Question
From the top of a light-house 60 metres high with its base at the sea-level, the angle of depression of a boat is 15°. The distance of the boat from the foot of the light house is
A.
$$\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)60\,{\text{metres}}$$
B.
$$\left( {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right)60\,{\text{metres}}$$
C.
$${\left( {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right)^2}{\text{metres}}$$
D.
none of these
Answer :
$$\left( {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right)60\,{\text{metres}}$$
Solution :
$$\eqalign{ & \tan {15^ \circ } = \frac{{60}}{x} \cr & \Rightarrow \,\,x = 60\cot {15^ \circ } \cr & = 60\left[ {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right] \cr} $$
$$\eqalign{ & \tan {15^ \circ } = \frac{{60}}{x} \cr & \Rightarrow \,\,x = 60\cot {15^ \circ } \cr & = 60\left[ {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right] \cr} $$