OBLIQUE TRIGONOMETRY. The solution of the first two CASES of Oblique Trigonom etry 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. PART II. NOTE. When an angle exceeds 90°, make use of its supplement, which is what it wants of 180°. Note, Def. 25. GEOM. The angles and one side given, to find the other sides. Fig. 46. :side AC, 156 CASE I. Fig. 46. C 720 In the triangle ABC, given the angle at B 48°, the angle at C 720, consequently the angle at A 60°, and the side AB 200, to find the sides AC and B C. 200 To find the side A C. To find the side B C. As sine A CB, 72° 9.978206 As sine A CB, 720 - 9.978206 :side AB, 200 2.301030: side A B, 200 - 2.301030 :sine A B C, 48° -9.871073: : sine B A C, 60° - 9.937531 12.172103 2.193897: side B C, 182 B 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 B A C 60o, 0.86617. Thus, 0.95115 200 : 0.74314 : 156. CASE I I. Two sides, and an angle opposite to one of them given, to find the other angles and side. Fig. 47. A4 Fig. 47. с 46°30' In the triangle AB C, given the side A B 240, the side B C 200, and the angle at A 46° 30′; to find the other angles and the side A C. 200 Angle at A B 46o 30' 60 30 107.00 12.240773 2.301030 Sum of the three angles 1800 Sum of two 1070 :sineACB,60°30' nearly 9.939743 Angle at B 73 The side A C 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. Two sides and their contained angle given, to find the other angles and side. Fig. 48. Fig. 48. с 180 36°40' A 240 B The solution of this CASE depends on the following PROP A B A C 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 SUBSTRACT THE HALF DIFFERENCE FROM THE HALF SUM, AND YOU WILL HAVE THE LESSER ANGLE. In the triangle A B C, given the side A B 240, the side A C 180, and the angle at A 36° 40', to find the other angles and side. Side A B 240 A C 180 Sum of the two sides 420 Difference The given angle BAC 36° 40′, substracted from 180°, leaves 143° 20′, the sum of the other two angles, the half of which 240 180 60 As the sum of two sides, 420 : their difference 60 :: tangent half unknown ang. 71° 40′ : tangent half difference, 23° 20′ nearly The half sum of the two unknown angles, Add, gives the greater angle ACB Substract, gives the lesser angle ABC The side B C may be found by CASE I. or IL. CASE IV. The three sides given, to find the angles. Fig. 49. 85 2.623249 1.778151 10.479695 12.257846 2.623249 9.634597 71° 40' 23 20 95 00 48 23 Fig. 49. 75 A The solution of this CASE depends on the following PROP OSITION. 30 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 substracted 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 A B C, given the side A B 105, the side A C 85, and the side B C 50, to find the angles. 85 A C Side A C 50 B C Sum of the two sides 135 As the longest side A B, 105 : sum of the other two sides, 135 : difference between the segments, 45 Half the side A B Difference 11.875061 85 50 Add, gives the greater segment A D Substract, gives the lesser segment BD 30.0 Thus the triangle is divided into two right angled triangles, A DC and B DC; in each of which the hypothenuse and one leg are given to find the angles. To find the angle DC A. To find the angle DC B. As hyp. A C, 85 1.929419 As hyp. B C, 50 : radius 10.000000: radius :: seg. A D, 75 1.875061: seg. B D, 30 35 2.021189 2.130334 1.544068 3.674402 2.021189 1.653213 52.5 22.5 75.0 1.698970 10.000000 1.477121 11.477121 1.698970 :sine DCA, 61° 56′ 9.945642: sine DCB, 36° 52′ 9.778151 The angle DCA 61° 56′, substracted from 90°, leaves the angle C A D 28° 4′. The angle DCB 36° 52', substracted from 90°, leaves the angle CBD 53° 16′. The angle DC A 61° 56′, added to the angle D C B 36° 52′, gives the angle A CB 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 |