Question

At a point on level ground, the angle of elevation of the top of a vertical tower is found to be such that its tangent is $\frac{5}{12}$. On walking $192$ metres towards the tower, the tangent of the angle is found to be $\frac{3}{4}$. Find the height of the tower.

Answer 1

Let OH be the tower

From $△HOA$

$\tan {\theta }_1=\frac{OH}{OA}=\frac{h}{OA}$

$\frac{5}{12}=\frac{h}{OA}$

$OA=\frac{12}{5}h$

From $△HOB$

$\tan {\theta }_2=\frac{3}{4}=\frac{OH}{OB}=\frac{h}{OB}$

$OB=\frac{4}{3}h$

$\therefore OA-OB=192$

$\therefore (\frac{12}{5}-\frac{4}{3})h=192$

$h\left(\frac{36-20}{15}\right)=192$

$h=192\times \frac{15}{16}$

$h=15\times 12$

$h=180\text{m}$

$\therefore$ Height of tower is $180\text{m}$