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 \[\Delta ABC\] |
\[\tan 60{}^\circ =\frac{\text{Perpendicular}}{\text{Base}}\] |
\[\sqrt{3}=\frac{BC}{AB}\]\[\Rightarrow \]\[\sqrt{3}=\frac{BC}{20}\]\[\Rightarrow \]\[BC=20\sqrt{3}\,\,m\] |
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