The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is $$45^\circ$$. Let R be a point on AQ and from a point B, vertically above R, the angle of elevation of P is $$60^\circ$$. If $$\angle BAQ = 30^\circ$$, $$AB = d$$ and the area of the trapezium PQRB is $$\alpha$$, then the ordered pair $$(d, \alpha)$$ is

Let Q be at the origin with the horizontal ground along the x-axis. The tower PQ has height 10, so P = (0, 10) and Q = (0, 0).

Point A is on the ground with the angle of elevation of P being 45°:

$$\tan 45° = \frac{10}{AQ} \implies AQ = 10$$

So A = (10, 0).

Point R lies on segment AQ, so R = (x, 0) for some 0 ≤ x ≤ 10, and B is vertically above R, so B = (x, h) for some height h. Given ∠BAQ = 30° and AR = 10 − x,

$$\tan 30° = \frac{h}{10 - x} \implies h = \frac{10 - x}{\sqrt{3}}$$

From B = (x, h), the angle of elevation of P = (0, 10) is 60°:

$$\tan 60° = \frac{10 - h}{x}$$ $$\sqrt{3} = \frac{10 - \frac{10-x}{\sqrt{3}}}{x} = \frac{10\sqrt{3} - 10 + x}{\sqrt{3} \cdot x}$$ $$3x = 10\sqrt{3} - 10 + x \implies 2x = 10(\sqrt{3} - 1) \implies x = 5(\sqrt{3} - 1)$$

Hence

$$AR = 10 - x = 10 - 5(\sqrt{3}-1) = 15 - 5\sqrt{3} = 5(3-\sqrt{3})$$ $$h = BR = \frac{AR}{\sqrt{3}} = \frac{5(3-\sqrt{3})}{\sqrt{3}} = 5(\sqrt{3}-1)$$

The distance AB is

$$d = AB = \sqrt{AR^2 + BR^2} = \sqrt{AR^2\left(1 + \frac{1}{3}\right)} = AR \cdot \frac{2}{\sqrt{3}} = \frac{2 \times 5(3-\sqrt{3})}{\sqrt{3}} = \frac{10(3-\sqrt{3})}{\sqrt{3}} = 10(\sqrt{3} - 1)$$

PQ and BR are vertical (parallel), with PQ = 10 and BR = 5($$\sqrt{3}$$-1). The horizontal distance between them is QR = x = 5($$\sqrt{3}$$-1). Therefore, the area of the trapezoid is

$$\alpha = \frac{1}{2}(PQ + BR) \times QR = \frac{1}{2}(10 + 5(\sqrt{3}-1)) \times 5(\sqrt{3}-1) = \frac{1}{2} \times 5(1+\sqrt{3}) \times 5(\sqrt{3}-1) = \frac{25}{2}(\sqrt{3}+1)(\sqrt{3}-1) = \frac{25}{2}(3-1) = 25$$

Therefore,

$$(d, \alpha) = \boxed{(10(\sqrt{3}-1),\ 25)}.$$

The answer is Option A.