question_answer

The shadow of a tower is found to be 60 metre shorter when the sun?s altitude changes from \[{{30}^{o}}\]to\[{{60}^{o}}\]. The height of the tower from the ground is approximately equal to  [Kerala (Engg.) 2005]

A) 62m

B) 301m

C) 101m

D) 75m

E) 52m

Correct Answer: E

Solution :

\[\tan {{30}^{o}}=\frac{h}{x+60}\], \[\frac{1}{\sqrt{3}}=\frac{h}{x+60}\] \[x+60=\sqrt{3}h\], \[x=\sqrt{3}h-60\] \[\tan {{60}^{o}}=\frac{h}{x}\], \[x=\frac{h}{\sqrt{3}}\] Þ \[\sqrt{3}h-60=\frac{h}{\sqrt{3}}\]Þ \[3h-60\sqrt{3}=h\] Þ \[h=\frac{60\sqrt{3}}{2}=30\sqrt{3}\] \[=51.96\approx 52m\].