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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