By def. 27. sect. 4. The sine of 96°=the sine of 84°, which is the supplement thereof ; therrtore instead of the sine of 96°, look in the tables for the sine of 84. Extend from 84° (which is the supplement of 969) to 46° 30' on the sines; that distance will reach from 230 to 168, on the line of numbers, for BC. Extend from 840 to 370.30', on the sines ; that extent will reach from 230 to 141, on the line of numbers, for AC. CASE III. Two sides and a contained angle given ; to find the other ana gles and side. PL. 5. fig. 16. In the triangle ABC, there is AR 240, the angle A 36° 404 and AC 180, given ; to find the angles C and B, and the side BC. Ist. B, Construction. Draw a blank line, on which from a scale of equal parts, lay AB 240; at the point A of the line AB, make an angle of 36° 40', by a blank line ; on which from A, lay AC 180, from the same scale of equal parts ; measure the angles C and B, and the the side BC, as before ; and you have the answers required. 2d. By Calculation. By cor 1. theo. 5. sect. 4. 180°—the angle A 36° 40' =143o. 20/ the sum of the angles C and B: therefore half of 143°. 20', will be half the sum of the two required angles, C and B. By theo. 2. of this sect. As the sum of the two sides AB and AC = 420 is to their difference, = 60 So is the tangent of half the sum of = 71° 404 the two unknown angles C and B to the tangent of half their difference - 23o 20% By theo. 4. To half the sum of the angles Cand B=71° 40 The sum is the greatest angle, or ang. C=95 00 Subtract, and you have the least angle, or B=48 20 The angle C and B being found ; BC is had, as before, by theo. l. of this sect. Thus, S. B: AC::S: A: BC. 3d. By Gunter's Scale. Because the two first terms are of the same kind, extend from 420 to 60 on the line of numbers ; lay that extent from 45° on the line of tangents, and keeping the left leg of your compasses fixed, move the right leg to 71o. 40' ; that distance laid from 45° on the same line will reach to 23o. 30', the half difference of the required angles. Whence the angles are obtained, as before. The second proportion may be easily extended, from what has been already said. CASE IV. Pl. 5. fig. 17. The three sides given, to find the angles. In the triangle ABC, there is given, AB 64, AC 47, BC 34: the angles A, B, C, are required. Ist. By Construction. The construction of this triangle must be manifest, from prob. I. sect. 4. 2d. By Calculation. From the point C, let fall the perpendicular CD on the base AB; and it will divide the triangle into two right angled oness ADC and CBD ; as well as the base AB, into the two segments, AD and DB. As the base or the longest side, AB 64 is to the sum of the other sides, AC and BC, 81 So is the difference of those sides 13 to the difference of the segments of 16.46 the base AD, DB. By theo. 4. of this sect. To half the base, or to half the sum of the segments AD and DB. Add half their difference, now found, } Subtract, and their difference will be the least segment DB, be} 23.77 In the right angled triangle ADC, there is AC 47, and AD 40. 28, given, to find the angle A. This is resolved by case 4. of right angled plane trigonometry, thus, AD: R::AC: Sec. A. Or it may be bad by finding the angle ACD, the complement of the angle A ; without a secant, thus, AC: R::AD: S. ACD. 90—58° 52' =319.08, the angle A. Then by theo. 1 of this sect. BC: S. A:: AC: S. B. By cor. 1. theo. 5. sect. 4. 180o--the sum of A and Bec. A 314.08' 180°-76. 45=1039.15', the angle C. |