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 $\cos \left(A\right)\cos \left(C\right)$ is

• A. $\frac{2(c^2–a^2)}{3ac}$
• B. $\frac{3(c^2–a^2)}{2ac}$
• C. $1$
• D. None of these

Answer 1

The correct option is A

$\frac{2(c^2–a^2)}{3ac}$

Explanation for the correct option:

Step 1: Interpret the given data

$ABC$ is a triangle

$BC,CA,$ and $AB$are of length $a,b,$and $c$, respectively.

$D$ is the mid-point of $BC$

$AD$ is perpendicular to $AC$

Step 2: Apply cosine rule

The cosine rule states that, in a triangle with sides $a$, $b$ and $c$ with opposite angles $A$, $B$ and $C$ respectively.
$c^2=a^2+b^2-2ab\cos \left(C\right)$

This applies for any side of the triangle. It is an extension of the Pythagoras theorem.

From, right $∆ACD$

$\cos \left(C\right)=\frac{b}{{\displaystyle \frac{a}{2}}}=\frac{2b}{a}\dots \left(1\right)$

By cosine rule

$\begin{array}{ccc}& & \\ \Rightarrow & c^2=a^2+b^2-2ab\cos \left(C\right)& \\ \Rightarrow & \cos \left(C\right)=\frac{a^2+b^2-c^2}{2ab}& \dots \left(2\right)\\ \Rightarrow & \frac{2b}{a}=\frac{a^2+b^2-c^2}{2ab}& \\ \Rightarrow & b^2=\frac{a^2-c^2}{3}& \dots \left(3\right)\end{array}$

Also, by cosine rule

$\Rightarrow \cos \left(A\right)=\frac{b^2+c^2-a^2}{2bc}$

Then,

$\begin{array}{rcl}\cos \left(A\right)\cos \left(C\right)& =& \left(\frac{b^2+c^2-a^2}{2bc}\right)\times \left(\frac{2b}{a}\right)\\ & =& \left(\frac{b^2+c^2-a^2}{ac}\right)\\ & =& \left[\frac{\left({\displaystyle \frac{a^2-c^2}{3}}\right)+c^2-a^2}{ac}\right]\left[\because frome{q}^n\left(3\right)\right]\\ & =& \left(\frac{a^2-c^2+3c^2-3a^2}{3ac}\right)\\ \therefore \cos \left(A\right)\cos \left(C\right)& =& \left(\frac{2c^2-2a^2}{3ac}\right)\end{array}$

Hence, option A is correct.