question_answer

The shadow of a tower, when the angle of elevation of the sun is \[{{45}^{o}}\] is found to be 15 metres longer than when it is \[{{60}^{o}}\]. Find the height of the tower.           

A)  \[35.92\text{ }m\]                  

B) \[36.59\text{ }m\]     

C)         \[35.49\text{ }m\]       

D)         \[37.49\text{ }m\]     

Correct Answer: C

Solution :

Let the length of the tower AB be x m. Angle of elevation at point \[C={{60}^{o}}\] Shadow of tower, \[(BC)=y\,m,\,CD=15\,m\] Now, in right                         \[\Delta ABC,\,\,\frac{AB}{AC}=\tan {{60}^{o}}\Rightarrow \frac{x}{y}=\sqrt{3}\] \[\Rightarrow \]            \[\sqrt{3}y=x\Rightarrow y=\frac{x}{\sqrt{3}}\]         ?..(1) In right  \[\Delta ABD,\,\,\frac{AB}{DB}=\tan {{45}^{o}}\] \[\Rightarrow \]  \[\frac{x}{15+y}=1\Rightarrow x=15+y\] Putting the value of y in (2), we get \[x=15+\frac{x}{\sqrt{3}}\Rightarrow x=\frac{15\sqrt{3}\left( \sqrt{3}+1 \right)}{2}\Rightarrow x=35.49\,m\]