Question

The sides $BC,CA$ and $AB$ of a $△ABC$ are of lengths $a,b$, and $c$ respectively. If $D$ is the mid-point of $BC$ and $AD$ is perpendicular to $AC$, then the value of $\cos A\cos C$ is

• A. $\frac{3\left(a^2-c^2\right)}{2ac}$
• B. $\frac{3\left(a^2-c^2\right)}{3bc}$
• C. $\frac{\left(a^2-c^2\right)}{3ac}$
• D. $\frac{2\left(c^2-a^2\right)}{3ac}$

Answer 1

Answer: (D)

Using sine rule in $△ACD$,
$\frac{\sin \left(90^{\circ }-C\right)}{b}=\frac{\sin 90^{\circ }}{\frac{a}{2}}$
$\Rightarrow \cos C=\frac{2b}{a}$
and in $\Delta ABD,$
$\frac{\sin \left(A-90^{\circ }\right)}{\frac{a}{2}}=\frac{\sin \left(90^{\circ }+C\right)}{c}$
$\Rightarrow -\frac{\cos A}{\frac{a}{2}}=\frac{\cos C}{c}\Rightarrow \cos A=-\frac{b}{c}$
$\Rightarrow \frac{b^2+c^2-a^2}{2bc}=-\frac{b}{c}\Rightarrow c^2-a^2=-3b^2$
$\therefore \cos A\cos C=-\frac{2b^2}{ac}=\frac{2}{3ac}\left(c^2-a^2\right)$