(a) (√3 + 1)h
Let PQ be the light house whose height = h metre. Let A and B be the position of the ships on opposite sides of the lighthouse such that angle of depression for A and B are 30° and 45° respectively.
Let AQ = x metre, QB = y metres. ∠PAQ = 30°, ∠PBQ = 45°.
Required distance between the ships = AB = AQ + QB = x + y
In rt. Δ PAQ,
tan 30° = \(\frac{h}{x}\) ⇒ \(\frac{1}{\sqrt3}\) = \(\frac{h}{x}\) ⇒ \(x\) = h√3
In rt. Δ PBQ,
tan 45° = \(\frac{h}{y}\) ⇒ 1 = \(\frac{h}{y}\) ⇒ y = h
∴ x + y = h√3 + h = h(√3 + 1)m.