Question

In a triangle $ABC$, $D$ is the middle point of $BC$. If $AD$ is perpendicular to $AC$, then $\cos A\cos C=$

• A. $\frac{-2b^2}{ac}$
• B. $\frac{2(c^2+a^2)}{3ca}$
• C. $\frac{2(c^2-a^2)}{3ca}$
• D. $\frac{2b^2}{ac}$

Answer 1

Answer: (A), (C)

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

In $\Delta ABC,$
$\cos A=\frac{b^2+c^2-a^2}{2bc}\cdots \left(i\right)$
In $\Delta CAD,$
$\cos C=\frac{AC}{CD}=\frac{b}{\left(a/2\right)}=\frac{2b}{a}\cdots \left(ii\right)$
In $△BCE,AD$ bisect both sides $BC$ and $EC$
So, $BE\parallel AD,BE=2AD$ and $\angle DAB=\angle ABE$
So,
$\angle ABE={180}^{\circ }-(\angle E+\angle EAB)={180}^{\circ }-({90}^{\circ }+{180}^{\circ }-\angle A)=\angle A-{90}^{\circ }$
In $\Delta ABD,$
$\frac{BD}{\sin (A-{90}^0)}=\frac{AB}{\sin \angle ADB}$
$\cos A=-\frac{b}{c}\cdots \left(iii\right)$[from Eq.$\left(ii\right)$].From $\left(i\right)$ and $\left(iii\right)$
$\frac{b^2+c^2-a^2}{2bc}=\frac{-b}{c}$
$b^2+c^2-a^2=-2b^2$ or $c^2-a^2=-3b^2$
$-b^2=\frac{c^2-a^2}{3}$
Now, $\cos A\cos C=\frac{-2b^2}{ac}$
$\cos A\cos C=\frac{2(c^2-a^2)}{3ca}$