NOTE 3. An equation involving trigonometric ratios is a trigonometric equation. Thus, for example, tan A = 1. One angle which satisfies this equation is the acute angle A= 45°. Other solutions can be found after Arts. 84-87 have been taken up. EXAMPLES. A few simple exercises are given below, the solution of which brings in the relations shown in this article. These exercises are algebraic in character; collections of exercises of this kind are given also in other places in this book. In the following examples, the positive values of the radicals are to be taken. The meaning of the negative values is shown in Art. 44. 1. Given that sin A =}, find the other trigonometric ratios of A by means of the relations shown in this article. These results may be verified by the method used in solving Exs. 1, 5–7, Art. 14. 2. Express all the ratios of angle A in terms of sin A. sec1 A - 1 = (sec2 A)2 − 1 = (1 + tan2 A)2 – 1 = 2 tan2 A + tan1 A. On taking the plus sign, one solution is the acute angle = 60°; other solutions will be found later. For the minus sign there is also a set of solutions; these will be found later. sine cannot exceed unity; hence sin = 2 does not afford any solution. 7. Given cos A=1, sin B=}, tan C=2, cot D=, sec E=3, cosec F=2.5 ; find the other trigonometric ratios of A, B, C, D, E, F, by the algebraic method. Verify the results by the method used in Art. 14. 8. Find by the algebraic method the ratios required in Exs. 1, 5–7, 10, 11, Art. 14. Arrange the 9. Express all the trigonometric ratios of an angle A in terms of: (a) cos A; (b) tan A; (c) cot A; (d) sec A; (e) cosec A. results and those of Ex. 2, neatly in tabular form. Prove the following identities: 10. (sec2 A-1) cot2 A = 1; cos A tan A = sin A; (1 − sin2 A) sec2 A = 1. sin20 sec2 = sec2 0 - 1 ; tan2 0 - cot2 0 = sec2 0 11. cosec2 0. 1 sec Atan A; sec1 A-sec2 A tan1 A + tan 2 A. = sec Atan A Solve the following equations : 15. 2 sin 02 - cos 0. 17. tan + 3 cot 0 = 4. 19. 8 sin2 - 10 sin 0 + 3 = 0. 16. tan + cot = 2. 19. Summary. In this chapter important additions have been made to the knowledge concerning angles that one gained in geometry. A process of measuring angles has been introduced. The close connection between angles and the ratios of lines has been emphasized. It has been shown that each (acute) angle has, associated with it, a definite set of six numbers, called trigonometric ratios; and it has been seen that the sets of numbers are different for different angles. It has also been shown that the seven quantities (namely, the angle and the six numbers) are so related, that, if one of the seven be given, then the remaining six can be determined. A few applications to the measurement of lines and angles have been made in some of the preceding articles. The next two chapters are taken up with a formal treatment of such applications. It should be stated, however, that any one who understands the contents of this chapter is in possession of all the principles which will be used in the next two chapters, and can proceed directly to the solution of the problems given there. The student is recommended to attack some of the exercises in Chapters III., IV., before reading the explanations given in the text. Attention may again be given to the first part of Art. 14. N.B. Questions and exercises suitable for practice and review on the subject-matter of Chapter II. will be found at pages 182, 183. CHAPTER III. SOLUTION OF RIGHT-ANGLED TRIANGLES. Before the solution of right-angled triangles is entered upon, a few remarks will be made on the solution of triangles in general. Some of the ideas expressed in Arts. 20-24 are applicable to practical problems throughout the book. 20. Solution of a triangle. Every triangle has three sides and three angles. These six quantities are called the parts or elements of a triangle. Sometimes one or several of the parts of a triangle are known; for instance, the three sides, two angles and a side, two sides, one side, three angles, and so on. In such cases the questions arise: Can the remaining parts be found or determined? and, if so, by what method shall this be done? The process of deducing the unknown parts of a triangle from the known, is called solving the triangle, or, the solution of the triangle. This Chapter and Chapter VII. are concerned with showing, in detail, methods of solving triangles. There are two methods which can be used to find (only approximately, in general) the unknown parts of a triangle when some of its parts are given. These methods are: (a) The graphical method; (b) The method of computation. 21. The graphical method. This method consists in drawing a triangle which has angles equal to the given angles, and sides proportional to, and thus representing the given sides, and then measuring the remaining parts directly from the drawing. For example, a triangle has two sides whose lengths are 10 ft., 5 ft., and the included angle is 28° 30'; the third side and the other angles are required. The graphical solution is as follows: Construct a triangle QPR having two sides, PQ, PR, representing 10 ft., 5 ft., respectively, on some con venient scale, and shown in Fig. 14. measure the side with their included angle, QPR, equal to 28° 30', as Measure the angles PRQ, PQR with the protractor; RQ and, by reference to the scale, find the length represented by RQ. [The results thus obtained may be compared with those obtainable by the method of computation explained in Arts. 54, 57. The latter results are R = 128° 26′ 46", Q23° 3' 14", RQ 6.092.] = P R 5 Ft. 28 30 10 Ft. FIG. 14. The conditions necessary and sufficient for constructing a triangle, and the methods of drawing triangles that satisfy given conditions, are shown in plane geometry and in geometrical drawing. It is obvious that the graphical method can be employed only when the values of the parts given are consistent with one another, and when the parts given are sufficient in number to determine a definite triangle. For instance, suppose that one is asked to find the remaining parts of a triangle one of whose sides is 10 inches long. In this case as many unequal triangles as one please, can be constructed, all of which will satisfy the given condition. Again, a given side and a given angle are insufficient data on which to proceed to find the remaining parts of a triangle, for there is an infinite number of unequal triangles which can have parts equal to the given parts. So also the method fails if three angles be given; for an infinite number of unequal triangles can be drawn whose angles are equal to the given angles. Again, let it be required to find the angles of a triangle whose sides are 10 feet, 40 feet, 60 feet. Such a triangle is impossible, since the length of one side (60 feet) is greater than the sum of the lengths of the other two. One more instance: let two given angles be 85° and 105° and the included side be 40 inches; this triangle is impossible, since the sum of the two given angles is greater than two right angles. 22. The method of computation. This method is applicable in precisely the same cases in which the preceding method can be employed; namely, in the cases in which the parts given are consistent with one another, and afford conditions sufficient to enable one to construct a definite triangle. This will be fully apparent later, when the various cases will be treated in detail. One of |