By means of this equation we can compute 4. We find 6 by the equation, b2 = c2 − a2 = (c + a) (c − a). Taking the logarithms of both sides, 2 log b = log (c + a) + log (c − a). EXAMPLES. 1. Given a = 13.2, and c = 127; find b, A, and B. 2. Given a = 3. Given a = c+ a = 140.2; c — a = = 113.8 log 140.22.1467480 512, and c = 1007; find the angles A and B. Ans. A 30° 33′ 36′′. B = 59° 26′ 24′′. 32.712, and c = 96.2; find the angles A and B. 4. Given a 123, and c = 157; find A, B, and b. Ans. A = 51° 34′ 35′′- b = 97.5704. 5. Given a = 576, and c = 880; find the angles A and B. Ans. A 40° 53′ 7′′. 6. Given a = 21.7, and c = 54.31; find A, B, and b. CASE III. 53". Ans. A 23° 33′ 2′′. B = 66° 26′ 58′′. b = 49.78640. Given a and A, it is required to find B, b, and c. By equation (1), 1. Given a = 13, and A = 35° 2′; find B, b, and c. Ans. B 54° 58′. b = 18.543. c = 22.646. 2. Given a = 1157, and B = 58° 3′ 27′′; find A, b, and c. Ans. A 31° 56′ 33′′. = b = 1855.7290. c = 2186.8648. 3. Given a = 825, and B = 36°; find A, b, and c. Ans. A = 54°. b = 599.397. 4. Given a = 1426, and A = 3° 21'; find B, b, and c. = Ans. B 86° 39. =24403.08988. C= 5. Given a = 28.75, and A = 17° 30′ 30′′; find B, b, and c. Ans. B 72° 29′ 30′′. b = 91.18333. C = 95.56433. 4. Ch 6-16, and B = 22° 3′ 56′′; find A. e, and h. Ans. A 6 55 2°. a = 1631.08835b = 661.17424 4 (#6 - 3%,141, and 4 = 18′ 5′ 12′′; find B, a, and b. Ans. B = 71° 54′ 48′′. 4. Given 6-897.3, and A = 31° 21′ 6′′; find B, a, and b. Ans. B = 58° 38′ 54′′. α= 466.8557. b = 766.2852. 1013, and B 10"; find A, a, and b. Ans. A 80°. a = 997.61. 8. Cliven - ge, and 46° 13' 40"; find B, a, and b. Ans. B 83° 46' 20". = CHAPTER IV. TRIGONOMETRICAL FORMULE. LET it be proposed to find the value of the sine and cosine of the sum of two angles, in terms of the sines and cosines of the angles themselves. Let the angles be AOB and BOC (fig. 10); let their numerical values be A and B respectively; on OC assume a point R; let fall the perpendiculars RP and RS; from S draw SQ perpendicular to, and ST parallel to OA; then, as the angles RSO and TSQ are each right, they are equal; take away TSO, which is common, and RST and QSO remain equal; the angles RTS and SQO are also equal, each being right; therefore the triangles RTS and SQO are similar, and the angle SRT equal to AOS = A. From the right-angled triangles RTS and SOQ we obtain, by Prop. I. Chap. III., ᎡᏢ = RS cos A + SO sin A. If OR = R we have (Prop. 1. Chap. III.) RP = R sin (A + B), RS = R sin B, SO= R cos B; substituting these values, we obtain R sin (A+B) = R sin B cos A+ R cos B sin A. Expunging R, which is common to both sides, and changing the order of the terms, sin (A+B) = sin A cos B+ cos A sin B. D 7 TAS FOR 204 BOC (82. 11) be equal to A and Bumperrey, on OC ajaumé a poff: R; let fall 116 porzando nan RP and RS; from S draw SQ proppende wat 90, and ST parallel to OA; then, as The anger 1850 and TSQ are each right, they are go take away RSQ, which is common, and Till remains equal to SOQ; the angles RTS and 800 we also equal, each being right; therefore the Fringles RCTS and SOQ are similar, and TRS = OBO A From the right-angled triangles RTS and SOQ, we obtain NQ-NO sin 4, RT RS cos A; and me RP SQ - RT, RP-SO sin A- RS cos A. |