Question

In the given figure, If $\angle AOB={125}^0$, then $\angle COD$ is equal to :

• A. ${62.5}^0$
• B. ${45}^0$
• C. ${125}^0$
• D. ${55}^0$

Answer 1

The correct option is D ${55}^0$

The quadrilateral angle's circle meet at the point of the supplementary angles and that is at the center.

$ABCD$ will meet at a point where it won't cut it but it will just pass it, and that would be the center at $O.$

$\angle AOB+\angle COD={180}^o$

$\Rightarrow$ ${125}^o+\angle COD={180}^o$

$\Rightarrow$ $\angle COD={180}^o-{125}^o$

$\Rightarrow$ $\angle COD={55}^o$