On a horizontal plane there is a vertical tower with a flag pole on the top of the tower. At a point 9 metres away from the foot of the tower the angle of elevation of the top and bottom of the flag pole are $ 60^{\circ} $ and $ 30^{\circ} $ respectively. Find the height of the tower and the flag pole mounted on it.
Given:
On a horizontal plane there is a vertical tower with a flag pole on the top of the tower.
At a point 9 metres away from the foot of the tower the angle of elevation of the top and bottom of the flag pole are \( 60^{\circ} \) and \( 30^{\circ} \) respectively.
To do:
We have to find the height of the tower and the flag pole mounted on it.
Solution:
Let $AB$ be the tower and $BC$ be the flag pole on top of it.
Let $D$ be the point 9 metres away from the foot of the tower.
From the figure,
$\mathrm{AD}=9 \mathrm{~m}, \angle \mathrm{CDA}=60^{\circ}, \angle \mathrm{BDA}=30^{\circ}$
Let the height of the tower be $\mathrm{AB}=h \mathrm{~m}$ and the height of the flag pole be $\mathrm{BC}=x \mathrm{~m}$.
This implies,
$\mathrm{AC}=x+h \mathrm{~m}$
We know that,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { AB }}{DA}$
$\Rightarrow \tan 30^{\circ}=\frac{h}{9}$
$\Rightarrow \frac{1}{\sqrt3}=\frac{h}{9}$
$\Rightarrow h=\frac{9}{\sqrt3} \mathrm{~m}$
$\Rightarrow h=3\sqrt3 \mathrm{~m}$...........(i)
Similarly,
$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$
$=\frac{\text { AC }}{DA}$
$\Rightarrow \tan 60^{\circ}=\frac{h+x}{9}$
$\Rightarrow \sqrt3=\frac{3\sqrt3+x}{9}$ [From (i)]
$\Rightarrow 9\sqrt3=3\sqrt3+x \mathrm{~m}$
$\Rightarrow x=(9-3)\sqrt3 \mathrm{~m}$
$\Rightarrow x=6\sqrt3 \mathrm{~m}$
Therefore, the height of the tower is $3\sqrt3 \mathrm{~m}$ and the height of the flag pole is $6\sqrt3 \mathrm{~m}$.
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