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 $$QR = 15 \text{ m}$$. If from a point $$A$$ on the ground the angle of elevation of $$R$$ is $$60°$$ and the part $$PR$$ of the tower subtends an angle of $$15°$$ at $$A$$, then the height of the tower is

A tower $$ PQ $$ stands on the ground with base $$ Q $$. Point $$ R $$ divides the tower such that $$ QR = 15 $$ m. From point $$ A $$ on the ground, the angle of elevation of $$ R $$ is $$ 60° $$ and the part $$ PR $$ subtends an angle of $$ 15° $$ at $$ A $$.

Let the height of the tower be $$ h = PQ $$, so $$ PR = h - 15 $$.

Let $$ d = AQ $$ (horizontal distance from A to the base).

The angle of elevation of $$ R $$ from $$ A $$ is $$ 60° $$, so:

$$\tan 60° = \frac{QR}{d} = \frac{15}{d}$$

$$d = \frac{15}{\tan 60°} = \frac{15}{\sqrt{3}} = 5\sqrt{3}$$

The angle of elevation of $$ P $$ from $$ A $$ is $$ 60° + 15° = 75° $$.

$$\tan 75° = \frac{h}{d} = \frac{h}{5\sqrt{3}}$$

$$\tan 75° = \tan(45° + 30°) = \frac{\tan 45° + \tan 30°}{1 - \tan 45° \cdot \tan 30°} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$

Rationalizing:

$$\tan 75° = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$

$$h = 5\sqrt{3} \cdot \tan 75° = 5\sqrt{3}(2 + \sqrt{3}) = 10\sqrt{3} + 5 \cdot 3 = 10\sqrt{3} + 15$$

$$h = 5(2\sqrt{3} + 3) \text{ m}$$

The height of the tower is $$ 5(2\sqrt{3} + 3) $$ m, which corresponds to Option A.