| In given figure, O is the centre of circle and \[AB=BC\] and \[\angle AOB=90{}^\circ ,\]then \[\angle \theta \]is |
 |
A) \[30{}^\circ \]
B) \[45{}^\circ \]
C) \[60{}^\circ \]
D) None of the above
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 |
| \[\angle B=\theta =\frac{90{}^\circ }{2}=45{}^\circ \] |
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