In the figure, 'O' is the centre of the circle. $∠ A O B = 100^{o}$ find $∠ A D B$ .

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Solution

In given figure ,O is the center of circle . $∠ A O B = 100^{0}$

Then external $∠ A O B = 360^{0} - 100^{0} = 260^{0}$

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 $∠ A D B = \frac{1}{2} E x ∠ A O B = \frac{1}{2} \times 260 = 130^{0}$

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In the figure, 'O' is the centre of the circle. ∠AOB=100o find ∠ADB.

In given figure ,O is the center of circle .∠AOB=1000Then external ∠AOB=3600−1000=2600We 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 ∠ADB=12Ex∠AOB=12×260=1300

In the figure, 'O' is the centre of the circle. $∠ A O B = 100^{o}$ find $∠ A D B$ .