the first three parts exceeds 90, the fourth is 90 (art. 336.). Or since the signs of the first three terms are +, for radius is always positive, that of the fourth must be so, and +tan B shows that B is 90. To find AB. The complement of AB being the middle part (335), AC and C are opposite parts; and R cos M = sin O⚫ sin o, or Rcos comp. AB sin AC sin C. = Or, as cos comp. AB sin AB, and AB is required, R sin AB sin C sin AC. = The sine for 9.810594 may be either that of 40° 17', or its supplement 139° 43′; but in the given triangle, the angle C opposite to the side AB is acute; hence AB is 90 (331.) To find angle A. When AC is the middle part, angles A and C are the adjacent parts, and R⋅ cos M=cot Ocot o. Or R cos AC cot A cot C; hence EXERCISES. 1. The hypotenuse is 75° 20′, and one of the oblique angles 57° 16'; what are the other parts ? Ans. The two sides 64° 10′ and 54° 28′, and the other angle 68° 30'. 2. The hypotenuse is 64° 40', and an angle 64° 38′ 11′′; find the other parts. Ans. The other angle 48° 0' 14", its opposite side 42° 12', and the other side 54° 41′ 28′′. 338. Case II.-Given the hypotenuse and a side. EXAMPLE.-Let the hypotenuse AC and the side BC of the triangle ABC be given equal to 70° 24′ and 65° 10′ respectively, to find the other parts. To find angle C. C being M, AC and comp. BC are A and a, and R⚫ cos Mcot A cot a. Or R⋅ cos C = cot AC · tan BC, and When AC is M, the complements of BC and AB are O and o; and R⋅ cos M = sin O⚫ sin o. Or R⚫cos AC = cos BC⋅ cos AB, and Cos BC: R= cos AC: cos AB. When Cos AB 37° To find angle A. comp. BC is M, AC and angle A are O and o; and R cos Msin Osin o. Or, R⚫sin BC= sin A C⚫ sin A, and Sin AC: R sin BC: sin A. 1. The hypotenuse is 75° 20′, and a side is 64° 10′; required the other parts. Ans. The other side 54° 28', its opposite angle 57° 16', and the other angle 68° 30'. 2. The hypotenuse AC is 50°, and the side BC = 44° 18′ 39′′; what are the other parts? Ans. AB= 26° 3′ 53′′, angle A = 65° 46′ 5′′, and angle C = 35°, 339. Case III.-Given the two sides. EXAMPLE. The side AB is 37°, and BC is 65° 10′; find the other parts. To find AC. When AC is M, the complements of AB and BC are O and o; and R⋅ cos M = sin O⚫ sin o. Or, R cos AC cos AB cos BC; hence The comp. of AB being M, A and the comp. of BC are A and a; and R⋅ cos Mcot A⚫ cot a. Or, R sin AB cot Atan BC; hence = Tan BC: R= sin AB: cot A. To find angle C. When comp. BC is M, C and comp. AB are A and a; and R cos Mcot A cot a. Or, R · sin BC = cot C · tan AB; hence Tan AB: R= sin BC: cot C. 1. The two sides are 54° 28' and 64° 10′; find the other parts. Ans. The angles are 57° 16′ and 68° 30′, and the hypotenuse 75° 20'. 2. The two sides are 42° 12′ and 54° 41′ 28′′; what are the other parts? Ans. The angles are 48° 0′ 14′′, and 64° 38′ 11′′, and the hypotenuse 64° 40'. 340. Case IV.-Given the two oblique angles. EXAMPLE. The angle C is 106° 24′, and angle A 32o 30'; required the other parts. To find AC. AC being M, angles A and C are A and a. Hence, R cos AC cot A cot C, and +R = R: cot A cot C: cos AC. +Cot A 32° 30' -Cot C 106° 24' (73° 36′) 10. 10.195813 9.468814 9.664627 In the tables, the cosine here belongs to an arc of 62° 29'; but by 329 or 336, AC is 90°, and therefore = 180° 62° 29′117°31′.. Or, since the sign of cot C, one of the terms, is negative, that of the fourth term cos AB must also be negative (336); and hence AB 90. To find AB. Angle C being M, A and comp. AB are O and o; and Rcos Csin A⋅ cos AB; hence Angle A being M, C and comp. BC are O and R cos Asin C cos BC; hence 0; and 1. The two angles are 39° 42′ and 74° 26'; find the other parts. Ans. The sides are 37° and 65° 10′, and the hypotenuse 70° 24'. 2. The angles A and C are respectively 138° 15′ 45′′ and 105° 52′ 39′′; what are the other parts? Ans. The sides AB and BC are 114° 15′ 54′′ and 140° 52′ 40′′, and AC 71° 24' 30". 341. Case V.-Given a side about the right angle and its adjacent angle. EXAMPLE. The side BC is 140° 53′, and angle C is 105° 53′; find the other parts. To find AC. Angle C being M, AC and comp. BC are A and a; and hence Rcos C cot AC tan BC, therefore Tan BC: R= cos C: cot AC. |