Question

In a $△ABC,$ if $D$ is the midpoint of $BC$ and $AD$ is perpendicular to $AC,$ then

• A. $\cos A=-\frac{b}{c}$
• B. $\cos A\cos C=-\frac{2b^2}{ac}$
• C. $\cos B=\left(\frac{b^2+c^2}{ca}\right)$
• D. $a^2-3b^2-c^2=0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\cos A=-\frac{b}{c}$
B $\cos A\cos C=-\frac{2b^2}{ac}$
C $\cos B=\left(\frac{b^2+c^2}{ca}\right)$
D $a^2-3b^2-c^2=0$

Draw $BE\perp CA$ produced.
Then, $BD=DC=\frac{a}{2}$ and $EA=AC=b$
From $△ADC,\cos C=\frac{b}{\frac{a}{2}}=\frac{2b}{a}$
From $△ABC,\cos (\pi -A)=\frac{b}{c}$
$\Rightarrow \cos A=\frac{-b}{c}$
So that $\cos A\cos C=\frac{-b}{c}\times \frac{2b}{a}=\frac{-2b^2}{ca}$
Also,$\cos A=\frac{-b}{c}$
$\frac{b^2+c^2-a^2}{2bc}=\frac{-b}{c}$
$\Rightarrow b^2+c^2-a^2=-2b^2$
$\Rightarrow a^2=3b^2+c^2$
and $\cos B=\frac{c^2+a^2-b^2}{2ca}$
substituting for $a^2=3b^2+c^2$
$\Rightarrow \cos B=\frac{a^2+3b^2+c^2-b^2}{2ca}$
$\Rightarrow \cos B=\frac{2b^2+2c^2}{2ca}=\frac{b^2+c^2}{ca}$