Given, $A B$ be a vertical wall of height $30 \mathrm{~m}$ and $P Q$ be a vertical tower. Such that $\angle B Q A=\angle T A Q=60^{\circ}$ (Alternate angles) and $\angle C P A=\angle T A P=15^{\circ}$ Now, $\tan 60^{\circ}=\frac{A B}{B Q}=\frac{30}{B Q}$ $\begin{array}{ll} &\Rightarrow \sqrt{3}=\frac{30}{B Q} \\\\ &\Rightarrow B Q=\frac{30}{\sqrt{3}}=10 \sqrt{3} \end{array}$ $\begin{aligned} & \text { and } \tan 15^{\circ}=\frac{A C}{P C} \\\\ & \Rightarrow 2-\sqrt{3}=\frac{A C}{10 \sqrt{3}} (\because P C=B Q) \\\\ & \Rightarrow A C=10 \sqrt{3}(2-\sqrt{3})=20 \sqrt{3}-30 \\\\ & \therefore B C=A B-A C=30-(20 \sqrt{3}-30)=60-20 \sqrt{3} \end{aligned}$ $\therefore$ Area of quadrilateral $B C P Q$ $\begin{aligned} & =B Q \times B C=10 \sqrt{3} \times(60-20 \sqrt{3}) \\\\ & =10(60 \sqrt{3}-60) \\\\ & =600(\sqrt{3}-1) \mathrm{m}^2 \end{aligned}$