Two chords AB and CO of a circle with center O, intersect each other at P. If ∠AOD = 120° and ∠BOC = 80°, then the value of ∠APC is ?

80°

∠AOD = 120°

⇒ ∠ACD = \(\frac{1}{2}\) ∠AOD = \(\frac{1}{2}\) (120°) = 60°      [angle made by chord AD at circumference]

∠BOC = 80°

⇒ ∠BAC = 40°    [angle made by chord BC at circumference)

Now, in ΔAPC

⇒ ∠ACP + ∠APC + ∠CAP = 180°

⇒ 60° + ∠APC + 40° = 180°

⇒ ∠APC = 80°