A) \[10\,m\]
B) \[\frac{10}{\sqrt{3}}m\]
C) \[10\sqrt{3}\,m\]
D) None of these
Correct Answer: C
Solution :
Let height of the tower AB is h m. In \[\Delta ABD.tan{{60}^{o}}=\frac{h}{x}\Rightarrow \sqrt{3}=\frac{h}{x}\] \[\Rightarrow \] \[x=\frac{h}{\sqrt{3}}\] ln \[\Delta ABC,\,\,\tan {{30}^{o}}=\frac{h}{20+x}\] \[\Rightarrow \] \[\frac{1}{\sqrt{3}}=\frac{h}{20+x}\] \[\Rightarrow \] \[\sqrt{3}h=20+\frac{h}{\sqrt{3}}\] [from Eq. (i)] \[\Rightarrow \] \[\left( \sqrt{3}-\frac{1}{\sqrt{3}} \right)h=20\] \[\therefore \] \[h=20\times \frac{\sqrt{3}}{2}\] \[=10\sqrt{3}m\]You need to login to perform this action.
You will be redirected in
3 sec