Question

If you're good at trig pls help meeeeeee

Answer 1

*flashbacks to honors trig in 9th grade*

...Anyway. We're given that the bearing from A to C is $330^\circ,$ so that means that the counterclockwise angle from the north to C is $30^\circ.$ Therefore, $\angle BAC=50^\circ+30^\circ=80^\circ.$

Now, we apply the law of cosines:

$BC^2&=AC^2+BC^2-2(AC)(BC)\cos\angle BAC$

$BC^2&=65^2+135^2-2(65)(135)\cos 80^\circ=22450-17550\cos 80^\circ$

$\boxed{BC\approx 139\text{ km}}.$