an aeroplane leaves an airport,flies due north for two hours at 500km/hr.it then flies 450km on a bearing 053.how far is the plane from the airport and what is its bearing from bearing from east

Answer 1

using the law of cosines, the distance z is
z^2 = 1000^2 + 450^2 - 2*1000*450 cos127°
If the plane started at (0,0) then it ended at (359.4,1271)
Now you can figure the angle θ:
tanθ = 1271/359.4

Answer 2

Too snappy,not self explanatory pls I need the answer in a more explained form.