Question

an aircraft flies from its base 200 km on a bearing 162 degree, then 350km on a bearing 260 degree and then returns directly to base. Calculate the length and bearing of the return journey

Answer 1

X=200cos(162+90)+350cos(260+90)

y=200sin(162+90)+350sin(260+90)

x≈282.88

y≈-250.99

So the length of the return trip is:

d^2=x^2+y^2

d^2≈143017.0745

d≈378.18

tanα=-250.99/282.88

α=arctan(-250.99/282.88)

α≈-41.58°  that's the standard angle relative to the positive x-axis...

We need to be going in the exact opposite direction when returning so...

180-41.58=138.42°  again this angle is standard notation.  To convert this to a bearing we need to subtract 90°...

So the bearing is 138.42-90=48.42°

So the return trip will be 378.18km on a bearing of 48.42°

(I have assumed that the bearing was relative to true north, or the positive y axis....)