Question

In the figure, 'O' is the centre of the circle. $\angle AOB={100}^o$ find $\angle ADB$.

Answer 1

In given figure ,O is the center of circle .$\angle AOB={100}^0$

Then external $\angle AOB={360}^0-{100}^0={260}^0$

We know that the angle subtended by an arc at the center of a circle is twice

the angle subtended by the same arc at any other point on the remaining part of the circle.

Arc $\angle ADB=\frac{1}{2}Ex\angle AOB=\frac{1}{2}\times 260={130}^0$