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\[23.66\text{ }m\].You need to login to perform this action.
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