Let CE = height of lighthouse, CD = horizontal distance, DE = h (height above D), so CE = CD + DE...
Setup: Let BC = h (height of lighthouse above observation deck level), so BE = h. Let CD = horizontal distance from Amrita to lighthouse base.
(i) From point B (observation deck), angle of elevation = 45°:
$$\tan 45° = \frac{BE}{BD} \Rightarrow 1 = \frac{h}{BD} \Rightarrow BD = h$$
$$\therefore CD = BD = h$$
(ii) From point B, angle of elevation = 45°:
$$\tan 45° = \frac{BE}{BD} = 1 \Rightarrow BD = BE = BC$$
$$\therefore BD = BC$$
(iii) From point A (ground level), angle of elevation = 60°. AB = 40 m, so AE = h + 40:
$$\tan 60° = \frac{h + 40}{CD} \Rightarrow \sqrt{3} = \frac{h + 40}{h}$$
$$\sqrt{3}\,h = h + 40 \Rightarrow h(\sqrt{3} - 1) = 40$$
$$h = \frac{40}{\sqrt{3}-1} = \frac{40(\sqrt{3}+1)}{2} = 20(\sqrt{3}+1)$$
$$h = 20(1.73 + 1) = 20 \times 2.73 = 54.6 \text{ m}$$
$$\therefore CE = h + 40 = 54.6 + 40 = \textbf{94.6 m}$$
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