Question

Let a vertical tower $AB$ have its end $A$ on the level ground. Let $C$ be the mid-point of $AB$ and $P$ be a point on the ground such that $AP=2AB$. If $\angle BPC=\beta$, then $\tan \beta$ is equal to.

• A. $1/4$
• B. $2/9$
• C. $4/9$
• D. $6/7$

Answer 1

The correct option is B $2/9$

Let $AB=x$. Then $AP=2AB=2x$

$△ABP$ is a right-angled triangle with $BP$ as hypotenuse.

By Pythagoras Theorem,

$B{P}^2=A{P}^2+A{B}^2$

$B{P}^2=(2x{)}^2+x^2$

$B{P}^2=5x^2$

$\Rightarrow BP=\sqrt{5}x$

$AC=\frac{x}{2}$

$\tan \alpha =\frac{\frac{x}{2}}{2x}=\frac{1}{4}$

Now,
$\tan (\alpha +\beta )=\frac{x}{2x}=\frac{1}{2}$

$\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{1}{2}$

$\Rightarrow 2(\tan \alpha +\tan \beta )=1-\tan \alpha \tan \beta$

$\Rightarrow 2(\frac{1}{4}+\tan \beta )=1-\frac{1}{4}\tan \beta$

$\Rightarrow \frac{1}{2}+2\tan \beta =1-\frac{1}{4}\tan \beta$

$\Rightarrow \frac{9}{4}\tan \beta =\frac{1}{2}$

$\Rightarrow \tan \beta =\frac{2}{9}$