Question

A tower stands vertically on the ground. From a point on the ground, which is $50\text{m}$ away from the foot of the tower, the angle of elevation to the top of the tower is ${30}^{\circ }$. Find the height of the tower.

Answer 1

Draw a figure to represent the given information.

Let $AB$ be the height $\left(h\right)$ of the tower, and $BC$ is the distance between the tower and the point of the viewer.
In $△ABC$, $\angle ACB={30}^o$ $\tan \theta =\frac{AB}{BC}$
$\tan {30}^o=\frac{h}{50}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{50}$
$\therefore h=\frac{50}{\sqrt{3}}\text{m}$