Question

In $△ABC$, if $a=4,b=3,\angle A=60^{\circ }$. Then, $c$ is the root of the equation

• A. $c^2-3c-7=0$
• B. $c^2+3c+7=0$
• C. $c^2-3c+7=0$
• D. $c^2+3c-7=0$

Answer 1

Answer: (A)

We have, $a=4,b=3$ and $\angle A=60^{\circ }$
$\because \cos A=\frac{b^2+a^2-a^2}{2bc}$
$\therefore \cos 60^{\circ }=\frac{9+c^2-16}{6c}$
$\Rightarrow \frac{1}{2}\times 6c=c^2-7$
$\Rightarrow c^2-3c-7=0$