Question

$AB$ is a vertical tower. The point $A$ is on the ground and $C$ is the middle point of $AB$ . The part $CB$ subtend an angle $\alpha $ at a point $P$ on the ground. If $AP=n \, AB$ , then the correct relation is

• A. $n=\left(n^{2} + 1\right)tan \, \alpha $
• B. $n=\left(2 n^{2} - 1\right)tan \, \alpha $
• C. $n^{2}=\left(2 n^{2} + 1\right)tan \, \alpha $
• D. $n=\left(2 n^{2} + 1\right)tan \, \alpha $

Answer 1

Answer: (D)

In $\Delta PAB, \, tan \beta =\frac{A B}{A P}$



In $\Delta PAC,tan \, \theta =\frac{A C}{A P}$
$∴ \, tan \, \alpha =tan ⁡ \left(\beta - \theta \right)$
$=\frac{tan \, \beta - tan \, ⁡ \theta }{1 + tan \, ⁡ \beta \, tan \, ⁡ \theta }$
$= \, \frac{\frac{A B}{A P} - \frac{A C}{A P}}{1 + \frac{A B}{A P} \cdot \frac{A C}{A P}} \, \, $ ...(i)
$∵ \, AP=n \, \left(A B\right)=n\left(2 A C\right)$ ( $\because C$ is the mid point of $BA$ )
From Eq. (i), we get
$tan \, \alpha =\frac{\frac{1}{n} - \frac{1}{2 n}}{1 + \frac{1}{2 n^{2}}}=\frac{n}{2 n^{2} + 1}$
$⇒ \, n=\left(2 n^{2} + 1\right)tan \, \alpha $