The angle A = 180° — (B + C) = 91° 22′ 2′′. Also, Hence, a:c:: sin A: sin C, or c = a sin C sin A log clog a + log sin C log sin A. Similarly, log blog a + log sin B — log sin A. The work will hence stand thus, remembering that sin A = sin (180 A) : 2. Given b other parts. or c = 197.6654 = 152.67, c = 163.18, and C = 50° 18' 32", to find the As the given angle is opposite to the greater side, the problem is not Hence, log sin Blog b + log sin Clog e; log a = log blog sin A log sin B: log 152 67= = 2.1837537 log sin 50° 18' = 9.8861519 12.0699615 (B+ C) = 83° 38′ 31′′. 210.756 3. Given a = 272·13, b = 252·69, and B = 64° 18' 20", to find A, C, and c. Since the given angle is opposite to the less side, the solution is ambiguous. Now, Hence, log sin A = log a + log sin Blog b, log clog a + log sin C - log sin A: B 4. Given a = 305 296, B=51° 15′ 35′′, and C = 37° 21′ 25′′, to find b, c, and A. Ans. b=238 1974, c = 185 3011, A = 91° 23′. other parts. 6. Given a= A, C, and c. = 5. Given c=195 265, b=203 162, and B = 45° 0′ 55′′, to find the Ans. A 92° 9′ 23′′, C= 42° 49′ 42′′, a = 287.035. =350·169, b = 236.291, and B = 38' 39' 15", to find Ans. A = 67° 45′ 58′′ or 112° 14′ 2′′, C = 73° 34′ 47′′ or 29° 6' 43", c = 362 8674 or 184 048. 7. Given BC=145·3, AC=178·3, and A = 41° 10′ to find the other parts. Ans. B 53° 52′ 36" or 126° 7′ 24′′, C= 84° 57′ 24′′ or 12° 42′ 36′′, c = 219·882 or 48 5656. CASE II. 26. When two sides and the included angle are given. 1. The sides B C, CA, of a plane triangle A B C, are 17.802 and 21.704, and the included angle C is 26° 12′ 16"; what are the values of A B, A and B ? = Hence B (B+ A) + (BA) 99 53' 33", A (B+ A) - (B-A)=53° 54′ 11′′. = As a verification of this result, and an illustration of (1) and (2), Art. 24, the value of c will now be found without finding the angles A and B. By the properties just referred to, log 4+ loga + log b + 2 log sin C a), logc = log (ba) + log sec 10. log 3.902 0.5912873 1.2504688 log sec 66° 21' 14" 16 = 10·3967629 log 4 = log 17 802 log 21 704 = 0.6020600 = 1.3365398 2 log sin 132 6'8" = 18.7108610 log c = 0.9880502 Hence c = 9.7286, the same as 21 8999296 before nearly. 2. Given a = 16.9584, b = 11.9613, and C = 60° 43′ 36", to find = 108° 12' cos C = Hence A = 22° 19' 53" 91, B = 49° 27' 30" 23, C 35" 86. Since cos C is negative, ·3185 = cos (π — C), by (14) of Art. 15; hence having found the angle whose cosine is 180° for C. 3125, it is taken from may be inferred that 2. Given AB = 17862, B C = 19876, A C = 18304, to find the As the angles A, B, C, together, make 180°, it the work is correct. angles. By (4) of Art. 23 (the symbol A.C is put for arithmetical complement), we have = = 5, to find the angles. = 6, А С = 44° 24' 55" 12, C 7, BC = = 78 27' 46" 93. 43628, c = 62984, to find A, B, and C. 11", B 40° 40' 8", = = C 109° 48' 41". 610 91, CA = 74° 32′ 7′′ 96, C = = 714.23, to find 49° 56′ 21" 76. 4. Given AB Ans. A 5. Given a = 32986, b = Ans. A 29° 31' 6. Given AB = 567.18, BC the angles. Ans. A = 55° 31' 30" 28, B = 7. Given a = 24804, b 57876, c = 74412, to find A, B, and C. Ans. A 16° 11' 42" 44, B = 40° 36' 3" 84, C = 123° 12' 13" 72. 8. Given the sides A B = 13 219, BC = 11.76, A C = 14.507, of a triangle A B C, to find the angles. Ans. A 49° 55' 43" 5, B 70° 44' 2" 32, 9. The three sides of a triangle are 352.96, 628 54, and 569·16; find the angles. Ans. 33° 49′ 6′′· 12, 82° 21′ 12′′ · 60, and 63° 49′ 41" 28. MISCELLANEOUS EXERCISES ON TRIANGLES. 1. Express each side of a plane triangle in terms of the remaining sides and their opposite angles. Ans. a = b cos C+c cos B, b = a cos C + c cos A, c = b cos A+acos B. 2. Prove the following properties of a right-angled triangle, C being the right angle: 3. In an isosceles triangle in which a = b, prove that 4. Prove that if 2 cos B sin C = sin A, the triangle is isosceles; and if tan A sin 'B = tan B sin A, it is either isosceles or right-angled at C. 5. Prove that in any plane triangle A B C, sin A: sin B :: a (s − b) : b (s − a). 6. Show that in any plane triangle, 7. If A, B, C, be the angles of a plane triangle, then cot A cot B+ cot B cot C + cot C cot A = 1, sin 2 A+ sin 2 B + sin 2 C = 4 sin A sin B sin C. 8. Prove that the lengths of the three straight lines drawn from the angles A, B, C, of a triangle ABC, to the points of bisection of the opposite sides, are (b c cos A + ‡ a2)1, (a c cos B + b2), and (a b cos C + |
