Question:
Mark the correct alternative in each of the following:
In a triangle $\mathrm{ABC}, a=4, b=3, \angle A=60^{\circ}$ then $c$ is a 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) $c^{2}+3 c-7=0$
Solution:
It is given that $a=4, b=3$ and $\angle A=60^{\circ}$.
Using cosine rule, we have
$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$
$\Rightarrow \cos 60^{\circ}=\frac{9+c^{2}-16}{2 \times 3 \times c}$
$\Rightarrow \frac{1}{2}=\frac{c^{2}-7}{6 c}$
$\Rightarrow c^{2}-7=3 c$
$\Rightarrow c^{2}-3 c-7=0$
Thus, $c$ is the root of $c^{2}-3 c-7=0$.
Hence, the correct answer is option (a).