**(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 = **\(\frac{d\,\text{tan}\,\alpha\,\text{tan}\,\beta}{\text{tan}\,\alpha\,+\text{tan}\,\beta}\)**