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In a triangle $\mathrm{ABC},$ if $\mathrm{a}=2, \mathrm{~B}=60^{\circ}$ and $\mathrm{C}=75^{\circ}$, then b equals
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Verified Answer
The correct answer is:
$\sqrt{6}$
$\mathrm{A}=180^{\circ}-60^{\circ}-75^{\circ}=180^{\circ}-135^{\circ}=45^{\circ}$
Now, $\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}$
$\Rightarrow \frac{2}{\sin 45^{\circ}}=\frac{\mathrm{b}}{\sin 60^{\circ}} \Rightarrow \mathrm{b}=\frac{2 \cdot(\sqrt{3} / 2)}{1 / \sqrt{2}}=\sqrt{6}$
Now, $\frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}$
$\Rightarrow \frac{2}{\sin 45^{\circ}}=\frac{\mathrm{b}}{\sin 60^{\circ}} \Rightarrow \mathrm{b}=\frac{2 \cdot(\sqrt{3} / 2)}{1 / \sqrt{2}}=\sqrt{6}$
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