Question

A church tower AB standing on a level plane is surmounted by a spire BC of the same height as the tower. D is a point on AB such that AD is equals to 1 by 3 AB. At a point on the plane 100 metres from the foot of the tower the angles subtended by AD and BC are equal. The height of the tower<br /> is ​

Answer 1

Height of the tower is 100 metre

Explanation:

In $\bigtriangleup AED$

$\tan \alpha = \frac{h}{300}$           ........................(i)

In $\bigtriangleup AEB$

$\tan (\alpha +\beta ) = \frac{h}{100}$   ...................(ii)

In $\bigtriangleup AEC$

$\tan \left (\alpha + (\alpha +\beta ) \right )= \frac{2h}{100}$

$</p><p>\frac{\tan \alpha +\tan (\alpha +\beta)}{1-\tan \alpha \tan (\alpha +\beta) }=\frac{2h}{100}$

On putting the value from equation (i) and (ii) , we get

$\Rightarrow \frac{\frac{h}{300}+\frac{h}{100}}{1-\frac{h}{300}\times \frac{h}{100}}=\frac{2h}{100}$

$\Rightarrow h^{2}=10000$

Hence, height of the tower $h= 100 metre$.