From the top A of a vertical wall AB of height 30m, the angles of...

Given that from the top $A$ of a vertical wall $AB$ of height $30m,$ the angles of depression of the top $P$ and bottom $Q$ of a vertical tower $PQ$ are ${15}^{∘}$ and ${60}^{∘}$ respectively,

$B$ and $Q$ are on the same horizontal level and $C$ is a point on $AB$ such that $CB = PQ .$

The required diagram will be

From the diagram,

$⇒ \tan 60 ° = \dfrac{AB}{BQ}$

$⇒ \dfrac{30}{x} = \sqrt{3};$

$⇒ x = 10\sqrt{3}m$

Also, $\tan 15 ° = \dfrac{AC}{CP}$

$⇒ \dfrac{30 - h}{x} = 2 - \sqrt{3}$

$⇒ 30 - h = 10\sqrt{3}\left(2 - \sqrt{3}\right)$

$⇒ 30 - h = 20\sqrt{3} - 30$

$⇒ h = 60 - 20\sqrt{3}$

Now area of the quadrilateral $BCPQ$ is $BC \times BQ$

$∴$ area$= hx$

$= ( 60 - 20\sqrt{3} ) 10\sqrt{3}$

$= 200\sqrt{3}\left(3 - \sqrt{3}\right)$

$= 600\left(\sqrt{3} - 1\right){m}^{2}$

Hence, this is the correct option.