Question:
If the sides of a $\triangle A B C$ are $a, b, \sqrt{a^{2}+a b+b^{2}}$, then the measure of the largest angle is _________________
Solution:
Let us suppose the greatest angle is c
Using cosine formula,
$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$
Since $C=\sqrt{a^{2}+a b+b^{2}}$
$\Rightarrow C^{2}=a^{2}+a b+b^{2}$
ie $\cos C=\frac{a^{2}+b^{2}-a^{2}-a b-b^{2}}{2 a b}=\frac{-1}{2}$
ie $\cos C=\frac{1}{2}$
ie $C=\frac{2 \pi}{3}$
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