Quantitative Aptitude Ques 1372
Question: 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
Options:
A) $ 20m $
B) $ 20\sqrt{3}m $
C) $ 30m $
D) $ 30\sqrt{3}m $
Show Answer
Answer:
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 $ $ [since,L\times AB=90{}^\circ ] $
$ \therefore $ In $ \Delta ABC $
$ \tan 60{}^\circ =\frac{Perpendicular}{Base} $
$ \sqrt{3}=\frac{BC}{AB} $
$ \Rightarrow $ $ \sqrt{3}=\frac{BC}{20} $
$ \Rightarrow $ $ BC=20\sqrt{3}m $
BETA
