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If the sides of a triangle are $6 \mathrm{~cm}, 10 \mathrm{~cm}$ and $14 \mathrm{~cm}$, then what is the largest angle included by the sides?
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The correct answer is:
$120^{\circ}$
We know that largest side has greatest angle opposite it. $\therefore \mathrm{a}=14 \mathrm{~cm}, \mathrm{~b}=10 \mathrm{~cm}$ and $\mathrm{c}=6 \mathrm{~cm}$
$\therefore \cos A=\frac{c^{2}+b^{2}-a^{2}}{2 b c}$
$=\frac{36+100-196}{2 \times 6 \times 10}=-\frac{1}{2}=\cos 120^{\circ}$
$\Rightarrow \angle \mathrm{A}=120^{\circ}$
$\therefore \cos A=\frac{c^{2}+b^{2}-a^{2}}{2 b c}$
$=\frac{36+100-196}{2 \times 6 \times 10}=-\frac{1}{2}=\cos 120^{\circ}$
$\Rightarrow \angle \mathrm{A}=120^{\circ}$
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