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In a triangle $\mathrm{ABC}, a=8, b=10$ and $c=12$. What is the angle C equal to?
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The correct answer is:
$2 \mathrm{~A}$
$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{164-144}{2(8)(10)}=\frac{1}{8}$
$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3}{4}$
$\operatorname{Sin} C=\frac{3 \sqrt{7}}{8}$ and $\operatorname{Sin} A=\frac{\sqrt{7}}{4}$
$\frac{\cos C}{\cos A}=\frac{1}{6} < 1 \Rightarrow \cos C < \cos A \Rightarrow \mathrm{C}>\mathbf{A}$
$\cos (C-A)=\cos C \cos A+\sin C \sin A$
$=\frac{1}{8} \times \frac{3}{4}+\frac{3 \sqrt{7}}{8} \times \frac{\sqrt{7}}{4}=\frac{3}{4}=\cos A$
$\Rightarrow \mathrm{C}-\mathrm{A}=\mathrm{A} \Rightarrow \mathrm{C}=2 A$
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