Show also that (i.) a = c. sin A, (ii.) b = c. sin B, (iii.) a = c. cos B, (iv.) b = c.cos A, (v.) sin A = cos B, (vi.) cos A = sin B, (vii.) tan A = cot B. 6. The sides of a right-angled triangle are in the ratio 5: 12:13; find the sine, cosine, and tangent of each acute angle of the triangle. 7. The sides of a right-angled triangle are in the ratio 1:2: √3; find the sine, cosine, and tangent of each acute angle of the triangle. 8. Prove that if A be either of the angles of the above two triangles, sin2 A+ cos2 A = 1. 9. ABC is a right-angled triangle, C being the right angle. AB is 2 feet and AC is 1 foot; find the length of BC, and thence find the value of sin A, cos A, and tan A. 10. ABC is a right-angled triangle, C being the right angle, AB = √2 ft. and AC 1 foot; prove that sin A = cos A : = sin B = = cos B. 11. ABC is a right-angled triangle, C being the right angle; BC = 1 foot, and AB = √3 feet; find AC and sin A and sin B. CHAPTER V ON THE TRIGONOMETRICAL RATIOS OF CERTAIN ANGLES 35. The trigonometrical ratios of an angle are numerical quantities simply, as their name ratio implies. They are in nearly all cases incommensurable numbers. Their practical value has been found for all angles between 0° and 90°, which differ by 1'; and a list of these values will be found in any volume of Mathematical Tables. The general method of finding trigonometrical ratios belongs to a more advanced part of the subject than the present, but there are certain angles whose ratios can be found in a simple manner. 36. To find the sine, cosine, and tangent of an angle of 45°. When one angle of a right-angled triangle is 45°, that is, the half of a right angle, the third angle must also be 45°. Hence 45° is one angle of an isosceles right-angled triangle. Let POM be an isosceles triangle such that PMO is a right angle, and OM MP. = Then POM= OPM = 45°. Let the measures of OM and of MP each be m. Let the measure of OP be x. 37. To find the sine, cosine, and tangent of an angle of 60°. Each angle in an equilateral triangle is 60°, because they are each one-third of 180°. If we draw a perpendicular from one of the angular points of the triangle to the side opposite, we get a right-angled triangle in which one angle is 60°. Let OPQ be an equilateral triangle. Draw PM perpendicular to OQ. Then OQ is bisected in M. Let the measure of OM be m; then that of OQ is 2 m, and therefore that of OP is 2 m. 38. To find the sine, cosine, and tangent of an angle of 30°. With the same figure and construction as above we have the angle OPM 30°, since it is half of OPQ, i.e. of 60°. = 39. To find the sine, cosine, and tangent of an angle of 0°. Let XOP be a small angle. Draw PM perpendicular to OX, and let OP be always of the same length, so that P lies on the circumference of a circle whose centre is 0. Then if the angle XOP be diminished, we can see that MP is diminished also, and that consequently which is sin XOP, is MP diminished. And by diminishing the angle XOP sufficiently, we can make MP as small as we please, and therefore we can make sin XOP smaller than any assignable number however small that number may be. This is what is meant when it is said that the value to which sin XOP approaches as the angle is diminished, is 0. expressed by saying, sin 0° = 0. This is (i.) Again, as the angle XOP diminishes, OM approaches OP in OM length; and cos XOP, which is OP' approaches in value to OP ОР i.e. to 1. while OM does not; .. tan XOP approaches 0. This is expressed by saying, 40. To find the sine, cosine, and tangent of 90°. Let XOY be a right angle = 90°. Draw XOP nearly a right angle; draw PM perpendicular to OX, and let OP be always of the same length, so that P lies on the circumference of a circle whose centre is 0. M FIG. 19. Then, as the angle XOP approaches to XOY, we can see that MP approaches OP, while OM continually diminishes. Hence when XOP approaches 90°, sin XOP, which is MP OP' (i.) Again, when XOP approaches 90°, tan XOP, which is MP OM' approaches in value to OP |