In a triangle ABC,with usual notations,if b=s | vedclass

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(D) Using the Law of Cosines: $\cos A = \frac{b^2+c^2-a^2}{2bc}$ $\cos 30^{\circ} = \frac{(\sqrt{3})^2 + 1^2 - a^2}{2 	imes \sqrt{3} 	imes 1}$ $\fra

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(D) Using the Law of Cosines: $\cos A = \frac{b^2+c^2-a^2}{2bc}$ $\cos 30^{\circ} = \frac{(\sqrt{3})^2 + 1^2 - a^2}{2 	imes \sqrt{3} 	imes 1}$ $\frac{\sqrt{3}}{2} = \frac{3 + 1 - a^2}{2\sqrt{3}}$ $3 = 4 - a^2$ $\Rightarrow a^2 = 1$ $\Rightarrow a = 1$ Now,using the Law of Cosines for angle $B$: $\cos B = \frac{a^2+c^2-b^2}{2ac}$ $\cos B = \frac{1^2 + 1^2 - (\sqrt{3})^2}{2 	imes 1 	imes 1} = \frac{1+1-3}{2} = -\frac{1}{2}$ Since $\cos B = -\frac{1}{2}$,we have $B = 120^{\circ}$.

In a triangle $ABC$,with usual notations,if $b=\sqrt{3}, c=1, \angle A=30^{\circ}$,then angle $B$ is (in $^{\circ}$)

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