question_answer

The shadow of a tower standing on a level ground is found to be 60 m longer when the sun's altitude is\[{{30}^{o}}\]than when it is\[{{45}^{o}}\]. The height of the tower is [EAMCET 2001]

A) 60 m

B) 30 m

C) \[60\sqrt{3}m\]

D) \[30(\sqrt{3}+1)m\]

Correct Answer: D

Solution :

\[\because \,AB=AM-BM\Rightarrow \frac{AB}{h}=\frac{AM}{h}-\frac{BM}{h}\] \[\frac{AB}{h}=\cot {{30}^{o}}-\cot {{45}^{o}}\Rightarrow h=\frac{60}{\sqrt{3}-1}\,=\frac{60(\sqrt{3}+1)}{3-1}\,\]  Þ \[h=30(\sqrt{3}+1)\,m\].