Question

The sides $BC,CA$ and $AB$ of a triangle $ABC$ are of lengths $a,b,c$ 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. $\frac{3(a^2-c^2)}{2ac}$
• B. $\frac{2(a^2-c^2)}{3ac}$
• C. $\frac{(a^2-c^2)}{3ac}$
• D. $\frac{2(c^2-a^2)}{3ac}$

Answer 1

Answer: (D)

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

Given $D$ is mid-point of $BC$ and $AD\perp AC$

From right angle $△DAC$

$\cos C=\frac{b}{a/2}=\frac{2b}{a}...\left(i\right)$

also from cosine formulae

$\cos C=\frac{a^2+b^2-c^2}{2ab}...\left(ii\right)$

From (i) and (ii) we get $\frac{2b}{a}=\frac{a^2+b^2-c^2}{2ab}$

$\Rightarrow 4b^2=a^2+b^2-c^2\Rightarrow b^2=\frac{a^2-c^2}{3}$

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

$=\frac{[\left(\frac{a^2-c^2}{3}\right)+c^2-a^2]}{ca}=\frac{2}{3}\left(\frac{c^2-a^2}{ca}\right)$