Question

In the given figure, $\Delta ABC$ is a right angled triangle at $A$. The hypotenuse is divided into $n$ equal parts. Let $PQ$ be the segment that contains the midpoint of the hypotenuse.


If $a$ and $h$ be the length of the hypotenuse and altitude to the hypotenuse of the triangle respectively. Then which of the following option is correct ?
(Given that $n$ is an odd integer and $\alpha =\angle PAQ$)

• A. $\tan \alpha =\frac{4nh}{(n^2-1)a}$
• B. $\tan \alpha =\frac{4an}{(n^2-1)h}$
• C. $\cot \alpha =\frac{4nh}{(n^2-1)a}$
• D. $\cot \alpha =\frac{4an}{(n^2-1)h}$

Answer 1

The correct option is A $\tan \alpha =\frac{4nh}{(n^2-1)a}$

Let coordinates be $A=(0,0)$, $B=(b,0)$, and $C=(0,c)$.


Then,
$P=\frac{n+1}{2}B+\frac{n-1}{2}C=(\frac{n+1}{2}b,\frac{n-1}{2}c)$ and
$Q=\frac{n-1}{2}B+\frac{n+1}{2}C=(\frac{n-1}{2}b,\frac{n+1}{2}c)$

So,
$\text{Slope of}PA=\tan \angle PAB=\frac{c}{b}\cdot \frac{n-1}{n+1}$
$\text{Slope of}QA=\tan \angle QAB=\frac{c}{b}\cdot \frac{n+1}{n-1}$

Now,
$\tan \alpha =\tan (\angle QAB-\angle PAB)=\frac{(\frac{c}{b}\cdot \frac{n+1}{n-1})-(\frac{c}{b}\cdot \frac{n-1}{n+1})}{1+(\frac{c}{b}\cdot \frac{n+1}{n-1})\cdot (\frac{c}{b}\cdot \frac{n-1}{n+1})}$
$=\frac{\frac{c}{b}\cdot \frac{4n}{n^2-1}}{1+\frac{c^2}{b^2}}$
$=\frac{4nbc}{(n^2-1)(b^2+c^2)}$
$=\frac{4nbc}{(n^2-1)a^2}$

Since, area of $△ABC=\frac{1}{2}bc=\frac{1}{2}ah$
$\Rightarrow bc=ah$
$\therefore \tan \alpha =\frac{4nh}{(n^2-1)a}$