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 \[x=15+\frac{x}{\sqrt{3}}\Rightarrow x=\frac{15\sqrt{3}\left( \sqrt{3}+1 \right)}{2}\Rightarrow x=35.49\,m\]You need to login to perform this action.
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