CASE II. Given one of the sides about the right angle and one of the acute angles, to find the remaining parts. 39. The other acute angle may be found by subtracting the given one from 90°. The hypothenuse may be found by formula (7), and the unknown side about the right angle by formula (8). Examples. = 1. Given c 56.293, and C 54° 27' 39", to find B, a, and b. Applying logarithms to formula (7), we have Applying logarithms to formula (8), we have 35° 32′ 21′′, a = 69.18, and b = 40.2114. 2. Given c = 358, and B 28° 47', to find C, a, and b. Ans. C 61° 13', a = 408.466, and b = 196.676. 33 3. Given b = 152.67 yds., and C 50° 18' 32", to find the other parts. Ans. B 39° 41′ 28′′, c = 183.95, and a = 239.05. 4. Given c 379.628, and C 39° 26' 16", to find B, a, and b. Ans. B 50° 33' 44", a = 597.613, and b = 461.55. CASE III. Given the two sides about the right angle, to find the remaining parts. 40. The angle at the base may be found by formula (12), and the solution may be completed as in Case II. Examples. 1. Given b = 26, and c = 15, to find C, B, and a. Operation. Applying logarithms to formula (12), we have 9.761118 .. C 29° 58′ 54′′. [From Art. 28, it is evident that log tan C here found corresponds to two angles, viz., 29° 58′ 54′′, and 180° 29° 58' 54", or 150° 1' 6". right-angled, the angle C is acute, and the smaller value As, however, the triangle is must be taken.] B = 90° C = 60° 01' 06". Ans. C 29° 58' 54", B = 60° 01' 06", and a 30.017. 2. Given b = 1052 yds., and c = 347.21 yds., to find B, C, and a. B = 71° 44′ 05′′, C = 18° 15′ 55′′, and a = 1107.82 yds. 3. Given b = 122.416, and c = 118.297, to find B, C, and a. 4. Given b = 103, and c = 101, to find B, C, and a. B = 45° 33′ 42′′, C 44° 26' 18", and a = 144.256. CASE IV. Given the hypothenuse and either side about the right angle, to find the remaining parts. 41. The angle at the base may be found by one of formulas (10) and 11), and the remaining side may then be found by one of formulas (7) and (8). Examples. 1. Given a = a = 2391.76, and b 385.7, to find C, B, and c. Operation. = Applying logarithms to formula (11), we have c = 2360.45. Ans. B 9° 16′ 49′′, C 80° 43' 11", and c 2360.45. = 2. Given a 127.174 yds., and c 125.7 yds., to find C, B, and b. Operation. From formula (10), we have .. b 19.8. Ans. B 8° 43' 54", C 81° 16' 6", and b = 19.3 yds. = 4. Given a = 60, to find B, C, and c. 53° 7' 49", and c = 80. = 19.209, and c 15, to find B, C, and b. B 38° 39' 30", C = 51° 20′ 30′′, b = 12. = SOLUTION OF OBLIQUE-ANGLED TRIANGLES. 42. In the solution of oblique-angled triangles, four cases may arise. We shall discuss these cases in order. CASE I. Given one side and two angles, to determine the remaining parts. 43. Let ABC represent any oblique-angled triangle. From the vertex C, draw CD perpendicular to the base, forming two rightangled triangles ACD and BCD. Assume the notation of the figure. From formula (1), we have CD b sin A, Α B Since a and b are any two sides, and A and B the angles lying opposite to them, we have the following principle: |