6. Given a = 1000, and A = 18°; find B, b, and c. Ans. B = 72°. 3077.68368. - 3236.06791. b = C= CASE IV. с Given c and A, it is required to find a, b, and B. By equations (1) and (2), a = c sin A, b = c cos A; from which, log a = log c + log sin A - 10, . EXAMPLES. a = a = 1. Given c = 100, and A = 39° 48'; find B, a, and b. Ans. B = 50° 12'. 64.01097 b = 76.82835. 2. Given c= 1760, and B = 22° 3' 56"; find A, a, and b. Ans. A = 67° 56' 4". 1631.088. b = 661.174. 3. Given c = 28.347, and A = 18° 5' 12"; find B, a, and b. Ans. B = 71° 54' 48". 8.80047 6 26.94632. 4. Given c = 897.3, and 4 = 31° 21' 6"; find B, a, and b. Ans. B = 58° 38' 54". 466.8557 b = 766.2852. 5. Given c = 1013, and B = 10°; find 4, a, and b. Ans. A = 80°. a= a = 997.61. b = 175.90. 6. Given c = 50, and A = 6° 13' 40"; find B, a, and b. Ans. B = 83° 46' 20". a= 5.424. b = 49.705. CHAPTER IV. TRIGONOMETRICAL FORMULÆ. LET it be proposed to find the values 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, 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 QOS = A. From the right-angled triangles RTS and SOQ we obtain, by Prop. 1. Chap. III., RT = RS cos A, SQ - SO sin A; , and as RP = RT + SQ, RP = RS cos A + SO sin A. If OR = R we have (Prop. I. 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. From the same triangles we obtain QO = SO cos A, and ST = RS sin A; and as OP = QO – ST, OP = SO cos A - RS sin A. Substitute for OP, SO, and RS, their equals R cos (A + B), R cos B, and R sin B (Prop. I. Chap. III.) we obtain R cos (A + B) = R cos B cos A R sin B sin A; therefore cos (A + B) = cos A cos B - sin A sin B. We proceed to find the values of the sine and cosine of the difference of two angles in a similar manner. Let AOB and BOC (fig. 11) be equal to A and B respectively; on OC assume a point R; let fall the perpendiculars RP and RS; from S draw SQ perpendicular, and ST parallel to OA; then, as the angles RSO and TSQ are each right, they are equal ; take away RSQ, which is common, and TSR remains equal to OSQ; the angles RTS and SQO are also equal, each being right; therefore the triangles RTS and SQO are similar, and TRS = SOQ = A. From the right-angled triangles RTS and SOQ, obtain SQ = SO sin A, RT = RS cos A; we and as RP-SQ - RT, = = and as If OR = R we have (Prop. I. Chap. III.) RP=R sin (A-B), SO=R cos B, and RS-R sin B; substituting these values, we obtain R sin (A - B) = R cos B sin A - R sin B cos A; therefore sin (A - B) = sin A cos B -cos A sin B. = From the same right-angled triangles we obtain QO = SO cos A, ST = RS sin A; OP = QO + ST, OP = SO cos A + RS sin A. Substituting for OP, SO, and RS, their respective values R cos (A - B), R cos B, and R sin B (Prop. I. Chap. III.), we obtain R cos (A - B) = R cos B cos A + R sin B sin A; therefore cos (A - B) = cos A cos B + sin A sin B. The values for the sine and cosine of the sum and difference of two angles are of the highest importance, as from them are immediately derived a great variety of useful formulæ : sin (A + B) = sin A cos B + cos A sin B, (1) cos (A + B) = cos A cos B – sin A sin B, (2) sin (A - B) = sin A cos B - cos A sin B, (3) cos (A - B) = cos A cos B + sin A sin B. (4) By adding equations (1) and (3), we obtain sin (A + B) + sin (A - B) = 2 sin A cos B. By subtracting equation (3) from (1), we obtain sin (A + B) - sin (A - B) = 2 cos A sin B. By adding equations (2) and (4), we obtain cos (A + B) + cos (A - B) = 2 cos A cos B. = By subtracting equation (4) from (2), we obtain cos (A + B) – cos (A - B) 2 sin A sin B. These results together form the following group: sin (A + B) + sin (A - B) = 2 sin A cos B, (5) sin (A + B) – sin (A - B) = 2 cos A sin B, (6) cos (A + B) + cos (A - B) = 2 cos A cos B, (7) cos (A + B) - cos (A - B) =- 2 sin Asin B. (8) = If we assume A + B = A', A - B = B, it follows by addition that A + B Substituting these values in equation (5), we obtain : sin A + sin B' = 2 sin 1:(A + B) cos } (A - B). Similar transformations may be had from substituting these values of A and B in equations (6), (7), and (8). Removing the accents from the letters A and B in order to preserve a uniform notation, we obtain the following group of formulæ : |