 A 2 , if A is the area of a plane triangle whose sides are a, b, c. (Plane Trigoii. Art. 41.) a- + I- c- 16 A 2 /. COS. C == - jr 7 --- TTT' 7-3- 2 ao 24 aor- Now if 5C is tlie correction to be applied to the triangle, in order that -we raay be enabled to reckon the triangle plane, so that C SC is the angle of a plane triangle whose sides are a, b, c, then Cos. (C 5C) = -' + 2 ^~ (Plane Trigon., Art. 37.) And Cos. (C 5C) = Cos. C + 8C. sin. C /. Cos. C 1 = Cos. C 4- SC, sin. C |- A " /. 8 C = 3 ab -r A- 1 A 2 ab sin. C. i~ 3 r 2 Since the area of the triangle is ab sin. C = A. a Now if E is the spherical excess, i.e. the excess of the sum of the three observed angles over two right angles, we have (Spherical Geometry, Prop, ix.) A : Area of hemisphere : : E : 2ir E being in circular measure. .'. 5C = E. 3 The same clearly holds good of either of the other angles ; hence the rule deter- mines the spherical excess into three equal parts, subtract one part from each of the angles, and the triangle can then be considered plane. If 5C is equal to n" then, as before, 2r- x -000004848 It is to be observed that n is very small, e.g. rarely more than 5 or 6, hence a small error in the area will produce no appreciable error. Hence A can be found on the supposition that the original triangle is plane. JOHN F. TWISDEN. University of California SOUTHERN REGIONAL LIBRARY FACILITY 305 De Neve Drive - Parking Lot 17 Box 951388 LOS ANGELES, CALIFORNIA 90095-1388 Return this material to the library from which it was borrowed. NOV 8 2002 UCLA URL/ILI - ,