Asked by Prisca.
An aeroplane leaves an airport, flies due north for 2hrs at 500km/hr.it then flies 450km on a bearing 053.how far is the plane from the airport and what is it's bearing from it.
Answers
Answered by
oobleck
the plane -- starting at (0,0) -- ends up at
(x,y) = (450 sin53°,2*500 + 450 cos53°)
and, as usual,
distance = √(x^2+y^2)
bearing is (90-θ) where tanθ = y/x
(x,y) = (450 sin53°,2*500 + 450 cos53°)
and, as usual,
distance = √(x^2+y^2)
bearing is (90-θ) where tanθ = y/x