Question

$AB$ is a vertical pole. The end $A$ is on the level ground. $C$ is the mid-point of $AB$. $P$ is a point on the level ground. The portion $CB$ subtends an angle $\beta$ at $P$. If $AP=nAB$, then $\tan \beta =$

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

Answer 1

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

Let $\angle APC=\alpha$

Let $AB=x$

Given , $AP=nAB$

$\Rightarrow AP=nx$

Since, C is mid-point of AB

So, $AC=CB=\frac{x}{2}$

In right angled triangle $\Delta CAP$

$\tan \alpha =\frac{CA}{AP}$

$\Rightarrow \tan \alpha =\frac{x}{2nx}$

or $\tan \alpha =\frac{1}{2n}$

In right angled triangle $\Delta BAP$

$\tan (\alpha +\beta )=\frac{AB}{AP}$

$\Rightarrow \tan (\alpha +\beta )=\frac{x}{nx}$

or $\tan (\alpha +\beta )=\frac{1}{n}$

$\Rightarrow \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{1}{n}$

$n\tan \alpha +n\tan \beta =1-\tan \alpha \tan \beta$

$\Rightarrow \tan \beta =\frac{1-n\tan \alpha }{(n+\tan \alpha )}$

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