Question on bearings This is the question:

Two points $P$ and $Q$ lying on the same level ground as $O$, the foot of the tower $OT$, $40$m high are on the bearing $284^o$ and $216^o$ respectively from O. The angle of elevation of the Top $T$ of the tower from $P$ and $Q$ are $36^o$ and $24^o$ respectively. Calculate correct to 2 decimal places the distance:
  (i) $|PO|$ (ii) $|QO|$ (iii) $|PQ|$ (iv) the bearing of $Q$ from $P$.

I attempted solving it and I got some answers (I do not know if they are wrong as there no answers in the textbook). I imagined the diagram as a 3D object, instead of a 2D object, with the tower sticking out of my book and lines joining $T$ to $P$ and $Q$ at the giving angles. I would like to know if this approach is good and if there’s another way of answering it.
This was how I solved mine: 
For $|PO|$ 
$$Tan 36 = \frac{40}{adj}$$
$$adj = \frac{40}{Tan 36} = 55.06m$$
For $|QO|$
$$Tan 24 = \frac{40}{adj} $$
$$adj = \frac{40}{Tan 24} = 89.84m$$
For $|PQ|$
Angle $POQ = 284^o-216^o = 68^o$,
Using cosine rule
$$o^2 =p^2+q^2-2pqCos O$$
$$o = \sqrt{89.84^2+55.06^2-2*89.84*55.06*Cos68}$$
$$o = 86.01m$$
For the bearing of $Q$ from $P$,
Using sign rule to get Angle $OPQ$
$$\frac{p}{sinP} = \frac{o}{sinO}$$
$$P = Sine^{-1}\left(\frac{pSinO}{o}\right)$$
$$P = 75.57^o$$
Counting from the north to the east ( since O is south east of P ) we have $90^o$
There’s an angle between $90^o$ and Angle $OPQ$ which is $$= 284^o-270^o = 14^o$$
So adding them together we get
$$90^o+75.57^o+14^o = 179.57^o$$
Here is a link to my diagram 
https://m.imgur.com/8cxzhAW
 A: Your approach seems reasonable.  As a check, I get the bearing from P to Q as $140.41^{\circ}$. If you need more help, show some work. I have a picture.
Well, starting from the same picture, we certainly did it differently.
I put the tower at $O=(0,0,0)$.  In the plane of $\angle TOP$ I used the law of sines to get length $OP=29.062$  Likewise, length $OQ=17.809$  Next I built a vector going from $O$ to $P$. $$\mathbf{p}=\left(\begin{array}{c}\overline{PO}cos(284^{\circ})\\\overline{PO}sin(284^{\circ})\\0\end{array}\right)$$ and similarly Q.  $$\mathbf{q}=\left(\begin{array}{c}\overline{QO}cos(216^{\circ})\\\overline{QO}sin(216^{\circ})\\0\end{array}\right)$$ The distance from Q to P is then just $\Vert Q-P \Vert$ and the direction vector is given by $$\mathbf{q-p}=\left(\begin{array}{c}-21.439\\17.73\\0\end{array}\right)$$ I used the arg() function on that last vector to get the bearing from P to Q.  I don't know if you are familiar with Geogebra, but if so, here is a link. https://www.geogebra.org/m/vukgrw2t 
