Question

$ABCD$ is a cyclic quadrilateral, where $BC$ is diameter and $\angle ADC={130}^o$. Find the value of $\angle ACB$?

Answer 1

Given : $ABCD$ lie on a circle, $\angle ADC={130}^{\circ },\text{and}BC$ is the diameter.
We know that opposite angles in a cyclic quadrilateral add up to ${180}^{\circ }$,
$\Rightarrow \angle ADC+\angle ABC={180}^{\circ }$
$\Rightarrow \angle ABC=180-\angle ADC=180-130={50}^{\circ }$
As $BC$ is the diameter,
$\Rightarrow \angle BAC={90}^{\circ }$,
$\Rightarrow \angle ACB=180-(\angle ABC+\angle BAC)=180-(50+90)={40}^{\circ }$

Alternatively, you can do it this way:
As $BC=$ diameter, $BDC={90}^{\circ }$,
$\Rightarrow \angle BDA=130-90={40}^{\circ }$
$\Rightarrow \angle ACB=\angle BDA={40}^{\circ }$