Let the height of the tower = h m, and distance of point P from the foot of the tower = d m.
Let AB = tower (height = h), BC = pole (height = 6 m), P = point on ground.
From the angle of depression of P from the top of the tower (45°):
$$\tan 45° = \frac{h}{d} \implies 1 = \frac{h}{d} \implies d = h \quad \text{...(1)}$$
From the angle of elevation of the top of the pole from P (60°):
$$\tan 60° = \frac{h + 6}{d} \implies \sqrt{3} = \frac{h + 6}{d} \quad \text{...(2)}$$
Substituting (1) into (2):
$$\sqrt{3} = \frac{h + 6}{h}$$
$$\sqrt{3}\, h = h + 6$$
$$h(\sqrt{3} - 1) = 6$$
$$h = \frac{6}{\sqrt{3} - 1} = \frac{6(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} = \frac{6(\sqrt{3} + 1)}{2} = 3(\sqrt{3} + 1)$$
$$h = 3(1.73 + 1) = 3 \times 2.73 = \mathbf{8.19 \text{ m}}$$
Distance of P from foot of tower:
$$d = h = \mathbf{8.19 \text{ m}}$$
∴ Height of tower = 8.19 m and distance of point P from the foot of the tower = 8.19 m.
Source: Chapter 9, Section 9.1 – Heights and Distances
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