f 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 TO EACH OTHER AS THE SINES OF THEIR OPPOSITE ANGLES. 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 OPPOSITE ANGLE, SO IS ANOTHER SIDE TO THE SINE OF ITS OPPOSITE ANGLE. Note Def. NOTE. When an angle exceeds 90° make use of its supplement, which is what it wants of 180°. 26. GEOM. CASE I. Fig.46. The angles and one side given to find the other sides. Fig. 46. 48 200 In the triangle ABC, given the angle at B 48°, the angle at C 72°, consequently the angle at A 60°, and the side AB 200, to find the sides AC and BC. BY NATURAL SINES. 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°, 0.74314; nat. sine of BAC 60° 0.86617. Thus, 0.95115: 200 :: 0.74314 : 156 CASE II. Fig. 47. Two sides, and an angle opposite to one of them given, to find the other angles and side. Fig. 47. 46°30' A 240 200 B 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. 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 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. As 200 0.72537 : : 240 : 0.87044-60° 30'. CASE III. Fig. 48. 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 PROP OSITION. 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 : tangent half difference, 23° 20′ nearly The half sum of the two unknown angles, Add, gives the greater angle ACB A 105 The solution of this CASE depends on the following PROP OSITION. IN EVERY PLANE TRIANGLE, AS THE LONGEST SIDE IS TO THE SUM OF THE OTHER TWO SIDES, SO IS THE DIFFERENCE BETWEEN THOSE TWO SIDES TO THE DIFFERENCE BETWEEN THE SEGMENTS OF THE LONGEST SIDE, MADE BY A PERPENDICULAR LET FALL FROM THE ANGLE OPPOSITE THAT SIDE. Half the difference between these segments, added to half the sum of the segments, that is, to half the length of the longest side, will give the greatest segment; and this half difference subtracted from the half sum will be the lesser segment. The triangle being thus divided, becomes two right angled triangles, in which the hypothenuse and one leg In the triangle ABC, given the side AB 105, the side AC 85, and the side BC 50, to find the angles. Thus the triangle is divided into two right angled triangles, ADC and BDC; in each of which the hypothenuse and one leg are given to find the angles. The angle DCA 61° 56' subtracted from 90° leaves the angle CAD 28° 4/ The angle DCB 36° 52′ subtracted from 90° leaves the angle CBD 53° 16'. The angle DCA 61° 56′ added to the angle DCB 36° 52′ gives the angle ACB 98° 48'. This CASE may also be solved according to the following PROPOSITION. IN EVERY PLANE TRIANGLE, as the PRODUCT OF ANY TWO SIDES CONTAINING A REQUIRED ANGLE IS TO THE PRODUCT |