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

Question

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 up to height $h$ of the tower with an angle of elevation $2\alpha$. When from $P$, he moves a distance $d$ in the direction of $\vec{AP}$, he can see the top $B$ of the tower with an angle of elevation $\alpha$. If $d=\sqrt7h$, then $	an\alpha$ is equal to:

Accepted Answer

Let $AP=x$. $\tan2\alpha = \dfrac hx$ \ldots  (i)

At the new point (distance $x+d$ from $A$): $\tan\alpha = \dfrac{2h}{x+d} = \dfrac{2h}{x+\sqrt7h}$ \ldots  (ii)

From (i): $h=x\tan2\alpha$. Substituting into (ii):
$$\tan\alpha = \frac{2x\tan2\alpha}{x+\sqrt7x\tan2\alpha} = \frac{2\tan2\alpha}{1+\sqrt7\tan2\alpha}$$

Let $t=\tan\alpha$, so $\tan2\alpha=\dfrac{2t}{1-t^2}$:
$$t = \frac{2\cdot\frac{2t}{1-t^2}}{1+\sqrt7\cdot\frac{2t}{1-t^2}} = \frac{4t}{1-t^2+2\sqrt7t}$$
$$1-t^2+2\sqrt7t = 4 \implies t^2-2\sqrt7t+3=0 \implies t=\sqrt7-2 \text{ (taking the valid root)}$$

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

Solutions · 1