Question

The area or $\Delta ABC$ is $\sqrt{3}$ sq. units, $\angle B={60}^{\circ }$ and $a^2,b^2,c^2$ are in A.P. then the length of the side AC is

• A. 1
• B. 2
• C. 3
• D. $\frac{3}{2}$

Answer 1

The correct option is B 2

$\frac{1}{2}ac\sin B=\sqrt{3}\Rightarrow ac=4$
$a^2,b^2,c^2$ are in $A.P.\Rightarrow 2b^2=a^2+c^2$
$a^2+c^2-b^2=b^2$
$\frac{a^2+c^2-b^2}{2ac}=\frac{b^2}{2ac}$
$\cos B=\frac{b^2}{2ac}$
$\frac{1}{2}=\frac{b^2}{2ac}$
$\Rightarrow b^2=ac=4\Rightarrow b=2$