In the figure, $ABCD$ is a cyclic quadrilateral.
Find $\angle BCA$, $\angle ABC$ and $\angle BAC$.
$ABCD$ is a cyclic quadrilateral.
$\angle BAD + \angle BCD = 180^\circ$ $\;\;\;$ [opposite angles of a cyclic quadrilateral are supplementary]
$\therefore \;$ $\angle BCD = 180^\circ - \angle BAD = 180^\circ - 103^\circ = 77^\circ$
But $\;$ $\angle BCD = \angle BCA + \angle ACD$
$\therefore \;$ $\angle BCA = \angle BCD - \angle ACD = 77^\circ - 50^\circ = 27^\circ$
$\angle ADC + \angle ABC = 180^\circ$ $\;\;\;$ [opposite angles of a cyclic quadrilateral are supplementary]
$\therefore \;$ $\angle ABC = 180^\circ - \angle ADC = 180^\circ - 75^\circ = 105^\circ$
In $\triangle ABC$, $\;$ $\angle BAC + \angle ABC + \angle BCA = 180^\circ$ $\;\;\;$ [sum of angles of a triangle equal $180^\circ$]
$\therefore \;$ $\angle BAC = 180^\circ - \angle ABC - \angle BCA = 180^\circ - 105^\circ - 27^\circ = 48^\circ$
$\therefore \;$ $\angle BCA = 27^\circ, \; \angle ABC = 105^\circ, \; \angle BAC = 48^\circ$