# A statue, $1.6\ m$ tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $60^o$ and from the same point the angle of elevation of the top of the pedestal is $45^o$. Find the height of the pedestal.

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Given:

A statue $1.6 \mathrm{~m}$ tall stands on the top of pedestal.

From a point on the ground, the angle of elevation of the top of the statue is $60^{\circ}$ and from the same point the angle of elevation of the top of the pedestal is $45^{\circ}$.

To do:

We have to find the height of the pedestal.

Solution:

Let $AB$ be the height of the pedestal and $BC$ be the height of the statue.

Point $D$ be the point of observation.

From the figure,

$\mathrm{BC}=1.6 \mathrm{~m}, \angle \mathrm{CDA}=60^{\circ}, \angle \mathrm{BDA}=45^{\circ}$

Let the height of the pedestal be $\mathrm{AB}=h \mathrm{~m}$ and the distance of the pedestal from the point $D$ be $\mathrm{DA}=x \mathrm{~m}$.

This implies,

$\mathrm{AC}=1.6+h \mathrm{~m}$

We know that,

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

$=\frac{\text { BA }}{DA}$

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

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

$\Rightarrow x(1)=h \mathrm{~m}$

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

Similarly,

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

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

$\Rightarrow \tan 60^{\circ}=\frac{1.6+h}{x}$

$\Rightarrow \sqrt3=\frac{1.6+x}{x}$              [From (i)]

$\Rightarrow x\sqrt3=1.6+x \mathrm{~m}$

$\Rightarrow x(\sqrt3-1)=1.6 \mathrm{~m}$

$\Rightarrow x=\frac{1.6}{\sqrt3-1)} \mathrm{~m}$

$\Rightarrow x=\frac{1.6\times(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)} \mathrm{~m}$

$\Rightarrow x=\frac{1.6(\sqrt3+1)}{3-1} \mathrm{~m}$

$\Rightarrow x=\frac{1.6(\sqrt3+1)}{2} \mathrm{~m}$

$\Rightarrow x=\frac{4(\sqrt3+1)}{5} \mathrm{~m}$

Therefore, the height of the pedestal is $\frac{4(\sqrt3+1)}{5} \mathrm{~m}$.

Updated on 10-Oct-2022 13:23:31