$ABCD$ is a cyclic quadrilateral in which $\angle BCD = 100^o$ and $\angle ABD = 70^o$, find $\angle ADB$.

Given:

$ABCD$ is a cyclic quadrilateral in which $\angle BCD = 100^o$ and $\angle ABD = 70^o$.

To do:

We have to find $\angle ADB$.

Solution:

$ABCD$ is a cyclic quadrilateral.

Join $BD$.

$\angle BCD = 100^o$ and $\angle ABD = 70^o$

$\angle A + \angle C = 180^o$              (Sum of the opposite angles of cyclic quadrilateral)

$\angle A+ 100^o = 180^o$

$\angle A= 180^o - 100^o$

$\angle A = 80^o$

In $\triangle ABD$,

$\angle A + \angle ABD + \angle ADB = 180^o$         (Sum of angles of a triangle)

$80^o + 70^o + \angle ADB = 180^o$

$150^o +\angle ADB = 180^o$

$\angle ADB = 180^o- 150^o$

$= 30^o$

Hence $ADB = 30^o$.

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Updated on: 10-Oct-2022

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