In the figure, if ∠BAC=60o and ∠BCA=20o, find ∠ADC
In ΔABC, ∠BAC+∠ABC+∠ACB=180o (Sum of angles of triangle)
⇒60o+∠ABC+20o=180o
⇒∠ABC+80o=180o
∴∠ABC=180o−80o=100o
∵ ABCD is a cyclic quadrilateral,
∴∠ABC+∠ADC=180o
⇒100o+∠ADC=180o
∴∠ADC=180o−100o=80o