question_answer

The angle of elevation of the top of a tower at a point on the ground is \[\text{3}{{\text{0}}^{\text{o}}}\text{.}\] If on walking 20 m toward the tower, the angle of elevation becomes \[\text{6}{{\text{0}}^{\text{o}}},\] then the height of the tower is

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)] \[\Rightarrow \]               \[\left( \sqrt{3}-\frac{1}{\sqrt{3}} \right)h=20\]  \[\therefore \]  \[h=20\times \frac{\sqrt{3}}{2}\] \[=10\sqrt{3}m\]