| In the given figure, BOC is a diameter of a circle with centre O. If \[\angle BCA=30{}^\circ ,\] then \[\angle CDA\] is equal to |
 |
A) \[30{}^\circ \]
B) \[45{}^\circ \]
C) \[60{}^\circ \]
D) \[50{}^\circ \]
Correct Answer: C
Solution :
| \[\angle BAC=90{}^\circ \] [angles in a semi-circle] |
| In \[\Delta ABC,\]\[\angle BAC+\angle ABC+\angle BCA=180{}^\circ \] |
| \[\Rightarrow \] \[90{}^\circ +\angle ABC+30{}^\circ =180{}^\circ \] |
| \[\Rightarrow \] \[\angle ABC=60{}^\circ \] |
| \[\therefore \] \[\angle CDA=\angle ABC=60{}^\circ \] |
| [angles in the same segment of a circle] |
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