Question

In a triangle $ABC$ if $2a=\sqrt{3}b+c$ then

• A. $c^2=a^2+b^2-ab$
• B. $a^2=b^2+c^2$
• C. $b^2=a^2+c^2-ac\sqrt{3}$
• D. $a=b=c$

Answer 1

The correct option is B $a^2=b^2+c^2$

$a=k\sin A$
$b=k\sin B$
$c=k\sin C$
$\sin A=\frac{\sqrt{3}}{2}\sin B+\frac{1}{2}\sin C$
Compare $\sin (B+C)=\frac{\sqrt{3}}{2}\sin B+\frac{1}{2}\sin C$
$\sin B\cos C+\cos B\sin C=\cos {30}^0\sin B+\cos {60}^0\sin C$
$\therefore \cos C=\cos {30}^0$
$\cos B=\cos {60}^0$
$\therefore C={30}^0$
$B={60}^0$
$A={90}^0$
$\therefore a^2=b^2+c^2$