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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Aicharya · Accepted Answer

$\begin{aligned} & \cos A=\frac{b^2+c^2-a^2}{2 b c} \ & \Rightarrow \cos 30^{\circ}=\frac{3+1-a^2}{2 	imes \sqrt{3} 	imes 1} \ & \Rightarrow \frac{\sqrt{3}}{2}=\frac{4-a^2}{2 \sqrt{3}} \ & \Rightarrow a=1\end{aligned}$ Now $\cos B=\frac{c^2+a^2-b^2}{2 c a}=\frac{1^2+1^2-(\sqrt{3})^2}{2 	imes 1 	imes 1}=-\frac{1}{2}$ $\Rightarrow B=120^{\circ}$

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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