Question

Question 2
In figure, if $\angle AOB={125}^{\circ }$. Then $\angle$ COD is equal to
(A) 62.5°
(B) 45°
(C) 35°
(D) 55°

Answer 1

We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle
$i.e\angle AOB+\angle COD={180}^{\circ }\angle COD={180}^{\circ }-\angle AOB={180}^{\circ }-{125}^{\circ }={55}^{\circ }$