CASE V. The perpendicular and hypothenuse given, to find the angles and base. Plate V. Fig. 8. In the triangle ABC there is BC 306, and AC 370 given; to find the angles A and C; and the base AB. Geometrically. Draw a blank line from any point, in which, at B, erect a perpendicular, on which lay BC 306, from a scale of equal parts: from the same scale, with AC 370, in the compasses, cross the first drawn blank line in A, and you have the triangle ABC constructed. Measure the angle A. (by prob. 17. sect. 1;) and also AB, from the same scale of equal parts the other sides were taken from, and you have the answer. The operations by calculation, the square root, and Gunter's scale, are here omitted, as they have been heretofore fully explained: the statings, or proportions must also be obvious, from what has already been said. Answer. The angle A 55°. 48'; therefore the angle C 34. 12', and AB 208. CASE VI. The base and perpendicular given ; to find the angles and hypothenuse. Plate V. fig. 9. In the triangle ABC, there is AB 225, and BC 272, given; to find the angle A and C, and the hypothenuse AC. Geometrically. Draw a blank line, on which lay AB 225, from a scale of equal parts; at B, erect a perpendicular; on which lay BC, 272, from the same scale; join A and C, and the triangle is constructed. As before, let the angle A, and the hypothenuse AC be measured; and you have the answer. By Calculation. 1. Making AB the radius. AB: R:: BC: T. A. R. AB:: Sec. A: AC. 2. Making BC the radius. BC: R:: AB : T. C. R.: BC:: Sec. C: AC. By calculation; the answer from the foregoing proportions is easily obtained, as before. But because AC, by either of the said proportions, is found by means of a secant; and since there is no line of secants on Gunter's scale; after having found the angles, as before, let us suppose AC the radius, and then 1. S. A: BC:: R.: AC. or 2. S. C: AB:: R.: AC. These proportions may be easily resolved, either by calculation or Gunter's scale, as before; and thus the hypothenuse AC may be found without a secant. From the two given legs, the hypothenuse may be easily obtained, from cor. 1. theo. 14. sect. 1. Thus the square of AB-50625 From what has been said on logarithms, it is plain, 1. That half the logarithm of the sum of the squares of the two sides, will be the logarithm of the hypothenuse. Thus, The sum of squares, as before, is 124609; its log. is 5.09554, the half of which is 2.54777 ; and the corresponding number to this, in the table, will be 353, for AC. 2. And that half of the logarithm of the difference of the squares of AC and AB, or of AC and BC, will be the logarithm of BC, or of AB. The following examples are inserted for the use of the learner. 1. Given,} the angle C 64° 40′ (AB AC 3876 BC required. The answers are omitted, that the learner may resolve them himself by the foregoing methods; by which means he will find and see more distinctly their mutual agreements: and become more expert, and the better acquainted with the subject. OBLIQUE ANGULAR PLANE TRIGONOMETRY. BEF EFORE we proceed to the solution of the four cases of Oblique angular triangles, it is necessary to premise the following theorems. THEOREM I. Plate V. In any plane triangle ABC, the sides are proportional to the sines of their opposite angles, i. e. S. C: AB :: S. A: BC, and S. C: AB : : S. B ; AC; also S. B: AC:: S. A: BC. Fig. 10. By theo. 10. sect. 1. the half of each side is the sine of its opposite angle; but the sines of those angles, in tabular parts, are proportional to the sines of the same in any other measure; and therefore, the sines of the angles will be as the halves of their opposite sides; and since the halves are as the wholes, it follows, that the sines of their angles are as their opposite sides, i. e. S. C: AB :: S. A: BC, &c. Q. E. D. THEOREM II. In any plane triangle ABC, the sum of the two given sides AB and BC, including a given angle ABC, is to their difference, as the tangent of half the sum of the two unknown angles A and C is to the tangent of half their difference. Fig. 11. |