Given a side and two angles of a triangle, to find the remaining parts. First, subtract the sum of the two angles from two right angles, the remainder will be the third angle. The remaining sides can then be found by Theorem III. To the angle A Add the angle B I. In the triangle ABC, there are given the angle A=58° 07′, the angle B=22° 37′, and the side c=408 yards: required the remaining angle and the two other sides. Their sum taken from 180° leaves the angle C As sine C Is to sine A So is side c So side a CASE I. As sine C To side b 99° 16' 58° 07' 408 351.024 This angle being greater than 90° its sine is found by taking that of its supplement 80° 44'. To find the side a. ar.-comp. log. CASE II. log. 0.005705 9.928972 2.610660 2.545337 0.005705 9.584968 2.610660 2.201333 2. In a triangle ABC, there are given the angle A-38° 25 B=57° 42', and the side c=400: required the remaining parts. Ans. Angle C-83° 53′, side a=249.974, side b-340.04. Given two sides of a triangle, and an angle opposite one of them to find the third side and the two remaining angles. The ambiguity in this, and similar examples, arises in consequence of the first proportion being true for both the triangles ACB, ACB'. As long as the two triangles exist, the ambiguity will continue. But if the side CB, opposite the given angle, be greater than AC, the arc BB' will cut the line ABB' on the same side of the point A, but in one point, and then there will be but one triangle answering the conditions. If the side CB be equal to the perpendicular Cd, the are BB' will be tangent to ABB', and in this case also, there will be but one triangle. When CB is less than the perpendicular Cd, the arc BB' will not intersect the base ABB', and in that case there will be no triangle, or the conditions are impossible. 2. Given two sides of a triangle 50 and 40 respectively, and the angle opposite the latter equal to 32° : required the remaining parts of the triangle. Ans. If the angle opposite the side 50 be acute, it is equal to 41° 28′ 59′′, the third angle is then equal to 106° 31′ 01′′, and the third side to 72.368. If the angle opp site the side 50 be obtuse, it is equal to 138° 31′ 01′′, the third angle to 9° 28′ 59', and the remaining side to 12.436. CASE III. Given two sides of a triangle, with their included angle, to find the third side and the two remaining angles. Let ABC be a triangle, B the given angle, and c and a the given sides. Knowing the angle B, we shall likewise know the sum of the other two angles C+A=180°—B, and their half sum (C+A)=90—B. We shall next A compute the half difference of these two angles by the proportion (Theorem V.), b C c+a: c—a :: tang } (C+A) or cot B : tang (C—A,) in which we consider c>a and consequently C>A. Having found the half difference, by adding it to the half sum (C+A), we shall have the greater angle C; and by subtracting it from the half-sum, we shall have the smaller angle A. For, C and A being any two quantities, we have always, As sine A So is side a C=} (C+A) +} (C—A) (CA). Knowing the angles C and A to find the third side b, we have the proportion. sın A: sin B::a:b Ex. 1. In the triangle ABC, let a=450, c=540, and the included angle B 80°: required the remaining parts. c+a=990, c-a-90, 180°-B=100°C+A. log. B As c+a 990 90 Is to c-a (C+A) 50° Hence, 50°+6° 11' 56° 11'=C; and 50°-6° 11′ 43° 49′ =A. 43° 49' 80° 450 ar.-comp. To find the third side b. ar.-comp. log. 7.004365 1.954243 10.076187 9.034795 0.159672 9.993351 2.653213 To side b 640.082 2.806236 Ex. 2. Given two sides of a plane triangle, 1686 and 960, and their included angle 128° 04': required the other parts. Ans. Angles, 33° 34′ 39′′, 18° 21′ 21′′ side 2400 CASE IV Given the three sides of a triangle, to find the angles. We have from Theorem IV. the formula, (p-b) (p-c sin ↓ A=R√(↓ in which bc p represents the half sum of the three sides. Hence (p—b) (p—c)), or bc 2 log. sin A=2 log. R+log. (p—b)+log. (p—c)—log. log. b. Here p 2 Log. R log. (p-b) 9.5 log. (p-c) 15.5 -log.c -log.b 34 40 2 log. sin Ex. 1. In a triangle ABC, let b=40, c=34, and a=25: required the angles. 40+34+25 2 = =49.5, p―b=9.5, and ar.-comp. A log. sinA 19° 12′ 39′′ Angle A 38° 25′ 18′′. C p-c=15.5. 20.000000 0.977724 1.190332 8.468521 8.397940 19.034517 9.517258 In a similar manner we find the angle B=83° 53′ 18′′ and the angle C 57° 41′ 24′′. APPLICATIONS. Ex. 2. What are the angles of a plane triangle whose sides are, a=60, b=50, and c=40? Ans. 41° 24′ 34′′, 55° 46′ 16′′ and 82° 49′ 10′′. Suppose the height of a building AB were required, the foot of it being accessible. As R Is to tang. C 41° 04′ 67.84 To EB 59.111 On the ground which we suppose to be horizontal or very nearly so, measure a base AD, neither very great nor very small in comparison with the altitude AB; then at D place the foot of the circle, or whatever be the instrument, with which we are to measure the angle BCE formed by the horizontal line CE parallel to AD, and by the visual ray direct it to the summit of the building. Suppose we find AD or CE 67.84 yards, and the angle BCE=41° 04′ in order to find BE, we shall have to solve the right angled_triangle BCE, in which the angle C and the adjacent side CE are known. A D = To find the side EB. ar.-comp. E Hence, EB-59.111 yards. To EB add the height of the instrument, which we will suppose to be 1.12 yards, we shall then have the required height AB=60.231 yards. If, in the same triangle BCE it were required to find the hypothenuse, form the proportion As cos C 41° 04' ar.-comp. Is to R So is CE 67.84 To CB 89.98 0.000000 9.940183 1.831486 1.771669 log. 0.122660 10.000000 1.831486 1.954146 Note. If only the summit B of the building or place whose height is required were visible, we should determine the distance CE by the method shown in the following example; this distance and the given angle BCE are sufficient for solv. ing the right angled triangle BCE, whose side, increased by the height of the instrument, will be the height required. |