Question

Let there exist a unique point $P$ inside a $△ABC$ such that $\angle PAB=\angle PBC=\angle PCA=\alpha .$ If $PA=x,PB=y,PC=z,\Delta =$ area of $△ABC$ and $a,b,c,$ are the sides opposite to the angle $A,B,C$ respectively, then $\tan \alpha$ is equal to

• A. $\frac{a^2+b^2+c^2}{4\Delta }$
• B. $\frac{a^2+b^2+c^2}{2\Delta }$
• C. $\frac{4\Delta }{a^2+b^2+c^2}$
• D. $\frac{2\Delta }{a^2+b^2+c^2}$

Answer 1

The correct option is C $\frac{4\Delta }{a^2+b^2+c^2}$

Given, $\angle PAB=\angle PBC=\angle PCA=\alpha$ and
$PA=x,PB=y,PC=z$
By cosine rule, in $△PCA$
$\cos \alpha =\frac{z^2+b^2-x^2}{2bz}$
$\Rightarrow x^2=z^2+b^2-2bz\cos \alpha \cdots \left(1\right)$
Similarly, $y^2=x^2+c^2-2cx\cos \alpha \cdots \left(2\right)$
and $z^2=y^2+a^2-2ay\cos \alpha \cdots \left(3\right)$
Adding the equestions $\left(1\right),\left(2\right)$ and $\left(3\right),$ we get
$2(cx+ay+bz)\cos \alpha =a^2+b^2+c^2$
$\Rightarrow \cos \alpha =\frac{a^2+b^2+c^2}{2(cx+ay+bz)}\cdots \left(4\right)$


Also, $\Delta =$ area of $△ABC$
$\Rightarrow \Delta =\text{area of}△PAB+\text{area of}△PBC+\text{area of}△PCA$
$\Rightarrow \Delta =\frac{1}{2}cx\sin \alpha +\frac{1}{2}ay\sin \alpha +\frac{1}{2}bz\sin \alpha$
$\Rightarrow \Delta =\frac{1}{2}(cx+ay+bz)\sin \alpha$
$\Rightarrow \sin \alpha =\frac{2\Delta }{(cx+ay+bz)}\cdots \left(5\right)$
From $\left(4\right)$ and $\left(5\right),$ we have
$\tan \alpha =\frac{4\Delta }{a^2+b^2+c^2}$