Question

In the figure, $AB$ is parallel to $DC$, $\angle BCD={80}^o$ and $\angle BAC={25}^o$. Then $\angle CAD$ is:

• A. ${55}^o$
• B. ${65}^o$
• C. ${75}^o$
• D. ${35}^o$

Answer 1

Answer: (C)

${75}^o$

Given, $ABCD$ is a cyclic quadrilateral. $AC$ is its diagonal. $\angle BAD={25}^o,\angle BCD={80}^o$.

Here, $ABCD$ is a cyclic quadrilateral.

$\therefore \angle BCD+\angle BAD=1{80}^o$ (since the sum of the opposite angles of a cyclic quadrilateral is ${180}^o$)

$\Rightarrow \angle BAD=1{80}^o-\angle BCD=1{80}^o-{80}^o=1{00}^o$.

But $\angle BAD=\angle BAC+\angle CAD$

$\therefore \angle BAC+\angle CAD=1{00}^o\Rightarrow \angle CAD=1{00}^o-\angle BAC=1{00}^o-{25}^o={75}^o$.

Hence, option $C$ is correct.