With usual notations, if the lengths of the sides of a triangle are $7 \mathrm{~cm}, 4 \sqrt{3} \mathrm{~cm}$ and $\sqrt{13} \mathrm{~cm}$, then the measures of the smallest angle is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } a=7, b=4 \sqrt{3}, c=\sqrt{13} \ 	herefore \quad & \cos C=\frac{a^2+b^2-c^2}{2 a b}=\frac{49+48-13}{2 	imes 7 	imes 4 \sqrt{3}}=\frac{\sqrt{3}}{2} \ 	herefore \quad & \angle C=\frac{\pi}{6}\end{aligned}$

With usual notations, if the lengths of the sides of a triangle are $7 \mathrm{~cm}, 4 \sqrt{3}…

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