From the top $$A$$ of a vertical wall $$AB$$ of height 30 m, 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. If $$C$$ is a point on $$AB$$ such that $$CB = PQ$$, then the area (in m$$^2$$) of the quadrilateral $$BCPQ$$ is equal to

Trigonometry:

1. Distance ($$d$$): From $$\triangle\ $$ABQ, $$\tan 60^\circ = \frac{30}{d} \implies d = \frac{30}{\sqrt{3}} = 10\sqrt{3}$$.
2. Tower Height ($$h$$): From top $$A$$, $$\tan 15^\circ = \frac{30 - h}{d}$$.

   Using $$\tan 15^\circ = 2 - \sqrt{3}$$:

   $$30 - h = 10\sqrt{3}(2 - \sqrt{3}) = 20\sqrt{3} - 30 \implies \mathbf{h = 60 - 20\sqrt{3}}$$.

3. Area $$BCPQ$$: Since $$CB = PQ = h$$, it is a rectangle of sides $$d$$ and $$h$$

   $$\text{Area} = d \times h = 10\sqrt{3}(60 - 20\sqrt{3}) = 600\sqrt{3} - 600$$.

   Result: $$600(\sqrt{3} - 1) \text{ m}^2$$.