Question

the angle of elevation of the top q of a vertical tower pq from a point x on the ground is 60. at a pt. y, 40m vertically above x, the angle of elevation is 45. find the height of tower pq and the distance xq.

Answer 1

Let the point on the tower at height 40m be R so YR is parallel and equal to XP.
Now, $tan 60=PQ/XP$
$PQ/XP= \sqrt{3}$
Also,
$tan 45=QR/YR=1$

Dividing these equations and putting $YR=XP$ and $QR=PQ-40$

$PQ/(PQ-40)= \sqrt{3}$

From this you can rearrange to find,
$PQ=40 \sqrt{3} /( \sqrt{3}-1)$

Now you know PQ so, 
$sin60=PQ/XQ= \sqrt{3}/2$

Put the values and you get 
$XQ= \frac{80}{ \sqrt{3}-1 }$