Question

In the given figure, AB is a diameter of the circle. CD||AB and $\angle LBAD={40}^{\circ }$, then $\angle$ACD is

• A. ${40}^{\circ }$
• B. ${90}^{\circ }$
• C. ${120}^{\circ }$
• D. ${130}^{\circ }$

Answer 1

The correct option is D ${130}^{\circ }$

In $△ADB,\angle ADB={90}^o$ as angle inscribed in a semi circle is a right angle.

So by angle sum property, we get $\angle ABD=180-90-40={50}^o$

In the quadrilateral $ABDC$, $\angle ABD+\angle ACD={180}^o$ as opposite angles of a cyclic quadrilateral are supplementary.

So, $\angle ACD=180-50={130}^o$

So none of the given options is the right answer.