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In a triangle $A B C$, with usual notations, if $b=\sqrt{3}, c=1, \angle A=30^{\circ}$, then angle $B$ is
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Verified Answer
The correct answer is:
$120^{\circ}$
$\begin{aligned} & \cos A=\frac{b^2+c^2-a^2}{2 b c} \\ & \Rightarrow \cos 30^{\circ}=\frac{3+1-a^2}{2 \times \sqrt{3} \times 1} \\ & \Rightarrow \frac{\sqrt{3}}{2}=\frac{4-a^2}{2 \sqrt{3}} \\ & \Rightarrow a=1\end{aligned}$
Now $\cos B=\frac{c^2+a^2-b^2}{2 c a}=\frac{1^2+1^2-(\sqrt{3})^2}{2 \times 1 \times 1}=-\frac{1}{2}$
$\Rightarrow B=120^{\circ}$
Now $\cos B=\frac{c^2+a^2-b^2}{2 c a}=\frac{1^2+1^2-(\sqrt{3})^2}{2 \times 1 \times 1}=-\frac{1}{2}$
$\Rightarrow B=120^{\circ}$
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