Trigonometry! #30

Geometry Level 4

In $\Delta ABC$ , $D$ is the midpoint of $BC$ . If $AD \perp AC$ , then $\cos A \cos C$ is equivalent to which of the following expressions?

This problem is part of the set Trigonometry .

\frac {2(c^2 - a^2 )}{3ac}

\frac {3(c^2 - a^2 )}{2ac}

\frac {2(a^2 - c^2 )}{3ac}

\frac {3(a^2 - c^2 )}{2ac}

1 solution

Omkar Kulkarni
Feb 9, 2015

By cosine rule,

$\cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc}$

$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}$

In $\Delta ADC$ , $\cos C = \frac{AC}{DC}=\frac{2b}{a}$

$\therefore\cos C=\frac{2b}{a}=\frac{a^{2}+b^{2}-c^{2}}{2ab}$

$\Rightarrow a^{2}-c^{2}=3b^{2}$

$\therefore \cos A = \frac{b^{2}-3b^{2}}{2bc}=-\frac{2b^{2}}{2bc}=-\frac{b}{c}$

$\therefore \cos A \cos C = \left(\frac{2b}{a}\right)\left(-\frac{b}{c}\right)=\frac{-2b^{2}}{ac}$

$\boxed{\cos A \cos C = \frac{2(c^{2}-a^{2})}{3ac}}$