Question

$ABCD$ is a cyclic quadrilateral with $AD$ parallel to $BC$. $\angle DCB={65}^o$. The magnitude of $\angle CBE$ is:

• A. ${100}^o$
• B. ${110}^o$
• C. ${115}^o$
• D. ${120}^o$

Answer 1

Answer: (C)

${115}^o$

Given, $AD\parallel BC$.

Since, $ABCD$ is a cyclic quadrilateral.

Then, $\angle OCB+\angle DAB={180}^{\circ }$ ...[Opposite angles of cyclic quadrilateral are supplementary]

$⟹\angle DAB={115}^{\circ }$.

Since, $AD\parallel BC$

$⟹$ $\angle DAB=\angle CBE={115}^{\circ }$ ...[Corresponding angles].

Hence, option $C$ is correct.