# From the top of a lighthouse AB, the angles of depression of two stations C and D on opposite sides at a distance d

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From the top of a lighthouse AB, the angles of depression of two stations C and D on opposite sides at a distance d apart are α and β. The height of the lighthouse is

(a) $\frac{d}{\text{cot}\,\alpha\,\text{cot}\,\beta}$

(b) $\frac{d}{\text{cot}\,\alpha\,-\text{cot}\,\beta}$

(c) $\frac{d\,\text{tan}\,\alpha\,\text{tan}\,\beta}{\text{tan}\,\alpha\,+\text{tan}\,\beta}$

(d) $\frac{d\,\text{cot}\,\alpha\,\text{cot}\,\beta}{\text{cot}\,\alpha\,+\text{cot}\,\beta}$

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(c) $\frac{d\,\text{tan}\,\alpha\,\text{tan}\,\beta}{\text{tan}\,\alpha\,+\text{tan}\,\beta}$

Let the height of the light house AB = h metres.

Given ∠ACB = α, ∠ ADB = β, CD = d.

tan α = $\frac{AB}{BC}$

⇒ tan α = $\frac{h}{BC}$ ⇒ $\frac{h}{\text{tan}\,\alpha}$               .....(i)

In rt. Δ ABD,

tan β = $\frac{AB}{BD}$ ⇒ $\frac{h}{BD}$ = tan β ⇒ BD = $\frac{h}{\text{tan}\,\beta}$      ......(ii)

∴ CD = BC + BD = $\frac{h}{\text{tan}\,\alpha}$ + $\frac{h}{\text{tan}\,\beta}$           (Adding eqn. (i) and (ii)

⇒ d = $\frac{h(\text{tan}\,\beta+\text{tan}\,\alpha)}{\text{tan}\,\alpha+\text{tan}\,\beta}$ = h = $\frac{d\,\text{tan}\,\alpha\,\text{tan}\,\beta}{\text{tan}\,\alpha\,+\text{tan}\,\beta}$