question_answer

From the top of a light house at a height 20 m above sea-level, the angle of depression of a ship is \[30{}^\circ .\]The distance of the ship from the foot of the light house is                                                           [SSC (CPO) 2014]

A) \[20\,\,m\]                     

B) \[20\sqrt{3}\,\,m\]

C) \[30\,\,m\]                     

D) \[30\sqrt{3}\,\,m\]

Correct Answer: B

Solution :

Let AB be the light house with top A and AB = 20 m.

Also, C be the ship.

Given, angle of depression of ship from \[A=30{}^\circ \]

\[\therefore \]\[\angle BAC=60{}^\circ \]              \[[\text{since},L\times AB=90{}^\circ ]\]

\[\therefore \] In \[\Delta ABC\]

\[\tan 60{}^\circ =\frac{\text{Perpendicular}}{\text{Base}}\]

\[\sqrt{3}=\frac{BC}{AB}\]\[\Rightarrow \]\[\sqrt{3}=\frac{BC}{20}\]\[\Rightarrow \]\[BC=20\sqrt{3}\,\,m\]