A tower of height 15 m stands vertically on the ground. From a point on the ground the angle of elevation of the top of the tower is found to be $30^{\circ}$. What is the distance of the point from the foot of the tower?

• Correct answer
  A

  $15 \sqrt{3} \mathrm{~m}$
• B

  $10 \sqrt{3} \mathrm{~m}$
• C

  $5 \sqrt{3} \mathrm{~m}$
• D

  30 m

Solution

Let $A B$ is tower
$\begin{aligned} \tan 30^{\circ} & =\frac{15}{x} \\ \frac{1}{\sqrt{3}} & =\frac{15}{x}\end{aligned}$ $x=15 \sqrt{3}$ $\therefore$ Distance of the point from the foot of the tower is $15 \sqrt{3} \mathrm{~m}$.

Hence, the correct answer is option: 'A'.