Question

$AB$ is a vertical pole. The end $A$ lies on the ground. $C$ is the midpoint of $AB.P$ is a point on the level ground such that the portion $BC$ subtends an angle $\varphi$ at $P$. If $AP=nAB$ then the value of $\cot \varphi$ is:

• A. $\frac{2n^2+1}{n}$
• B. $\frac{n}{2n^2+1}$
• C. $\frac{2n^2+1}{2n}$
• D. $\frac{2n}{2n^2+1}$

Answer 1

The correct option is A $\frac{2n^2+1}{n}$

Let $BC=AC=x$ and $\angle APC=\theta$

$\because C$ is the mid point

$\Rightarrow AB=2x$

$\Rightarrow AP=nAB=2nx$

Now, In $△APC,$

$\Rightarrow \tan \theta =\frac{AC}{PA}=\frac{x}{2nx}$

$\Rightarrow \tan \theta =\frac{1}{2n}⟶\left(1\right)$

In $△BPA,$

$\Rightarrow \tan (\theta +\varphi )=\frac{AB}{AP}=\frac{2x}{2nx}=\frac{1}{n}$

$\Rightarrow \frac{\tan \theta +\tan \varphi }{1-\tan \theta \tan \varphi }=\frac{1}{n}$

From $\left(1\right),$ using $\tan \theta =\frac{1}{2n}$

$\Rightarrow \frac{\frac{1}{2n}+\tan \varphi }{1-\frac{1}{2n}\tan \varphi }=\frac{1}{n}$

$\Rightarrow n(\frac{1}{2n}+\tan \varphi )=1-\frac{1}{2n}\tan \varphi$

$\Rightarrow \frac{1}{2}+n\tan \varphi =1-\frac{1}{2n}\tan \varphi$

$\Rightarrow \tan \varphi (n+\frac{1}{2n})=1-\frac{1}{2}$

$\Rightarrow \tan \varphi \left(\frac{{2n}^2+1}{2n}\right)=\frac{1}{2}$

$\Rightarrow \tan \varphi =\frac{2n}{2({2n}^2+1)}=\frac{n}{{2n}^2+1}$

$\Rightarrow \cot \varphi =\frac{1}{\tan \varphi }=\frac{{2n}^2+1}{n}$

Hence, the answer is $\frac{{2n}^2+1}{n}.$