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 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

• A. √5-2
• B. √3-1
• C. √7-2
• D. √7-√3

Answer 1

The Correct Option is C

The correct answer is (C):

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
⇒ \(t =\frac{ 2\frac{2t}{1-t^2}}{1+√7.\frac{2t}{1-t^2}}\)
\(⇒ t^2-2\sqrt7t+3 = 0\)
\(t = \sqrt{7}-2\)