Question

A tower $PQ$ stands on a horizontal ground with base $Q$ on the ground. The point $R$ divides the tower in two parts such that $QR =15 \,m$. If from a point $A$ on the ground the angle of elevation of $R$ is $60^{\circ}$ and the part $PR$ of the tower subtends an angle of $15^{\circ}$ at $A$, then the height of the tower is :

• A. $5(2 \sqrt{3}+3) m$
• B. $5(\sqrt{3}+3) m$
• C. $10(\sqrt{3}+1) m$
• D. $10(2 \sqrt{3}+1) m$

Answer 1

Answer: (A)

$\frac{15}{ AQ }=\tan 60^{\circ}$
$\frac{15+x}{ AQ }=\tan 75^{\circ}$
$\frac{(1)}{(2)} \Rightarrow x=10 \sqrt{3}$So, $P Q=5(2 \sqrt{3}+3) m$