Let a vertical tower $$AB$$ of height $$2h$$ stands on a horizontal ground. Let from a point $$P$$ on the ground a man can see upto height $$h$$ of the tower with an angle of elevation $$2\alpha$$. When from $$P$$, he moves a distance $$d$$ in the direction of $$\overrightarrow{AP}$$, he can see the top of the tower with an angle of elevation $$\alpha$$. If $$d = \sqrt{7}h$$, then $$\tan \alpha$$ is equal to

Let

$$AP=x$$

Initially, the man sees the point on the tower at height

$$h$$

with angle of elevation

$$2\alpha$$.

Therefore,

$$\tan 2\alpha=\frac{h}{x}$$

After moving a distance

$$d=\sqrt7\,h$$

away from the tower, his distance from the foot of the tower becomes

$$x+\sqrt7\,h$$

and he sees the top of the tower of height

$$2h$$

with angle of elevation

$$\alpha$$.

Hence,

$$\tan\alpha=\frac{2h}{x+\sqrt7\,h}$$

Let

$$t=\tan\alpha$$

Then

$$x=\frac{h}{\tan2\alpha}$$

Using

$$\tan2\alpha=\frac{2t}{1-t^2}$$

we get

$$x=\frac{h(1-t^2)}{2t}$$

Substituting into

$$t=\frac{2h}{x+\sqrt7\,h}$$

gives

$$t=\frac{2}{\frac{1-t^2}{2t}+\sqrt7}$$

Multiplying through,

$$\frac{1-t^2}{2t}+\sqrt7=\frac{2}{t}$$

Multiplying by

$$2t$$,

$$1-t^2+2\sqrt7\,t=4$$

$$t^2-2\sqrt7\,t+3=0$$

Solving,

$$t=\frac{2\sqrt7\pm\sqrt{28-12}}{2}$$

$$=\frac{2\sqrt7\pm4}{2}$$

$$=\sqrt7\pm2$$

Since

$$\alpha$$

is acute and

$$2\alpha<90^\circ$$,

we must have

$$\tan\alpha<1$$.

Therefore,

$$t=\sqrt7-2$$

Hence

$$\tan\alpha=\sqrt7-2$$

Final Answer :

$$\sqrt7-2$$