Quantitative Aptitude Ques 2275

Question: In a circle with centre O, AOC is a diameter of the circle, BD is a chord and OB and CD are joined. If $ \angle AOB=130{}^\circ , $ then $ \angle BDC $ is equal to

Options:

A) $ 30{}^\circ $

B) $ 25{}^\circ $

C) $ 50{}^\circ $

D) $ 60{}^\circ $

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ \because $ $ AOB=130{}^\circ $ and $ \angle AOB+\angle BOC=180{}^\circ $

$ \therefore $ $ \angle BOC=180{}^\circ -130{}^\circ =50{}^\circ $ Again, $ \angle BDC=\frac{1}{2}\angle BOC $ [angle at circle is half the angle at centre]

$ \Rightarrow $ $ \angle BOC=\frac{1}{2}\times 50{}^\circ =25{}^\circ $