The angles of elevation of the top of a rock from the top and foot of a $ 100 \mathrm{~m} $ high tower are respectively $ 30^{\circ} $ and $ 45^{\circ} $. Find the height of the rock.

Given:

The angles of elevation of the top of a rock from the top and foot of a \( 100 \mathrm{~m} \) high tower are respectively \( 30^{\circ} \) and \( 45^{\circ} \). 

To do:

We have to find the height of the rock.

Solution:  

Let $AB$ be the high tower and $CD$ be the height of the rock.

Let $B, A$ be the top and bottom of the tower.

From the figure,

$\mathrm{AB}=\mathrm{CE}=100 \mathrm{~m}, \angle \mathrm{DBE}=30^{\circ}, \angle \mathrm{DAC}=45^{\circ}$

Let the height of the rock be $\mathrm{CD}=h \mathrm{~m}$ and the distance between the tower and the rock be $\mathrm{AC}=\mathrm{BE}=x \mathrm{~m}$.

This implies,

$\mathrm{DE}=h-100 \mathrm{~m}$

We know that,

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

$=\frac{\text { DC }}{AC}$

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

$\Rightarrow 1(x)=h$

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

Similarly,

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

$=\frac{\text { DE }}{BE}$

$\Rightarrow \tan 30^{\circ}=\frac{h-100}{x}$

$\Rightarrow \frac{1}{\sqrt3}=\frac{h-100}{h}$                              [From (i)]

$\Rightarrow h=(h-100)\sqrt3 \mathrm{~m}$

$\Rightarrow 100\sqrt3=h(\sqrt3-1) \mathrm{~m}$

$\Rightarrow 100(1.73)=h(1.73-1) \mathrm{~m}$

$\Rightarrow h=\frac{173}{0.73} \mathrm{~m}$

$\Rightarrow h=236.5 \mathrm{~m}$

Therefore, the height of the rock is $236.5 \mathrm{~m}$.   

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Updated on: 10-Oct-2022

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