Mathematics Question on Angle between a Line and a Plane

Question

Let a vertical tower AB of height 2 h 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 2α.
When from P, he moves a distance d in the direction of $\overrightarrow{AP}$ .
he can see the top B of the tower with an angle of elevation α. if $d = \sqrt7$ h, then tan α is equal to

Answer 1

Solution

The correct answer is (C):

Let a vertical tower AB of height 2h stands on a horizontal ground. Let from a point P
ΔAPM gives tan2α = $\frac{h}{x}$ ....(i)
ΔAQB gives tanα = $\frac{2h}{x+d }= \frac{2h}{x+h√7}$ ...(ii)
From (i) and (ii)
$tanα = \frac{2.tan2α}{1+√7.tan2α}$
Let t = tanα
⇒ $t =\frac{ 2\frac{2t}{1-t^2}}{1+√7.\frac{2t}{1-t^2}}$
$⇒ t^2-2\sqrt7t+3 = 0$
$t = \sqrt{7}-2$