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