Note. 1. The proportions in trigonometry are worked by logarithms; thus, from the sum of the logarithms of the second and third terms, subtract the logarithm of the first term, and the remainder will be the logarithm of the fourth term. 2. The logarithmic sine of a right angle or 90° is 10.00000, being the same as the logarithm of the radius. EXAMPLES. 1. In the triangle ABC, there are given the angle A 32° 15', the angle B 114° 24', and consequently the angle C33° 21', and the side AB 98*; required the sides AC and BC. By Construction, Fig. 41. Make AB equal to 98 by a scale of equal parts, and draw AC, making the angle A-32° 15'; also make the angle B=114° 24', and produce BC. AC, till they meet in C, then is ABC the triangle required; and AC, measured by the same scale of equal parts, is 162, and BC is 95. the sines of the angles A and B to the equal radii AC and BF. Now the triangles BDC and BEF being similar, we have CD: FE :: BC: BF or AC, that is sin. A: sin. B:: BC: AC. In like manner it is proved, that sin. A: sin. C:: BC AB. When one of the angles is obtuse the demonstration is the same. Hence it appears, that in any plane triangle, the sides are to one another as the sines of their opposite angles. This 98 may express so many feet, or yards, &c., and the other sides will be of the same denomination as the given side. 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 hy. pothenuse AC = 480, and the angle A 53° 8'. To find 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 given angle, To the sine of the angle opposite the other given side.* 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 BC7912 by a scale of equal parts, and draw CA, making the angle C33° 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 26o 21′ and |