Question

At apoint on the ground yhe angle of elevations of a tower is such that its tangents is $\frac{4}{9}$. On walking $60m$ towards the tower, the tangent of the angle of elevation changes to $\frac{2}{3}$. find the height of the tower.

Answer 1

$AB=60$m $BC=x$m (say)

$\tan {\theta }_1=\frac{h}{60+x}=\frac{4}{9}\Rightarrow 9h=240+4x-\left(i\right)\tan {\theta }_2=\frac{h}{x}=\frac{2}{3}\Rightarrow 3h=2x-\left(ii\right)$

Using $\left(ii\right)$ in $\left(i\right)$ we get

$9h=240+6h$

$\Rightarrow h=80$m

$x=120$m

$\therefore$ Height of the tower is $80$m.