The hypotenuse is immediately deduced from which gives h2 = b2+p2, h = √b2+p2. (3.) Since this last formula is not well adapted to the use of logarithms, we will remark that after having found the angles, by means of equations (1) or (2), we may then determine the hypotenuse by Case II., equation (1), which gives log. h = log. p — log. sin. A. (4.) As an example, suppose the base to be 83-5 and the perpendicular to be 62-25. Required the other parts. b = 83·5; p = 62·25. BY NATURAL NUMBERS. (TABLE III.) 83.5) 62-25 (0·74550 = tang. A = cotan. C. BY LOGARITHMS. log. 62.25=1·794139 log. 83.5 1.921686 log. tang. A= log. cotan. C=9-872453 Again, log. 62.25 = 1.794139 log. sin. 36° 42′ 18′′ = 9.776480 Hence, h 104.15. log. h=2.017659 847. 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 opposite 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. The other angle = 43° 57'. Ans. The other side 109.44. The hypotenuse = 152.01. 4. Given, the hypotenuse equal 111.1, and one of the sides equal 37.5, to solve the triangle. The other side = 104.58. 5. Given, the two sides equal 29-37 and 37-29, to solve the The hypotenuse = 47·467. 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. § 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 A shall have (§ 8) CD CD sin. A sin. B= a C a B C D dividing the first equation by the second, we have By equation (9), § 16, we have sin. B+sin. C __tang. (B+C) or, b+c_tang. (B+C) C tang. (B-C)' (2.) b+c:b−c:: tang. (B+C): tang. (B–C). (3.) or, Subtracting (2) from (1), we have b2-a3 = AD3-BD2, (b + a) (b − a) = (AD + BD) (AD — BD), (3.) 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 A shall have c : α, sin. C: sin. A : : c which give a = c sin. A a C B sin. C (1.) c sin. B b = sin. C (2.) Using logarithms, we have log. a = log. c + log. sin. A-log. sin. C. (3.) log. b = log. c + log. sin. B−log. sin. C. (4.) |
