The angles of elevation of the top of a rock from the top and foot of a 100 m high tower are respectively 30° and 45°. Find the height of the rock.
Let be the height of Rock which is m. and makes an angle of elevations 45° and 30° respectively from the bottom and top of tower whose height is m.
Let $A E=h \mathrm{~m}, B C=x \mathrm{~m}$ and $C D=100 \cdot \angle A C B=45^{\circ}, \angle A D E=30^{\circ}$
We have to find the height of the rock
We have the corresponding figure as
So we use trigonometric ratios.
In,
$\tan 45^{\circ}=\frac{A B}{B C}$
$\Rightarrow \quad 1=\frac{100+h}{x}$
$\Rightarrow \quad x=100+h$
Again in $\triangle A D E$
$\tan 30^{\circ}=\frac{A E}{D E}$
$\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{x}$
$\Rightarrow 100+h=\sqrt{3} h$
$\Rightarrow \quad h=136.65$
$H=100+136.65$
$\Rightarrow \quad H=236.65$
Hence the height of rock is $236.65 \mathrm{~m}$.
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