$ABCD$ is a cyclic quadrilateral in which $BC \| AD, \angle ADC =110^o$ and
$\angle BAC = 50^o$. Find $\angle DAC$.

Given:

$ABCD$ is a cyclic quadrilateral in which $BC \| AD, \angle ADC =110^o$ and
$\angle BAC = 50^o$. 

To do:

We have to find $\angle DAC$.

Solution:

$ABCD$ is a cyclic quadrilateral.

This implies,

$\angle B + \angle D = 180^o$              (Sum of opposite angles)

$\angle B + 110^o = 180^o$

$\angle B = 180^o - 110^o = 70^o$

In $\triangle ABC$,

$\angle CAB + \angle ABC + \angle BCA = 180^o$      (Sum of angles of a triangle)

$50^o + 70^o + \angle BCA = 180^o$

$120^o + \angle BCA = 180^o$

$\angle BCA = 180^o - 120^o$

$= 60^o$

$\angle DAC = \angle BCA$             (Alternate angles)

$\angle DAC = 60^o$

Hence $\angle DAC = 60^o$.

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Updated on: 10-Oct-2022

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