Question

From the top of a light-house 60 m high with its base at the sea-level, the angle of depression of a boat is \(15\degree\). The distance of the boat from the foot of the light-house is
 

• A. \((\frac{\sqrt3-1}{\sqrt3-1})\times(60m)\)
• B. \((\frac{\sqrt3+1}{\sqrt3-1})\times(60m)\)
• C. \((\frac{\sqrt3+1}{\sqrt3-1})\times(30m)\)
• D. \((\frac{\sqrt3-1}{\sqrt3+1})\times(30m)\)

Answer 1

The Correct Option is B

The correct option is (B):
Given AB = \(60m\) \(\angle ACB=15\degree\) To find BC: 
Using identity \(tan(A-B)=\frac{tanA-tanB}{1-tanAtanB}\)

\(tan15\degree= (tan60\degree-tan45\degree)\)

\(\frac{(tan60\degree-tan45\degree)}{1-tan60\degree tan45\degree}\)

\(\frac{\sqrt3-1}{1+\sqrt3}\)……..(i)

\(In \bigtriangleup ABC\)
\(tan(15\degree)=\frac{AB}{BC}\)

\(\frac{\sqrt3-1}{\sqrt3+1}=\frac{60}{BC}\)

BC=\(\frac{\sqrt3+1}{\sqrt3-1}60\)