Obs. A subsequent chapter would enable us to state the other value of 0, but chapters I.-III. do not. A similar remark will apply to the results of most of the equations for exercise which immediately follow. EXAMPLES. 1. What are the complements of the angles 12°, 25° 4′, 56° 50′ 44′′, 88° 58′ 58′′.89 ? 2. Show that 45°+A is the complement of 45° 7 A. Prove the formulæ in 3-6. 3. sin A cosec (90°-4) = cot (90°-A). 7. If tan A 1 tan (-) + cot (-) cot 4, what is the numerical value of A? 8. When sin 5A = cos A, find A. 9. When sec 34 = cosec 24, find A. 10. Write out the numerical values of the secants, cosecants, cotangents, and versed sines of the angles 30°, 45°, 60°; and prove geometrically that 11. Show that sin° 30°: sin2 45°: sin2 60°: 1: 2: 3. Also prove geometrically that cot 30° = √3. Obtain values for 0 satisfying the equations in 18-27. 29. If 25 sin2 A 30. 3 sin2 a + 2 sin 31. tan A+ cot A 32. tan2 0 == 16, find cos A. a = 1; required sin a. = 4; required tan A. 3 tan 0+4√tan 0+3; required tan 0. 33. Solve the simultaneous equations, sin A = √2 sin B, tan A √3 tan B. 34. Solve tan (A+B) = √√/3, cot (A−B) = √√/3. 37. If m sin A = n cos A, find sin A. 38. Given tan a + sin a = m, tan a-sin a = n; find sec a. 39. Given cos e cos A and cos (90°-0) = = cos B -; show that CHAPTER IV. SOLUTION OF RIGHT-ANGLED TRIANGLES. 22. The angles and sides of any triangle, six in number, are spoken of as the parts of that triangle. It is usual to denote the three angles by the capitals A, B, C, and the three sides respectively opposite to A, B, C by the small letters a, b, c. When the triangle contains a right angle, we shall most commonly call it C, and hence the hypothenuse c. In the case of the right-angled triangle two of the parts being given besides the right angle, the other three, or one or two as may be required, can always be found. The solution of problems in which the parts of a triangle are required is what is meant by the "solution of triangles." The mode adopted in Art. 13, of expressing the trigonometrical ratios for acute angles will now be found useful, and should be effectually impressed upon the memory. In addition, Euc. I. 47. will be required. EXERCISE II.- (1.) ABC being a right-angled triangle, write out the functions of the acute angles A and B in terms of a, b, c. Ans. Sin A = a b cos A = &c.; sin B = , C (2.) Prove the truth of the following equations :- b &c. 23. Of the solution of problems, of whatever kind, much is most easily learnt by the inspection of examples worked out, in the first place; and in the second, by observation and experience. Accordingly we shall give but few hints in this part of our subject otherwise than by actual solution of triangles. Those few, however, will be worthy of attention. 24. The application of Euc. I. 47, is at all times so obvious as to require no separate illustration. With the aid of it, any two sides of a right-angled triangle being given, we find the remaining side; for c2=a+b2, or ca2+b2; .. a2 = c2-b2, or a = √c2 —b2 ; and b2 = c2—a2, b= or √c2-a2. The same may be said of the formula A+B 90°, by which the value of one acute angle determines that of the other. The operations which require exemplification in the cases of the solution of triangles will be treated of under two heads. 25. To find a side, when an angle and a side are given. Forming a fraction having the required side as numerator and the given side as denominator, we find that we have one of the trigonometrical ratios of the given angle. Equate the fraction and the ratio; then a simple equation results, from which we may determine the unknown side. For instance, it is required to find a, when the values of B and c are known. The fraction formed as described is, and this, by с Art. 13, is cos B. Equating, and clearing the fraction, |