\(1 \times \sin 60^{\circ}=\sqrt{3} \times \sin \theta\)

\(\frac{\sqrt{3}}{2}=\sqrt{3} \sin \theta\)

\(\sin \theta=\frac{1}{2} \Rightarrow \theta=30^{\circ}\)

So at \(B:\)

\(\theta+60^{\circ}+\alpha=180^{\circ}\)

\(30^{\circ}+60^{\circ}+\alpha=180^{\circ}\)

\(\alpha=90^{\circ}\)