An airplane left an airport and flew east 174 miles. Then it turned northward flying along a bearing of N22E. When it was 254 miles from the airport, there was an engine problem and it turned back to take the shortest route possible back to the airport. Find theta, the angle through which the plane turned
Is the angle 39.43?
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I see a triangle with sides 174 and 254 and the angle between those sides as 112º.
Let the third side be x. (The return path)
By Cosine Law:
x^2 = 174^2 + 254^2 - 2(174)(254)cos112º
x = 357.637
Let the angle at the turn be ß
By Sine Law:
sin ß/174 = sin 112/357.637
sin ß = .4510998
ß = 26.8º
So the plane must turn an angle of (180-26.8) or 153.2º
I see a triangle with sides 174 and 254 and the angle between those sides as 112º.
Let the third side be x. (The return path)
By Cosine Law:
x^2 = 174^2 + 254^2 - 2(174)(254)cos112º
x = 357.637
Let the angle at the turn be ß
By Sine Law:
sin ß/174 = sin 112/357.637
sin ß = .4510998
ß = 26.8º
So the plane must turn an angle of (180-26.8) or 153.2º
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