Extend the compasses, on the line of sines, from. 33° 21' to 65° 36 the supplement of the angle B; that. extent will reach, on the line of numbers, from 98 to 162, the side AC. Extend the compasses from 33° 21' to 32° 15′ on the line of sines; that extent will reach, on the line of numbers, from 98 to 95, the side BC. 2. In the right angled triangle ABC, are given the hypothenuse AC = 480, and the angle A= 53o 8. To From 90° subtract the angle A= 53° 8'; the remainder 36° 52′ will be the angle C. The angle B, being a right angle is 90°. By Construction, Fig. 42. This may be constructed as in the preceding example; or otherwise thus, Draw the line AB of any length, and draw AC making the angle A = 53° 8'; make AC = 480 by a scale of equal parts, and from C draw CB perpendicular to AB, then ABC is the triangle required. AB, measured by the same scale of equal parts, will be 288, and BC will be 384. By Gunter's Scale. Extend the compasses, on the line of sines, from 90* to 53° 8', that extent will reach, on the line of numbers, from 480 to 384 the perpendicular BC. Extend the compasses, on the line of sines, from 90° to 36° 52', the complement of the angle A; that extent will reach, on the line of numbers, from 480 to 288, the base AB. 3. In the triangle ABC, are given the angle A = 79° 23′, the angle B 54° 22′, and the side BC= 125; required AC and AB. Ans. AC = 103.4, and AB = 91.87. * 4. In a right-angled triangle, there are given the angle A 56° 48', and the base AB = 53.66, to find the perpendicular BC and hypothenuse AC. Ans. BC=82 and AC 98. 5. In the right-angled triangle ABC, are given the angle A 39° 10′, and the perpendicular BC= 407.37, to find the base AB and hypothenuse AC. Ans. AB = 500.1, and AC = 645. CASE 2. Two sides and an angle opposite one of them being given, to find the other angles and side. RULE. As the side opposite the given angle, Is to the other given side, So is the sine of the angle opposite the former, To the sine of the angle opposite the latter.* * This is evident from the demonstration of the rule in the preced Add the angle thus found to the given angle, and subtract their sum from 180°, the remainder will be the third angle. After finding the angles, the other side may be found by Case 1. Note.-The angle found by this rule is sometimes ambiguous, for the operation only gives the sine of the angle, not the angle itself; and the sine of every angle is also the sine of its supplement. When the side opposite the given angle is equal to, or greater than the other given side, then the angle opposite that other given side is always acute; but when this is not the case, that angle may be either acute or obtuse, and is consequently ambiguous. EXAMPLES. 1. In the triangle ABC, are given the angle C = 33° 21', the side AB = .98 and the side BC= .7912; required the angles A and B, and the side BC. By Construction, Fig. 43. Make BC= .7912 by a scale of equal parts, and draw CA, making the angle C = 33° 21'; with the side. AB.98, in the compasses, taken from the same scale of equal parts, and B as a centre, describe the arc ab, cutting AC in the point A, and join BA; then is ABC the triangle required: the side AC, measured by the scale of equal parts will be 1.54, and the angles A and B, measured by a scale of chords will be 26° 21′ and G Here the arc ab cuts AC in one point only, because AB is greater than BC; therefore the angle A is acute, and no ambiguous. To the angle C = 33° 21' add the angle A = 26° 21', and the sum is 59° 42′ which subtracted from 180°, leaves the angle B = 120° 18'. 1. Extend the compasses from .98 to .79 on the line of numbers, that extent will reach from 33° 21′ to 26° 21', the angle A, on the line of sines. 2. Add the angle A= 26° 21' to the angle C33° 21'′, and the sum will be 59° 42'; then extend the compasses from 33° 21′ to 59° 42′, on the line of sines, that extent will reach from .98 to 1.54, the side AC, on the line of |