Question

$AB$ is a vertical pole. The end $A$ is on the level ground, $C$ is the middle point of $AB$ and $P$ is a point on the level ground. The portion $CB$ subtends an angle $\beta$ at $P$. If $AP=nAB$; then show that
$\tan \beta =n/(2n^2+1)$.

Answer 1

Let $AB=2x$ so that $AC=CB=x$
By given condition
$AP=n.AB=n.2x,$
and let $\angle CPA=\alpha$
then $\tan \alpha =\frac{x}{2nx}=\frac{1}{2n}$
$\tan (\alpha +\beta )=\frac{2x}{2nx}=\frac{1}{n}.$
$\therefore \tan \beta \tan \{\alpha +\beta )-\alpha \}$
$=\frac{\tan (\alpha +\beta )-\tan \alpha }{1+\tan (\alpha +\beta )\tan \alpha }$
$=\frac{\frac{1}{n}-\frac{1}{2n}}{1+\frac{1}{n}.\frac{1}{2n}}=\frac{n}{2n^2+1}$