§47. Additional Examples of Right Triangles. 1. Given, the hypotenuse, equal 365, and one of the acute angles, equal 33° 12', to solve the triangle. 2. Given, one of the sides equal to 33-33, and the angle oppo site this side = 83° 33′, to solve the triangle. 3. Given, one of the sides equal to 105.5, and the acute angle adjacent this side equal to 46° 3', to solve the triangle. 4. Given, the hypotenuse equal 111·1, and one of the sides equal 37.5, to solve the triangle. The other side = 104.58. The angles = { 5. Given, the two sides equal 29-37 and 37-29, to solve the triangle. Ans. { The hypotenuse = 47:467. 38° 13′ 28′′. 51°46′ 32′′ The angles = { The above examples have all been wrought by the use of logarithms, which is, as a general thing, more simple than by the use of natural numbers only. It will be well, however, for the pupil to apply both methods, as was done in the examples which followed immediately after each case; by so doing, he will be the better prepared to comprehend the true nature of the trigonometrical values as well as their logarithmic values. § 48. Hereafter, when we use the word sine, cosine, tangent, or cotangent, in connection with logarithmic calculations, we wish to have it understood as meaning the logarithms of those values, increased by 10, as given in Table II. In the usual analytical investigations of trigonometrical formulas, we make use of their real or natural values, as given in Table III. CHAPTER IV. SOLUTION OF OBLIQUE TRIANGLES. §49. ALL the different cases of oblique triangles may be solved by the aid of the following theorems: THEOREM I. The sides of any plane triangle are to each other as the sines of their opposite angles. And conversely, the sines of the angles of any plane triangle are to each other as their opposite sides. In the triangle ABC denote the sides opposite the angles A, B, C, respectively by the letters a, b, c. Draw CD perpendicular to AB, and we shall have (§ 8) sin. A = CD A -, sin. B = C CD a dividing the first equation by the second, we have a C D B sin. A which gives a:b::sin. A : sin. B. In the same way it may be shown that a:c::sin. A : sin. C. hence, by composition and division, we have b+c:b-c:: sin. B+ sin. C: sin. B-sin. C, b+c:b-c::tang. (B+C): tang. (B-C). (3.) i ! or, Subtracting (2) from (1), we have b-a2 = AD2-BD, (3.) (b + a) (b − a) = (AD + BD) (AD - BD), which, converted into a proportion, will give, observing that AD + BD = c, c:b+a::b-a: AD-BD. (4.) If the triangle has an obtuse angle, then the perpendic ular should be drawn from this obtuse angle, otherwise it will not meet the opposite side. The theorem might be so modified as to embrace this case; but as it is used only in the solution of a triangle when the three sides are given, such modification is unnecessary. § 50. The solution of all oblique triangles may be included in four cases, as follows: CASE I. Given, a side and two angles, to find the other parts. Since the sum of the three angles of any plane triangle is equal to 180°, the third angle may be found by subtracting the sum of the two given angles from 180°. Having all the angles, the two remaining sides may be found by Theorem I. Thus, if we suppose the side e to be given, we shall have A C 4 B By using arithmetical complements (§ 32), we shall have log. a= ar. co. log. sin. C + log. sin. A + log. c log. b = ar. co. log. sin. C + log. sin. B + log. c (5.) (6.) As an example, suppose the angle A = 30° 20', the angle B = 50° 10', and consequently the angle C = 99° 30′. If the side c is 93-37, what will be the lengths of the other sides? Using equations (5) and (6), we have ar. co. sin. 99° 30′ (80° 30') =0.005997 sin. 30° 20′=9-703317; sin. 50°10′=9-885311 log. 93-37=1-970207 log. a=1.679521 Hence, a = 47.81; b = 72-697. 0.005997 1-970207 log. b=1-861515 NOTE.-Since the angle C = 99° 30' is an obtuse angle, we subtract it from 180, and obtain its supplement 80° 30', the sine of which is 9-994003. Now subtracting this from 10, we have 0-005997 for the arithmetical complement of the logarithmic sine of 99° 30′. The arithmetical complement may be readily taken from the table, by beginning at the left hand, and subtracting each figure from 9, except the last significant figure on the right, which must be subtracted from 10. (§ 32.) CASE II. Given, two sides and an angle opposite one of them, to find the other parts. Suppose the sides a and b to be given, and the angle A op posite the side a. By Theorem I., we have a:b:: sin. A : sin. B, or, = a sin. B-6 sin. A Using logarithms, we have log. sin. B = ar. co. log. a+log. b + log. sin. A. Having thus found the second angle, the third angle at C may be obtained by subtracting the sum of these two from 180°. (1.) (2.) |