A) \[48\sqrt{3}m\]
B) \[16\sqrt{3}m\]
C) \[24\sqrt{3}m\]
D) \[\frac{16}{\sqrt{3}}m\]
Correct Answer: B
Solution :
Let \[AE\] be the tower and \[BD\] be the house. Then, \[BD=h\,\,m\] \[AE=H+h\] \[\angle ABC={{60}^{o}}\] \[\angle BED={{30}^{o}}\] and \[DE=BC=12\,\,m\] In\[\Delta BDE\], \[\tan {{30}^{o}}=\frac{BD}{DE}=\frac{h}{12}\] \[\Rightarrow \] \[h=12\tan {{30}^{o}}\] \[\Rightarrow \] \[h=\frac{12}{\sqrt{3}}\] In\[\Delta ACB\], \[\tan {{60}^{o}}=\frac{AC}{BC}=\frac{H}{12}\] \[\Rightarrow \] \[H=12\sqrt{3}\]. \[\therefore \]Height of tower\[=H+h\] \[=12\sqrt{3}+\frac{12}{\sqrt{3}}\] \[=16\sqrt{3}m\]You need to login to perform this action.
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