question_answer

At a point 20 m away from the foot of a tower the angle of elevation of the top of the tower is\[30{}^\circ \]. The height of the tower is

A)  \[20\sqrt{3}\,\,m\]                     

B)  \[\frac{20}{\sqrt{3}}\,\,m\]

C)         \[\frac{\sqrt{3}}{20}\,\,m\]                   

D) None of these

Correct Answer: B

Solution :

AB = height of tower. From \[\Delta ABC,\]             \[\tan 30{}^\circ =\frac{AB}{BC}\] \[\Rightarrow \]   \[\frac{1}{\sqrt{3}}=\frac{AB}{20}\] \[\Rightarrow \]   \[AB=\frac{20}{\sqrt{3}}m\]