SOLUTION: An aeroplane leaves an airport and flies due north for 1 and half hours at 500 km per hour. It then flies 400 km on the bearing 053 degree. Calculate it's final distance and bearin

Question 1201881: An aeroplane leaves an airport and flies due north for 1 and half hours at 500 km per hour. It then flies 400 km on the bearing 053 degree. Calculate it's final distance and bearing from the airport. Leave your answer to the nearest km and 0.1 respectively.

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
An aeroplane leaves an airport and flies due north for 1 and half hours at 500 km per hour. It then flies 400 km on the bearing 053 degree. Calculate it's final distance and bearing from the airport. Leave your answer to the nearest km and 0.1 respectively.
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Using triangles:
2 sides given, 750 kms and 400 kms
The included angle is 127 degs.
Use the Cosine Law.
c^2 = a^2 + b^2 - 2ab*cos(127)
c^2 = 903044.5
c = 950.3 kms from the airport
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Use the Law of Sines to find the angle from the airport (angle A)
400/sin(A) = 950.3/sin(127)
sin(A) = 400*sin(127)/950.3 = 0.33616
A = 19.6 degs (from the x_axis)
Heading = 90 - 19.6 = 70.4 degs from the airport
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If there are 2 "legs" triangles work well.
If there are more than 2, vectors work better.
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PS it's = it is
"Leave your answer to the nearest km and 0.1 respectively." is not clear
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