Machinery's Handbook, 31st Edition
SPHERICAL TRIGONOMETRY 117 The side and angle labels in examples that follow refer to those of the right spherical triangle in Fig. 1. Example 1: Find the length of arc A of a right spherical triangle on the surface of a sphere where radius K = 30.00 inches and angle F = 10 ° . Solution: A K π 180 ----- F × 30 π 180 ----- 10 × 5.2359 in. = = = × × Example 2: Find length of arc B on a sphere of radius J = 11.20 inches if angle G = 10 ° . Solution: B J π 180 ----- G × 11.20 π 180 ----- 10 × 1.9547 in. = = = × × Example 3: A right spherical triangle is to be constructed on the surface of a sphere 22.400 inches in diameter. Side A is 7.125 inches and angle E is 57 ° 59 ′ 19 ″ . Determine the lengths of sides B and C , and the measure of angle D , and the area of the triangle. Solution: The radius of the sphere, J = K = 11.200, and the length of side A is used to find the value of angle F . Angle E is converted to decimal degree format for simplicity; then angles E and F are used to solve the equation for angle tan G . Side B and angle D can then be found. Angle H can be calculated using either of the two equations given for cos H , and finally the length of side C can be found. Notice that the sum of angles D + E + 90 ° is not equal to 180 ° , but 194.98 ° . Calculation details are as follows: F ° 180 π ----- A K -- × 180 π ----- 7.125 11.200 -------- 36.449324 ° = = = E 57 ° 59 ′ 19 ″ 57 59 60 --- 19 3600 ------ + + 57.988611 ° = = = G tan F sin E tan 36.449324 ° ( ) sin 57.988611 ° ( ) tan 0.950357 = = = G tan –1 (0.950357) 43.541944 ° = = B J π 180 ----- G × × ° 11.200 π 180 ----- 43.541944 × 8.511443 = = = D tan F tan csc G × 36.449324 ° ( ) tan csc 43.541944 ° ( ) × 1.0721569 = = = D 180 π ----- tan –1 1.0721569 ( ) 46.994354 ° = = H cos G cos F cos × 43.541944 ° ( ) cos 36.449324 ° ( ) cos × 0.58307306 = = = H 180 π ----- cos –1 0.58307306 ( ) 54.333023 ° = = C J π 180 ----- H × ° 11.200 π 180 ----- 54.333023 × × ° 10.62085 = = = Angles D E 90 ° + + ( ) 46.994354 ° 57.988611 ° 90 ° + + 194.98297 ° = = Area 11.200 2 194.98297 – 180 ( ) × 50.142591 in 2 = = × × × × × × × Example 4: A right spherical triangle on a 20-mm diameter sphere has two 90 ° angles, and the distance B between the 90 ° angles is 1 ⁄ 3 the circumference of the sphere. Find angle E, the area of the triangle, and check using the conventional formula for area of a sphere. Solution: By inspection, angle G is 360 ° /3 = 120 ° . Because angles D and G are known, angle E can be calculated using E cos G cos D sin × = . Therefore, E cos G cos D sin × 120 ( °) cos 90 ( °) sin × –0.5 = = = E cos –1 –0.5 ( ) 120 ° = = Area 10 2 π 180 ----- 120 ° 90 ° 90 ° 180 ° – + + ( ) × 100 2.0943951 × 209.4 mm 2 = = = Check: Total area of 20-mm diameter sphere/6 4 π R 2 6 ------- 4 π 100 ( ) 6 ----------- 209.4 mm 2 = = =
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