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 $
BETA
