Question

In the given figure, if $\angle AOB={125}^{\circ }$, then $\angle COD$ is equal to :

• A. ${62.5}^{\circ }$
• B. ${45}^{\circ }$
• C. ${125}^{\circ }$
• D. ${55}^{\circ }$

Answer 1

Answer: (D)

${55}^{\circ }$

Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle. Hence, $\angle COD={180}^{\circ }-{125}^{\circ }={55}^{\circ }$