Two ships A and B leaves the some harbour at the same time in directions 152° and 142° respectively calculate the distance between the two ships when A has travelled 10km and B has travelled 12km. What is the bearing of A from B at this time

Answer 1

Draw a diagrem, as always.
use the law of cosines for the distance
d^2 = 10^2+12^2-2*10*12cos10° = 7.646
d = 2.765

To find the bearing, note that the locations are
A:(4.695,-8.829)
B:(7.388,-9.456)
So the bearing is 270+θ such that
tanθ = (9.456-8.829)/(7.388-4.695)=0.233
θ = 13.1°
So, the bearing of A from B is 283.1°

Answer 2

283.1