Two explorers leave a camp at the same time. one walks at 5km/h on a bearing 039°. the other walks at 7.5km/h on a bearing 265°. after two hours how far are they and what is the bearing of the second from the first.?
Use the law of cosines to find the distance z:
z^2 = 10^2 + 15^2 - 2*10*15 cos134°
To find the bearing, note that the two explorers are at positions
P=(10 sin39°,10cos39°) and Q=(-15cos5°,-15sin5°)
Thus, the bearing of #2 from #1 is 270-θ such that
tanθ = (Py-Qy)/(Px-Qx)