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In a triangle $\mathrm{ABC}, \mathrm{c}=2, \mathrm{~A}=45^{\circ}, \mathrm{a}=2 \sqrt{2}$, than what is $\mathrm{C}$
equal to ?
Options:
equal to ?
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Verified Answer
The correct answer is:
$30^{\circ}$
According to sine rule,
$\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}$
$\therefore \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{c}}{\sin \mathrm{C}}$
$\Rightarrow \sin C=\frac{\mathrm{c} \cdot \sin \mathrm{A}}{\mathrm{a}}=\frac{2 \cdot \sin 45^{\circ}}{2 \sqrt{2}}$
$=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}=\frac{1}{2}=\sin 30^{\circ}$
$\mathrm{C}=30^{\circ}$
$\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}$
$\therefore \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{c}}{\sin \mathrm{C}}$
$\Rightarrow \sin C=\frac{\mathrm{c} \cdot \sin \mathrm{A}}{\mathrm{a}}=\frac{2 \cdot \sin 45^{\circ}}{2 \sqrt{2}}$
$=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}=\frac{1}{2}=\sin 30^{\circ}$
$\mathrm{C}=30^{\circ}$
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