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 $AP=nAB$ , then the correct relation is

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

Answer 1

Answer: (D)

In $\Delta PAB,tan\beta =\frac{AB}{AP}$



In $\Delta PAC,tan\theta =\frac{AC}{AP}$
$\therefore tan\alpha =tan\left(\beta -\theta \right)$
$=\frac{tan\beta -tan\theta }{1+tan\beta tan\theta }$
$=\frac{\frac{AB}{AP}-\frac{AC}{AP}}{1+\frac{AB}{AP}\cdot \frac{AC}{AP}}$ ...(i)
$\because AP=n\left(AB\right)=n\left(2AC\right)$ ( $\because C$ is the mid point of $BA$ )
From Eq. (i), we get
$tan\alpha =\frac{\frac{1}{n}-\frac{1}{2n}}{1+\frac{1}{2n^2}}=\frac{n}{2n^2+1}$
$\Rightarrow n=\left(2n^2+1\right)tan\alpha$