question_answer

The angle of elevation of the top of a tower at a point on level ground is\[45{}^\circ \].When moved 20 m towards the tower, the angle of elevation becomes\[60{}^\circ \]. What is the height of the tower?

A)  \[10(\sqrt{3}-1)m\]                    

B)  \[10(\sqrt{3}+1)m\]

C)  \[10(3-\sqrt{3})m\]                    

D)  \[10(3+\sqrt{3})m\]

Correct Answer: D

Solution :

Let AB is the height of tower. TRICK: AB (Height of Tower) \[=\frac{CD}{\cot \,45{}^\circ -\cot 60{}^\circ }=\frac{20}{1-\frac{1}{\sqrt{3}}}\] \[=\frac{20\sqrt{3}}{\sqrt{3}-1}=10\sqrt{3}\,\,(\sqrt{3}+1)\] \[=\,\,\mathbf{10}\,\mathbf{(3+}\sqrt{\mathbf{3}}\mathbf{)}\,\mathbf{m}\]