Question

Two ships are sailing in the sea on either side of the light-house, the angles of depression of two ships as observed from the top of the light-house are \( 60^{\circ} \) and \( 45^{\circ} \) respectively. If \( t \) ir distance between the ships is \( 200\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right) m \), find the height of the light-house.

Answer 1

We have CD = \(\frac {200 (\sqrt3+1)}{\sqrt3}\)

Let Ac = x m

∴ In right △CAB,

tan 60° = \(\frac {AB}{AC}\)

= AC = \(\frac {AB}{tan 60°} = \frac {AB}{\sqrt3}\).......(i)

In right △ DAB

tan 45° = \(\frac {AB}{AD}\)

= AB = AD + tan 45° = AD

= AC + CD

= \(\frac {AB}{\sqrt3} + \frac {200(\sqrt3+1)}{\sqrt3}\) (∵ CD = \( \frac {200(\sqrt3+1)}{\sqrt3}\))

= AB - \(\frac {AB}{\sqrt3} + \frac {200(\sqrt3+1)}{\sqrt3}\)

= \((\frac {\sqrt3-1}{\sqrt3})\) AB = \( \frac {200(\sqrt3+1)}{\sqrt3}\)

= Ab = \( \frac {200(\sqrt3+1)}{\sqrt3-1}\)

= \( \frac {200(\sqrt3+1)^2}{\sqrt3-1}\)

= 100 (3+1 + 2√3)

= 200 (2+√3) m.

Hence, height of the light house is 200 (2+√3)m