Question

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

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