ABC is a triangle, where BC = 2AB, \( \Large \angle B\) = \( \Large 30 ^{\circ} \) and \( \Large \angle A\) = \( \Large 90 ^{\circ} \). The magnitude of the side AC is


A) \( \Large \frac{2 BC}{3} \)

B) \( \Large \frac{3 BC}{4} \)

C) \( \Large \frac{BC}{\sqrt{3}} \)

D) \( \Large \frac{\sqrt{3 BC}}{2} \)

Correct Answer:
D) \( \Large \frac{\sqrt{3 BC}}{2} \)

Description for Correct answer:

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)]


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