question_answer

From the top of a tower, the angle of depression of a point on the ground is \[\text{6}{{\text{0}}^{\text{o}}}\text{.}\] 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)         \[\frac{\sqrt{3}}{2}m\]

Correct Answer: C

Solution :

Let \[h\] be the height of the tower. In \[\Delta \Alpha \Beta C,\] \[\tan {{60}^{o}}=\frac{h}{\left( \frac{1}{\sqrt{3}+1} \right)}\] \[\Rightarrow \]               \[\frac{\sqrt{3}}{\sqrt{3}+1}=\frac{h}{1}\] \[\Rightarrow \]               \[h=\frac{\sqrt{3}(\sqrt{3}-1)}{3-1}\] \[\Rightarrow \]               \[=\frac{3-\sqrt{3}}{2}\]