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Sum of Pair of Opposite Angles in Quadrilateral

In the figure...

Question

In the figure $,$ $O$ is the centre of the circle angle $A O B = 100^{0}$ find angle $A D B$

130^{o}

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120^{o}

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45^{o}

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90^{o}

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Solution

The correct option is A $130^{o}$
Angle $A O B = 100$
angle $A C B = 50 ($ angle made at centre is double the angle made at any other point $)$
now,ACBD is a cyclic quadrilateral $,$ so angle $A D B +$ angle $A C B = 180 ($ sum of opposite angles of a cyclic quad. is $180)$
therefore $,$ angle $A D B = 180 - 50 = 130$
Answer IS $130 °$
hence, option $(A)$ is correct answer.

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In the figure, O is the centre of the circle angle AOB=1000 find angle ADB

In the figure $,$ $O$ is the centre of the circle angle $A O B = 100^{0}$ find angle $A D B$