The angle of elevation of the top of a vertical tower $ P Q $ from a point $ X $ on the ground is $ 60^{\circ} $. At a point $ Y, 40 $ m vertically above $ X $, the angle of elevation of the top is $ 45^{\circ} $. Calculate the height of the tower.

Given:

The angle of elevation of the top of a vertical tower \( P Q \) from a point \( X \) on the ground is \( 60^{\circ} \).

At a point \( Y, 40 \) m vertically above \( X \), the angle of elevation of the top is \( 45^{\circ} \).

To do:

We have to calculate the height of the tower.

Solution:

From the figure,

$\mathrm{XY}=40 \mathrm{~m}, \angle \mathrm{QXP}=60^{\circ}, \angle \mathrm{QYR}=45^{\circ}$

Let the height of the tower be $\mathrm{PQ}=h \mathrm{~m}$ and the distance between the tower and point $X$ be $\mathrm{XP}=x \mathrm{~m}$.

This implies,

$\mathrm{YR}=x \mathrm{~m}$

$\mathrm{PR}=40 \mathrm{~m}$

$\mathrm{QR}=h-40 \mathrm{~m}$

We know that,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { QR }}{YR}$

$\Rightarrow \tan 45^{\circ}=\frac{h-40}{x}$

$\Rightarrow 1=\frac{h-40}{x}$

$\Rightarrow x=h-40 \mathrm{~m}$...........(i)

Similarly,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { PQ }}{XP}$

$\Rightarrow \tan 60^{\circ}=\frac{h}{x}$

$\Rightarrow \sqrt3=\frac{h}{h-40}$ [From (i)]

$\Rightarrow (h-40)\sqrt3=h \mathrm{~m}$

$\Rightarrow h(\sqrt3-1)=40\sqrt3 \mathrm{~m}$

$\Rightarrow h=\frac{40(1.732)}{1.732-1} \mathrm{~m}$

$\Rightarrow h=\frac{69.28}{0.732} \mathrm{~m}$

$\Rightarrow h=94.64 \mathrm{~m}$

Therefore, the height of the tower is $94.64 \mathrm{~m}$ .

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