In the figure, ABCD is a cyclic quadrilateral.
Find ∠BCA, ∠ABC and ∠BAC.
ABCD is a cyclic quadrilateral.
∠BAD+∠BCD=180∘ [opposite angles of a cyclic quadrilateral are supplementary]
∴ \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