Question

In Fig.10.114, if $\angle AOB={125}^{°}$ , then $\angle COD$ is equal to
figure

(a) 45⁰ (b) 35⁰ (c) 55⁰ (d) $62{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{°}$

Answer 1

We know that opposite sides of a quadrilateral circumscribing a circle subtend supplementary
angles at the centre.
Quadrilateral ABCD circumscribe a circle with centre O.
$$\begin{gathered} \angle AOB+\angle COD=180° \\ \Rightarrow 125°+\angle COD=180° \\ \Rightarrow \angle COD=180°-125° \\ \Rightarrow \angle COD=55° \end{gathered}$$
Hence, the correct answer is option (c).