question_answer

From the top of a tower, the angle of depression of a point on the ground is\[60{}^\circ \]. If the distance of this point from the tower is\[\frac{1}{\sqrt{3}+1}m,\]then the height of the tower is

A)  \[\frac{4\sqrt{3}}{2}m\]                              

B)  \[\frac{\sqrt{3}+3}{2}m\]

C)  \[\frac{3-\sqrt{3}}{2}m\]            

D)         \[\frac{\sqrt{3}}{2}m\]

E)  \[\sqrt{3}+1m\]

Correct Answer: C

Solution :

Let h be the height of the tower. In \[\Delta ABC,\tan 60{}^\circ =\frac{h}{1/\sqrt{(\sqrt{3}+1)}}\] \[\Rightarrow \]               \[\frac{\sqrt{3}}{\sqrt{3}+1}=\frac{h}{1}\] \[\Rightarrow \]               \[h=\frac{\sqrt{3}(\sqrt{3}-1)}{3-1}\]                                 \[=\frac{3-\sqrt{3}}{2}m\]