Given that, \( \Large \angle A = 90 ^{\circ} and \angle B = 30 ^{\circ} \)



In \( \Large \triangle ABC, \)

\( \Large \angle A + \angle B + \angle C = 180 ^{\circ} \)

=> \( \Large \angle C = 180 ^{\circ} - 90 ^{\circ} -30 ^{\circ} \)

=> \( \Large \angle C = 60 ^{\circ} \)

and BC = 2AB ...(i)

From Pythagoras theorem

\( \Large BC^{2} = AC^{2} + AB^{2} \)

=> \( \Large \left(2AB\right)^{2} = AC^{2} + AB^{2} \)

=> \( \Large AC^{2} = 4AB^{2}-AB^{2}=3AB^{2} \)

=> \( \Large AC = \sqrt{3}.AB=\frac{\sqrt{3}}{2}. \left(2AB\right) \)

=> \( \Large AC = \frac{\sqrt{3}}{2}.BC \) [from Eq....(i)]