Q57 / 150 · MHT CET 2020 (13 Oct Shift 1) · Mathematics

In $\Delta \mathrm{ABC}$ with usual notations $\mathrm{a}=4, \mathrm{~b}=3, \angle \mathrm{A}=60^{\circ}$, then $\mathrm{c}$ is a root of the equation

  • Correct answer
    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

$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$ $\therefore \cos 60=\frac{1}{2}=\frac{9+c^{2}-16}{2 \times 3 c} \Rightarrow 1=\frac{c^{2}-7}{3 c} \Rightarrow 3 c=c^{2}-7$ $\therefore c^{2}-3 c-7=0$

Hence, the correct answer is option: 'A'.

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