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

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Aicharya · Accepted Answer

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

In $\Delta \mathrm{ABC}$ with usual notations $\mathrm{a}=4, \mathrm{~b}=3, \angle \mathrm{A}=60^{\circ}$,…

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