$ABCD$ 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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