ABCD is a cyclic quadrilateral where BC is diameter and angle ADC = 130 degree . Find the value of angle ACB?

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Feb 1, 2017

$\angle A C B = {40}^{\circ}$

Explanation:

Given : $A B C D$ lie on a circle, $\angle A D C = {130}^{\circ} , \mathmr{and} B C$ is the diameter.
We know that opposite angles in a cyclic quadrilateral add up to ${180}^{\circ}$,
$\implies \angle A D C + \angle A B C = {180}^{\circ}$
$\implies \angle A B C = 180 - \angle A D C = 180 - 130 = {50}^{\circ}$
As $B C$ is the diameter,
$\implies \angle B A C = {90}^{\circ}$,
$\implies \angle A C B = 180 - \left(\angle A B C + \angle B A C\right) = 180 - \left(50 + 90\right) = {40}^{\circ}$

Alternatively, you can do it this way:
As $B C =$ diameter, $B D C = {90}^{\circ}$,
$\implies \angle B D A = 130 - 90 = {40}^{\circ}$
$\implies \angle A C B = \angle B D A = {40}^{\circ}$

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