Given: ∠𝐴 = 60° and ∠𝐴𝐵𝐶 = 80° and ABCD is cyclic quadrilateral.
∠BAD+∠BCD = 180º
(The sum of the opposite angles of a cyclic quadrilateral is 180º)
60º+∠BCD=180º
∠BCD=120º ...eq(1)
∠ABC+∠CDA=180º
(The sum of the opposite angles of a cyclic quadrilateral is 180º)
80º+∠CDA=180º
∠CDA=100º ...eq(2)
ADP is a line and forming pair of linear angle
∠CDA+∠CDP=180º (using eq(2))
100º+∠CDP=180º
∠CDP=80º ...eq(3)
BCP is a line and forming pair of linear angle
∠BCD+∠DCP=180º using eq(1)
120º+∠DCP=180º
∠DCP=60º ...eq(4)
In
△DCP,
∠CDP+∠DCP+∠DPC=180º
(angle sum property of triangle)
Using eq(3) and eq(4)
80º+60º +∠DPC=180º
∠DPC=40º
ABQ is a line and forming pair of linear angle
∠ABC+∠CBQ=180
80+∠CBQ=180
∠CBQ=100 ...eq(5)
∠DCP=∠QCB=60... using eq(4)
(vertically opposite angle)
Similarly, in
△CBQ,
∠CBQ+∠QCB+∠BQC=180º
(angle sum property of triangle)
100º+60º+∠BQC=180º
∠BQC=20º