Question

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

• A. ${100}^o$
• B. ${80}^o$
• C. ${65}^o$
• D. ${55}^o$

Answer 1

The correct option is B ${80}^o$

Given, $ABCD$ is a cyclic quadrilateral and $AC$ is its diagonal.

Also, $\angle BAD={25}^o,\angle BCD={80}^o$

Since, $ABCD$ is a cyclic quadrilateral.

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

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

Again, $AB\parallel CD$...[Alternate interior angles]

$\Rightarrow \angle BAC=\angle ACD={25}^o$.

$\therefore \angle ACB=\angle BCD-\angle ACD={80}^o-{25}^o={55}^o$.

Now in $\Delta ABC$,

$\angle ABC={180}^o-(\angle BAC+\angle BCA)$ ....(angle sum property of triangles)

$\Rightarrow \angle ABC={180}^o-({25}^o+{55}^o)={100}^o$

Again $ABCD$ is a cyclic quadrilateral.

$\therefore \angle ABC+\angle ADC={180}^o$ ....(since the sum of the opposite angles of a cyclic quadrilateral is ${180}^o$)

$\therefore \angle ADC={180}^o-\angle ABC={180}^o-{100}^o={80}^o$.

Hence, option $B$ is correct.