ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110° and angle BAC = 50°. Find angle DAC and angle DCA.

Given, ABCD is a cyclic quadrilateral in which AD || BC

With, ∠ADC = 110°, ∠BAC = 50°.

We know that,

⇒ ∠B + ∠D = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]

⇒ ∠B + 110° = 180°

⇒ ∠B = 180° - 110°

⇒ ∠B = 70°.

Now in ∆ABC, we have

⇒ ∠BAC + ∠ABC + ∠ACB = 180° [By angle sum property of triangle]

⇒ 50° + 70° + ∠ACB = 180°

⇒ ∠ACB = 180° - 120° = 60°

As, AD || BC we have

∠DAC = ∠ACB = 60° [Alternate angles]

Now in ∆ADC,

⇒ ∠DAC + ∠ADC + ∠DCA = 180°

⇒ 60° + 110° + ∠DCA = 180°

⇒ ∠DCA = 180° - 170° = 10°

Hence, ∠DAC = 60° and ∠DCA = 10°.