So I’m new to this spherical trig and great circle route stuff. The question says:
There is a ship departing from point A (84° W, 15°S) on a great circle route. If its initial bearing is 70°, then what is its bearing when it crosses the 34° W longitude line?
From what I understand, you need two points to find a final bearing, but this only gives one point and I don’t know how to find the latitude when the ship is on the 34° longitude.
The answer to the question is apparently 66° if that helps anyone with reverse checking their math or anything.
I think I've got it...
Start with the isosceles triangle intersecting the North Pole, the initial point (15,84) and the point (15,34). The top of that triangle (call it $A'$) is
$$A' = 84 - 34 = 50^\circ.$$
The side $c$ is easily computed using the initial latitude
$$c = 90 - (-15) = 105^\circ.$$
We can then use the identity $$\cos a = \cos b' \cos c + \sin b' \sin c \cos A'$$ where I've labeled $b'$ as the equivalent of $c$ but on the opposite side. So now that we have $A', c$, and $a$, we can compute $C=B'$ via
$$\frac{\sin C}{\sin c} = \frac{\sin A'}{\sin a}.$$
We can then compute $B$ using the initial heading $$B = C - 70.$$
Finally, we can compute $A$ (no prime) which is the initial heading via
$$\cos A = -\cos B \cos C + \sin B \sin C \cos a.$$
Doing all that, I end up with $a = 66.04^\circ .$
I think that's correct. Sorry for the terrible picture.