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 |
A) \[30{}^\circ \]
B) \[25{}^\circ \]
C) \[50{}^\circ \]
D) \[60{}^\circ \]
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] |
\[\Rightarrow \] \[\angle BOC=\frac{1}{2}\times 50{}^\circ =25{}^\circ \] |
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