PROBLEM IV. Given the three angles of a plane triangle, and the radius of the inscribed triangle, to find the sides of the triangle. Let A, B, C, be the three given angles, r the radius, Given the three angles of a plane triangle, and the radius of the circumscribing circle, to find the sides of the triangle. As in Problem III. So, CQ. CD=AC. CB AC 2 R sin. B AB=2 R sin. C P C CHAPTER X. PROBLEMS IN TRIGONOMETRICAL SURVEYING. THE DETERMINATION OF TOPOGRAPHICAL DATA BY GEOMETRICAL CONSTRUCTION AND TRIGONOMETRICAL ANALYSIS. PROBLEM 1. To determine the height of an inaccessible object. Let AB be the object, and in a straight line A towards it measure any distance DC, and observe the angles of elevation ADB, ACB at the stations D, C. Put CD=h, ACB=a, ADB b: then DAC-a-b; hence we have AC sin. b CD sin. (a-b) and, multiplying these two equations, we have AB sin. a sin. b CD sin. (a a or, AB=k sin. a sin. b cosec, (a — b) ..log.AB=log.h+log.sin.a+log.sin.blog.cosec.(a−b)-30 (1) Cor. Since DB=AB cot. b, and CB=AB cot. a; therefore, by subtraction, CD=AB(cot.b-cot.a,)or AB= Ex. Let DC=h=200, BDA=b=31°, BCA=a=46°; to find AB and CB. = log. h log. sin. a log. sin. b = log. AB log. 200 = 2.3010300 = log. sin. 46° 9.8569341 log.cosec.(a—b)=log. cosec. 15°=10.5870038 =log. 286.29 = 2.4568072 ... AB=286.29 Also, BC AB cot. a, .. log. BC=log. AB+log. cot. a-10. PROBLEM II. To determine the height af an inaccessible object, which has no level ground before it. D Let AB be the object, and C, D, two stations in a vertical plane passing through AB; measure the distance CD, and at C take the angles of elevation or depression of the station D, and the top and bottom of the object. Also at D take the eleva- H of the top of AB. h K Put CD=h, DCH-a, BCK=b, ACK=c, ADG=d; then ACB=c-b, ADC=a+d, and CAD=c-d. Now, equations, AB sin. ACB sin. (c-b)_sin. (c—b) | AC sin. ADC_sin. (a+d) cos. b CD sin. CAD sin. (c-d); hence, multiply these we have AB sin. (c-b)sin. (a+d) b) sin. (a+d), and, therefore, CD cos. b sin. (cd) AB=h sin. (c-b) sin. (a+d) sec. b cosec. (c-d) ..log. AB=log. h+log. sin. (c—b)+log. sin.(a+d)+log. sec. b+log.cosec. (cd)-40 (1) Cor. When a=90°, and b=0°, then we have log.AB=log.h+log.sin.c+log.cos.d+log.cosec.(c-d)—30 (2) Ex 1. Let h=18 feet; c=40°; d=37°30', a=90° and b=0°; to find AB. log. h. log. sin. c. log. cos. d. =log. 18. 1.2552725 =log. sin. 40° = 9.8080675 log.cosec.(c-d)=log. cosec.2° 30'=11.3603204 Ex. 2. The angle of elevation of the top of a tower, standing on a hill, was 33° 45', and measuring on level ground 300 yards directly towards the tower, the angles of elevation of the top and bottom of the tower were 51° and 40° respectively. What is the height of the tower? Ans. 140 yds. Remark. When the station D is higher than A, the top of the tower, then the angle d must be considered negative, and therefore we should have AB= h sin. (c-b)sin. (a–d) sec. b cosec. (c+d) PROPOSITION I. LEMMA. If straight lines be drawn from any point, either within, or out of, a polygon, to all the angular points, the continued products of the sines of the alternate angles, made by the sides of the polygon, and the lines so drawn, will be equal. Let angle CDP=ƒ, and PEA=i; PA then But PA.PB.PC. PD.PE=PB.PC.PD. PE. PA; that is, the product of the numerators=product of the denominators, in the first members of these equations; hence, this being true in the second members also, sin. b sin. d sin. ƒ sin. h sin. k=sin. a sin. c sin. e sin. g sin. i. PROBLEM III. Given AB, and the angles a, b, c, d, to find C x, and thence CD. Put BCD+ADC=b+c=2s BCD-ADC=... 2x Then BCD = s+x, ADC=s-x; also, sin. ADB = sin. (b+c+d), and sin. ACB = sin. (a+b+c); hence, by Lemma, (I) we have sin. a sin c sin. (b+c+d) sin. (s+x)=sin. b sin. d sin. (a+b+c) sin. (s-x), or sin. a sin. c sin. (b+c+d) {sin. s cos. x+cos. s sin. x = sin. b sin. d sin. (a+b+c) {sin. s cos. x-cos. s sin. x}. sin. a sin. c sin. (b+c+d) (tan. s+tan. x)=sin. b sin. d sin. (a+b+c) (tan. s-tan. x), .. tan. x= tan.s. sin.b sin. d sin.(a+b+c)-sin.a sin.c sin.(b+c+d) sin.b sin.d sin.(a+b+c)+sin.a sin.c sin.(b+c+d) Dividing numerator and denominator by sin. a sin. c sin. sin. b sin. d sin. (a+b+c) (b+c+d), and putting sin. a sin. c sin. (b+c+d) = tan. ẞ and 1 tan. 45°; then hence x, s+x, s-x are CD For BD sin. d all known, and thence CD is known. sin. b and BD = BD ̄ sin. (s+x) AB sin. (b+c+d); therefore, CD=AB sin. b sin. d cosec. (s+x) cosec. (b+c+d). Cor. When CD is given, and the same angles, to find AB, we have AB=CD sin. (b+c+d) sin. (s+x) cosec. b cosec. d. EXAMPLE. Given AB=600 yards, a=37°, b=58° 20′, c=53° 30′, d= 45° 15', to find CD. Here, tan. 6=cosec. a sin. b cosec. c sin. d sin. (a+b+c) |