you fly on a heading, not a bearing.
Draw the diagram, then use the law of cosines to find the distance z:
z^2 = 500^2 + 400^2 - 2(500)(400) cos 127°
Relative to the airport, the plane is now at (x,y) where
y = 500 + 400cos57°
x = 400sin53°
So the plane's bearing is now (90-θ)°, where tanθ = y/x
An aeroplane leaves an international airport and flies due north for 1hour at 500km/hr. It then flies for 30 minutes at 800 kilometres/ hour on a bearing of North 53 degrees East. Calculate it's distance and bearing from the airport
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