Quantitative Aptitude Ques 2284
Question: In given figure, O is the centre of circle and $ AB=BC $ and $ \angle AOB=90{}^\circ , $ then $ \angle \theta $ is
Options:
A) $ 30{}^\circ $
B) $ 45{}^\circ $
C) $ 60{}^\circ $
D) None of the above
Show Answer
Answer:
Correct Answer: B
Solution:
- AB = BC OA = OC radius OB = BO common
$ \therefore $ $ \Delta OAB\cong \Delta OCB $
$ \Rightarrow $ AC is the diameter.
$ \Rightarrow $ $ \Delta ABC $ is an isosceles right angled triangle with $ \angle B=90{}^\circ $
$ \Rightarrow $ BO is the angle bisector of $ \angle B=\theta =\frac{90{}^\circ }{2}=45{}^\circ $
BETA
