Question

The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is ${30}^{\circ }$. On advancing 150 m towards the foot of the tower, the angle of elevation becomes ${60}^{\circ }$. Show that the height of the tower is 129.9 metres. [Take $\sqrt{3}=\mathrm{1.732.}$]

Answer 1

Use trigonometric ratios in the $△$ ABD,

$tan{60}^o=\frac{h}{a}\sqrt{3}=\frac{h}{a}h=\sqrt{3}a----\left(1\right)$

Similarly using trigonometric ratios in the $△$ ABC, we get,

$tan{30}^o=\frac{h}{a+150}\frac{1}{\sqrt{3}}=\frac{h}{a+150}\sqrt{3}h=150+a----\left(2\right)$

Use equation(1) in equation (2),

a = 75 -----(3)

Put the value of a in equation(1) to get the value of h,

$h=75\times \sqrt{3}=75\times 1.73=129.9m$

Therefore the height of the tower is 129.9 m.

Hence PROVED