The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°.

Solution:

We have,

$\mathrm{XY}=40 \mathrm{~m}, \angle \mathrm{PXQ}=60^{\circ}$ and $\angle \mathrm{MYQ}=45^{\circ}$

Let $\mathrm{PQ}=h$

Also, $\mathrm{MP}=\mathrm{XY}=40 \mathrm{~m}, \mathrm{MQ}=\mathrm{PQ}-\mathrm{MP}=h-40$

In $\Delta \mathrm{MYQ}$

$\tan 45^{\circ}=\frac{\mathrm{MQ}}{\mathrm{MY}}$

$\Rightarrow 1=\frac{h-40}{\mathrm{MY}}$

$\Rightarrow \mathrm{MY}=h-40$

$\Rightarrow \mathrm{PX}=\mathrm{MY}=h-40 \quad \ldots \ldots(\mathrm{i})$

Now, in $\triangle \mathrm{MXQ}$,

$\tan 60^{\circ}=\frac{\mathrm{PQ}}{\mathrm{PX}}$

$\Rightarrow \sqrt{3}=\frac{h}{h-40} \quad[$ From $(\mathrm{i})]$

$\Rightarrow h \sqrt{3}-40 \sqrt{3}=h$

$\Rightarrow h \sqrt{3}-h=40 \sqrt{3}$

 

$\Rightarrow h(\sqrt{3}-1)=40 \sqrt{3}$

$\Rightarrow h=\frac{40 \sqrt{3}}{(\sqrt{3}-1)}$

$\Rightarrow h=\frac{40 \sqrt{3}}{(\sqrt{3}-1)} \times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}$

$\Rightarrow h=\frac{40 \sqrt{3}(\sqrt{3}+1)}{(3-1)}$

$\Rightarrow h=\frac{40 \sqrt{3}(\sqrt{3}+1)}{2}$

$\Rightarrow h=20 \sqrt{3}(\sqrt{3}+1)$

$\Rightarrow h=60+20 \sqrt{3}$

$\Rightarrow h=60+20 \times 1.73$

$\Rightarrow h=60+34.6$

 

$\therefore h=94.6 \mathrm{~m}$

So, the height of the tower PQ is 94.6 m.