Sine and cosine of angles. 30. EXAMPLEs. 1. Given the sine of 23° 28' equal to 0.39822, to find the sine of 23° 29'. Solution. We find the cosine of 23° 28' by (10) to be cos. 23° 28' = 0.91729. Hence, by (18), making M = 23° 28′ sin. 23° 29′ sin, 23° 28' + 0.00029 cos. 23° 28', = 0.39822 + 0.00026, =0.39848. Ans. sin. 23° 29′ = 0.39848. 2. Given the sine and cosine of 46° 58′ as follows, sin. 46° 58' 0.73096, cos. 46° 58′ find the sine and cosine of 46° 59' 0.68242, 3. Given the sine and cosine of 11° 10' as follows, sin. 11° 10′ = 0.19366, cos. 11° 10′ = 0.98107, find the sine and cosine of 11° 11'. 31. By the formulas here given, a complete table of sines and cosines might be calculated. Such tables have been actually calculated; and table XXIV. of the Navigator is such a table; their logarithms are given in table XXVII. of the Navigator. Natural and artificial cosines. Radius of table. The sines, cosines, &c. of table XXIV. are called natural, to distinguish them from their logarithms, which are sometimes called their artificial sines, cosines, &c. The radius of table XXIV. is 105 = 100000, so that this table is, by § 21, reduced to the present system by dividing each number by 100000, that is, by prefixing the decimal point to each of the numbers of the table. The radius of table XXVII. is 101010000000000, so that this table is reduced to the present system by subtracting from each number the logarithm of this radius, which is 10, that is, by subtracting 10 from each characteristic. The method of using these two tables is fully explained in pp. 33-35, and p. 390, of the Navigator. Hypothenuse and an angle given. CHAPTER III. RIGHT TRIANGLES. 32. Problem. To solve a right triangle, when the hypothenuse and one of the angles are known. [B. p. 38.] Solution. Given (fig. 4) the hypothenuse h and the angle A, to solve the triangle. First. To find the other acute angle B, subtract the given angle from 90°. Secondly. To find the opposite side a, we have by (1) Thirdly. To find the side b, we have by (4) (21) (22) Leg and an angle given. 33. Problem. To solve a right triangle, when a leg and the opposite angle are known. [B. p. 39.] Solution. Given (fig. 4.) the leg a, and the opposite angle A, to solve the triangle. First. The angle B is the complement of A. Secondly. To find the hypothenuse h, we have by (21) a = h sin. A, which, divided by sin. A, gives by (6) log. h log. a + (ar. co.) log. sin. A =log. a+ log. cosec. A. Thirdly. To find the other leg b, we have by (4) 34. Problem. To solve a right triangle, when a leg and the adjacent angle are known. [B. p. 39.] Solution. Given (fig. 4.) the leg a and the angle B, to solve the triangle. First. The angle A is the complement of B. Hypothenuse and a leg given. Secondly. The other parts may be found by (23) and (24), or from the following equations, which are readily deduced from equations (4) and (6), log. h = log. a+ log. sec. B, log. blog. a+ log. tang. B. 35. Problem. To solve a right triangle, when the hypothenuse and a leg are known. [B. p. 49.] Solution. Given (fig. 4.) the hypothenuse h and the leg a, to solve the triangle. First. The angles A and B are obtained from equation (4), sin. A cos. B = a (27) or, by logarithms, log. sin. A = log. cos. B = log. a + (ar. co.) log. h. Secondly. The leg b is deduced from the Pythagorean property of the right triangle, which gives b = √(h2 a2)=√ by logarithms, (29) log. b.log. (h2— a2)= [log. (h+a) +log. (h—a)]. |