In the given figure, if  $\angle DAB={60}^{°},\angle ABD={50}^{°}$ then $\angle ACB$ is equal to

 

Concept: The angle subtended by an arc at the centre is twice the angle subtended by it at any other point on the circle.

Here, $\angle DAB={60}^{°},\angle ABD={50}^{°}$

By using the angle sum property in the triangle ABC ,

$\angle DAB+\angle ABD+\angle ADB={180}^{°}$

$\Rightarrow {60}^{°}+{50}^{°}+\angle ADB={180}^{°}$

$\angle ADB={180}^0-{110}^{°}$

$\angle ADB={70}^{°}$

It is known that the angles subtended at the circumference by the same arc are equal

$\angle ADB=\angle ACB$

We know that the angles subtended at the circumference by the same arc are equal.

Hence $\angle ACB = 70°$