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In a $\triangle \mathrm{ABC}$, if $\mathrm{c}=2, \mathrm{~A}=120^{\circ}, \mathrm{a}=\sqrt{6}$, then what is $\mathrm{C}$ equal
to ?
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to ?
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Verified Answer
The correct answer is:
$45^{\circ}$
Let $c=2, \angle A=120^{\circ}$ and $a=\sqrt{6}$ in $\Delta A B C$,
$\therefore$ By Sine rule, we have
$\begin{aligned} & \frac{a}{\sin A}=\frac{c}{\sin C} \Rightarrow \frac{\sqrt{6}}{\sin 120^{\circ}}=\frac{2}{\sin C} \\ \Rightarrow & \sin C=\frac{2 \times \sqrt{3}}{\sqrt{6} \times 2}=\frac{1}{\sqrt{2}} \\ \Rightarrow \sin C=\sin 45^{\circ} & \Rightarrow \quad \angle C=45^{\circ} \end{aligned}$
$\therefore$ By Sine rule, we have
$\begin{aligned} & \frac{a}{\sin A}=\frac{c}{\sin C} \Rightarrow \frac{\sqrt{6}}{\sin 120^{\circ}}=\frac{2}{\sin C} \\ \Rightarrow & \sin C=\frac{2 \times \sqrt{3}}{\sqrt{6} \times 2}=\frac{1}{\sqrt{2}} \\ \Rightarrow \sin C=\sin 45^{\circ} & \Rightarrow \quad \angle C=45^{\circ} \end{aligned}$
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