Question

In a triangle $ABC,a=4,b=3,\angle A=60^{\circ },$ 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)

Key Idea : $\cos A=\frac{b^2+c^2-a^2}{2bc}$
We have $a=4,b=3$ and $\angle A=60^{\circ }$
$\therefore \cos 60^{\circ }=\frac{c^2+9-16}{2\times 3\times c}$
$\Rightarrow \frac{1}{2}=\frac{c^2-7}{2\times 3c}$
$\Rightarrow c^2-7=3c$
$\Rightarrow c^2-3c-7=0$
Thus. $c$ is the root of above equation.