Three Dimensional Problems: Angles of Elevation (question below)

Question

The question goes like this:

The angle of elevation of the top of a tower is 38 degrees from point A due south of it. The angle of elevation of the top of the tower from another point B, due east of the tower is 29 degrees. Find the height of the tower if the distance AB is 50 m.

I tried to draw a diagram for this question (as seen in the image attached) and in order to get the answer, I have come to understand that POA and BOP are 90 degrees instead of OPA and OPB since I thought they look like right angles. However, if I consider OA or OB as the hypotenuse, I get the incorrect answer. So, I wanted to know how POA and BOP are right angles, along with whether there is an alternative way to find the solution for the problem (with diagrams if possible).

Answer 1

$PA \tan 38 = h$

$PB \tan 29 = h$

$PA^2+PB^2 = 2500$

$PB = \sqrt{2500 - PA^2}$

$\sqrt{2500 - PA^2} \tan 29 = PA \tan 38$

$\tan^2 29 (2500 - PA^2) = PA^2 \tan^2 38$

$\frac{2500}{PA^2} - 1 = \frac{\tan^2 38}{\tan^2 29}$

$\frac{2500}{PA^2} = 2.986624$

$PA = \sqrt{\frac{2500}{2.986624}} = 28.9321$

$h = 28.9321\tan 38 = 22.6042$

Answer 2

Use some imagination in your head (of the tower thing), you can reduce this to: $$\tan(29)=\frac{h}{OE}$$ $$\tan(38)=\frac{h}{OS}$$ $$\sin(\alpha)=\frac{OS}{50}$$ $$\cos(\alpha)=\frac{OE}{50}$$ where $O$ is the base of the tower, $S$ and $E$ are the points at South and East, $h$ is the height of the tower, and $\alpha$ is $\angle OES$.

Then you get (after some calculation) $$\tan(38)=\frac{h}{50\sin\alpha}$$ and $$\tan(29)=\frac{h}{50\cos\alpha}$$

At that point you can now compute $h$ (I'll put it up once I've done it)