Find distance AB given bearings and angles of elevation I'm having issues with the following problem, which I think are due to how I am modelling the problem. Seems like I have to think in 2D and 3D which is where I get confused. Could someone show me how they would approach modelling this?
Question:
From a point A the bearing of the base of a tower 60m high is 53° and the angle of elevation of the top of the tower is 16°. From a point B the bearing is 300° and the angle of elevation is 20°. Find the distance AB.
 A: What does the elevation tell you? Let's say the base of the tower is $C$, the top is $D$. From point $A$, assumed to be at the same height as the base of the tower, the tangent of the elevation angle gives you the ratio $\frac{DC}{AC}$. You know $DC=60$, so you can find $AC$. Similarly, find the length of the $BC$ line. Now we draw the $ABC$ triangle. Let's choose the tower at origin. We know that from $A$ the bearing is $53^\circ$. That means that from $C$ you see $A$ at $53^\circ\pm 180^\circ$. Using a similar reasoning, the $C$ point is at the $300^\circ\pm 180^\circ$. Now you know $AC$, $BC$, and the angle $BCA$, so just apply the generalized Pythagoras's theorem to find $AB$ 
A: From the given information we know that
$$\frac{60}{|AT|}=\tan16^\circ  $$
$$\frac{60}{|BT|}=\tan20^\circ  $$
The horizontal line through $B$ will intersect line $BT$ at an angle $60^\circ$ so $\beta=120^\circ$. So $\gamma=60^\circ+53^\circ=113^\circ$.
Now, $\triangle ABC$ is of type SAS and side $|AB|$ can be found by the Law of Cosines.

