Let a vertical tower A B of height 2 h stands on a horizontal ground.

Question

Let a vertical tower $A B$ of height $2 h$ 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 $$\overrightarrow{A P}$$, he can see the top $B$ of the tower with an angle of elevation $$\alpha$$.  If $$d=\sqrt{7} h$$, then $$\tan \alpha$$ is equal to

Accepted Answer

$$\Delta{APM}$$ gives

$$\tan 2\alpha  = {h \over x}$$ ..... (i)

$$\Delta{AQB}$$ gives

$$\tan 2\alpha  = {{2h} \over {x + d}} = {{2h} \over {x + h\sqrt 7 }}$$ ...... (ii)

From (i) and (ii)

$$\tan 2\alpha  = {{2\,.\,\tan 2\alpha } \over {1 + \sqrt 7 \,.\,\tan 2\alpha }}$$

Let $$t = \tan \alpha $$

$$ \Rightarrow t = {{2{{2t} \over {1 - {t^2}}}} \over {1 + \sqrt 7 \,.\,{{2t} \over {1 - {t^2}}}}}$$

$$ \Rightarrow {t^2} - 2\sqrt 7 t + 3 = 0$$

$$t = \sqrt 7  - 2$$

Solutions · 1