question_answer

The shadow of a tower, when the angle of elevation of the sun is \[{{45}^{o}},\] is found to be 10 metres longer than when the angle of elevation is \[{{60}^{o}}\]. Find the height of the tower. [Given : \[\sqrt{3}=1.732\]]

A)  \[22.66\text{ }m\]                  

B)  \[23\text{ }m\]           

C)  \[23.66\text{ }m\]     

D)  \[22.16\text{ }m\]

Correct Answer: C

Solution :

Let AB be the tower and let AC and AD be its shadows when the  angles of elevation of  the sun are \[{{60}^{o}}\] and \[{{45}^{o}}\]respectively.                   Let AB = h metres and AC = x metres.            In right \[\Delta CAB,\] \[\frac{AB}{AC}=\tan {{60}^{o}}\Rightarrow \frac{h}{x}=\sqrt{3}\Rightarrow x=\frac{h}{\sqrt{3}}\]    ?(i)           In right \[\Delta DAB,\frac{AB}{AD}=\tan {{45}^{o}}\Rightarrow \frac{h}{10+x}=1\] \[\Rightarrow \] \[10+x=h\,\Rightarrow \,x=(h-10)\]         ...(ii) From (i) and (ii), we get   \[\frac{h}{\sqrt{3}}=h-10\Rightarrow (\sqrt{3}-1)h=10\sqrt{3}=23.66\]                      Hence, the height of the tower is\[23.66\text{ }m\].