Question

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

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

Answer 1

Answer: (A)

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