Question

A man observes the angle of elevation of the top of a building to be ${30}^o$. He walks towards its in a horizontal line through its base. On covering $60$ m, the angle of elevation changes to $60$, Find the height of the building correct to the nearest meter.

Answer 1

Let from $D$ the angle of elevation of $AB$ is ${30}^0$ after moving $60m$ the angle of elevation is ${60}^0$

Let $CB=yAB=x$

$\tan 30=\frac{AB}{BD}=\frac{x}{6D+y}$

$\tan 60=\frac{AB}{BC}=\frac{x}{y}$

$\sqrt{3}=\frac{x}{y}x=y\sqrt{3}$ i.e $\frac{y\sqrt{3}}{60+y}=\frac{1}{\sqrt{3}}$

$3y=60+y$

$y=30$

$x=30\sqrt{3}$

Height of the building $=30\sqrt{3}$