CASE VI. The base and perpendicular given; to find the angles and hypothenuse. PL. 5. fig. 9. In the triangle ABC, there is AB 225, and BC 272, given; to find the angles A and C, and the hypothenuse AC. 1st. By Construction. 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; in order to find the an swers. 2d. By Calculation. 1. Making AB the radius. 2. Making BC the radius. BCR::AB: T. C. R. BC:: Sec. C: AC. By calculation; the answers from the foregoing proportions are easily obtained, as before. But because AC, by either of the said propor tions 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 sides, the hypothenuse may be easily obtained, from cor. 1. theo. 14. sect. 4. Thus the square of AB 50625 73984 124609 (353 AC 9 65)346 325 703)2109 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,* * Demonstration. The square of the hypothenuse of a right-angled triangle is equal to the sum of the squares of the sides, (theo. 14.) hence the Log. of (ACZAB2): the Log, of BC2, and by the nature of Logarithms, the Log. of BC is equal to the Log. of BC2 divided by 2; and in like manner the Log. of (AB2+BC2) the Log. of AC2, hence dividing the Log. of AC2 by 2 gives the Log. of AC. Q. E. D. The sum of squares, as before, is 124609; its log. is 5.095549, the half of which is 2.547774; and the corresponding number to this, in the tables, 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 exercise of the learner. Ex. 1. In the right-angled triangle ABC. the hypothenuse AC Given, 540 perches,4A33°45' Ans. To find the other two sides. { BC 300 AB 449 Ex. 2. In the right-angled triangle ABC. Given, the base AB 162 chains, Ans. To find the other two sides. AC 270 Ex. 3. In the right-angled triangle ABC. the perpendicu Given, lar BC 180 links, Ans. 2 C 62° 40′ To find the other two sides. { SAC 392.0146 AB 348.2464 Ex. 4. In the right-angled triangle ABC. the hypothenuse Given, AC 392 poles, the Ans. {AC 392 poles, the base AB 180 poles To find the angles and perpendicular. LA 62° 40' 4C 27° 20′ BC 348.25 Ex. 5. In the right-angled triangle ABC. Ex. 6. In the right-angled triangle ABC. the base AB 735.9 Given, links, the perpendi-Ans. cular BC 320 To find the angles and hypothenuse. 4C 66° 30' LA 23 30 AC 802.5 OBLIQUE ANGLED PLANE TRIGONOMETRY. BEFORE EFORE we proceed to the solution of the four cases of Oblique angled triangles, it is necessary to premise the following theorems. THEO. I. PL. 5. fig. 10. In any plane triangle ABC, the sides are proportional to the sines of their opposite angles; that is, S. C: AB:: S. A: BC, and S. C: AB:: S. B: AC; also S. B : AC : ; S. A : BC. By theo. 10. sect. 4. 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; that is, S. C: AB: S. A: BC, &c. Q. E. D. THEO. II Fig. 11. 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 tengent of half the sum of the two unknown angles A and C is to the tangent of half their difference. Produce AB, and make HB=BC, and join HC: let fall the perpendicular BE, and that will bisect the angle HBC (by theo. 9. sect. 4.) through B draw BD parallel to AC, and make HF DC, and join BF; take BI= BA, and draw IG rallel to BD or AC. pa It is then plain that AH will be the sum, and HI the difference of the sides AB and BC; and since HB = BC, and BE perpendicular to HC, therefore HEEC (by theo, 8. sect. 4.); and since BABI, and BD and IG parallel to AC, therefore GD = DC = FH, and consequently HG = FD, and HGFD or ED. Again, EBC being half HBC, will be also half the sum of the angles A and C (by theo. 4. sect. 4.) also, since HB, HF, and the included angle H, are severally equal to BC, CD, and the included angle BCD: therefore (by theo. 6. sect. 4.) HBF = DBC BCA (by part 2. theo. 3. sect. 4.) and since HBD = A (by part 3. theo. 3. sect. 4.) and HBF BCA: therefore FBD is the difference, and EBD, half the difference of the angles A and C: then making BE the radius, it is plain, that EC will be the tangent of half the sum, and ED the tangent of half the difference |