Question

The angle of elevation of the top $Q$ of a vertical tower $PQ$ from a point $X$ on the ground is ${60}^o$. At a point $Y$, $40m$ vertically above $X$, the angle of elevation is ${45}^o$ Find the height of the tower $PQ$ and the distance$XP$

Answer 1

In $△RYQ,x\tan 45°=QR$ and $PR=40$

$x=QR,PR=40$

And $PQ=PR+QR\Rightarrow h=40+x$

Also, In $△P\times Q,x\tan 60°=h\Rightarrow h=x\sqrt{3}$

So, $40+x=x\sqrt{3}\Rightarrow x=\frac{40}{\sqrt{3}-1}$ & $h=\frac{40\sqrt{3}}{\sqrt{3}-1}$