THE SQUARE OF THE HYPOTHENUSE, THE REMAINDER WILL BE THE SQUARE OF THE REQUIRed leg. As in the preceding EXAMPLE; the square of the leg AB 40 is 1600; this subtracted from the square of the hypothenuse 50 which is 2500, Icaves 900, the square of the leg BC, the square root of which is 30, the length of leg BC as found by logarithms. CASE IV. Fig. 45. The legs given to find the angles and hypothenuse. Fig. 45. In the triangle ABC, given the leg AB 78.7 and the leg BC 89; to find the angles and hypothenuse. Making the leg AB radius, the proportion to find the angle The angle ACB is consequently 41° 29'. Making the leg BC radius, the proportion to find the angle BCA will be similar, with the obvious differences. The angles being found, the hypothenuse may be found by BY THE SQUARE ROOT. In this case the hypothenuse may be found by the square root, without finding the angles; according to the following PROPOSITION. IN EVERY RIGHT ANGLED TRIANGLE, THE SUM OF THE SQUARES OF THE TWO LEGS IS EQUAL TO THE SQUARE OF THE HYPOTHENUSE. In the above EXAMPLE, the square of AB 78.7 is 6193.69, the square of BC 89 is 7921; these added make 14114.69 the square root of which is nearest 119. BY NATURAL SINES. The hypothenuse being found by the square root, the angles may be found by nat. sines, according to the preceding CASE. Hyp. 119) Leg. BC. 570 476 940 833 1070 952 Nat. Sine. (74789 The nearest degrees and minutes corresponding to the above nat. sine are 48° 24', for the angle BAC. The difference between this and the angle as found by logarithms is occasioned by dividing by 119, which is not the exact length of the hypothenuse, it being a fraction too much. 1180 1071 109 PART II. OBLIQUE TRIGONOMETRY. The solution of the first two CASES of Oblique Trigonometry depends on the following PROPOSITION. IN ALL PLANE TRIANGLES, THE SIDES ARE IN PROPORTION THAT IS, AS THE SINE OF ONE ANGLE IS TO ITS OPPOSITE SIDE, SO IS THE SINE OF ANOTHER ANGLE TO ITS OPPOSITE SIDE. OR, AS ONE SIDE IS TO THE SINE OF ITS OPPOSITF ANGLE, SO IS ANOTHER SIDE TO THE SINE OF ITS OPPOSITE ANGLE. [NOTE. When an angle exceeds 90° make use of its supplement, which is what it wants of 180°. GEOM.] Note Def. 26. CASE I. Fig. 46. c The angles and one side given to find the other sides. Fig. 46. In the triangle ABC, given the angle at B 48°, the angle at C 72°, consequently the angle at A 60°, and the side ÅB 200, to find the sides AC and BC. AS THE NAT. SINE OF THE ANGLE OPPOSITE THE GIVEN SIDE IS TO THE GIVEN SIDE, SO IS THE NAT. SINE OF THE ANGLE OPPOSITE EITHER OF THE REQUIRED SIDES TO THAT REQUIRED SIDE. Given side 200; nat. sine of 72°, its opposite angle, 0.95115; nat. sine of ABC 48°, 0.74334; nat. sine of BAC 0.95115: 200 :: 0.74334: 156 0.95115: 200:: 0.86617: 182. CASE II. Two sides, and an angle opposite to one of them given, to find the other angles and side. Fig. 47. In the triangle ABC, given the side AB 240, the side BC 200, and the angle at A 46° 30'; to find the other angles and the side AC. To find the angle ACB. 9.860562 2.380211 107.00 The side AC will be found by CASE I. to be nearest 253. NOTE. If the given angle be obtuse, the angle sought will be acute; but if the given angle be acute, and opposite a given lesser side, then the angle found by the operation may be either obtuse or acute. It ought therefore to be mentioned which it is, by the conditions of the question. BY NATURAL SINES. AS THE SIDE OPPOSITE THE GIVEN ANGLE IS TO THE NAT. SINE OF THAT ANGLE, SO IS THE OTHER GIVEN SIDE TO THE NAT. SINE OF ITS OPPOSITE ANGLE. One given side 200, nat. sine of 46° 30', its opposite angle, 0.72537, the other given side 240. CASE III. Two sides and their contained angle given, to find the other angles and side. Fig. 48. The solution of this CASE depends on the following PROPOSI TION. IN EVERY PLANE TRIANGLE, AS THE SUM OF ANY TWO SIDES IS TO THEIR DIFFERENCE, SO IS THE TANGENT OF HALF THE SUM OF THE TWO OPPOSITE ANGLES TO THE TANGENT OF HALF THE DIFFERENCE BETWEEN THEM. ADD THIS HALF DIFFERENCE TO HALF THE SUM OF THE ANGLES AND YOU WILL HAVE THE GREATER ANGLE, AND SUBTRACT THE HALF DIFFERENCE FROM THE HALF SUM AND YOU WILL HAVE THE LESSER ANGLE. In the triangle ABC, given the side AB 240, the side AC 180, and the angle at A 36° 40′ to find the other angles and side. The given angle BAC 36° 40′, subtracted from 180°, leaves 143° 20' the sum of the other two angles, the half of which is 71° 40'. As the sum of two sides, 420 : their difference 60 2.623249 1.778151 10.479695 12.257846 2.623249 : tangent half difference, 23° 20' nearly The half sum of the two unknown angles, Add, gives the greater angle ACB 9.634597 71° 40' 23 20 95 00 Subtract, gives the lesser angle ABC 48 23 |
