Question

If p and q be perpendiculars from the angular points A and B on any line passing through the vertex C of the $△ABC$, prove that
$a^2{p}^2+b^2-2abpqcosC=a^2{b}^{2}si{n}^{2}C.$

Answer 1

We are given AM = p, BN = q.

Let $\angle ACM=\theta$ and $\angle BCN=\varphi$

Then $sin\theta =p/b$ and $cos\theta =\sqrt{1-p^2/b^2}$;

$sin\varphi =q/a$ and $cos\varphi =\sqrt{1-q^2/a^2}$

Now $C=\pi (\theta +\varphi )$

$cosC=cos[\pi -(\theta +\varphi \left)\right]=-cos(\theta +\varphi )$

or $cosC=-cos\theta cos\varphi +sin\theta sin\varphi$

Hence $cosC=-\sqrt{(a-\frac{p^2}{b^2})}\sqrt{(}1-\frac{q^2}{a^2})+\frac{p}{q}.\frac{q}{a}$

or $\sqrt{(1-\frac{p^2}{b^2})}\sqrt{(1-\frac{q^2}{a^2})}=\frac{pq}{ab}-cosC.$

Squaring,

$1-\frac{p^2}{b^2}-\frac{q^2}{a^2}+\frac{p^2{q}^2}{a^2{b}^2}=\frac{p^2{q}^2}{a^2{b}^2}-\frac{2pq}{ab}cosC+co{s}^{2}C.$

or $a^2{p}^2+b^2{q}^2-2abpqcosC$

= $a^2{b}^2(1-co{s}^{2}C)=a^2{b}^{2}si{n}^{2}C$.