Two chords AB and CD of a circle with center O intersect at P. If ∠APC = 50°. Then the value of ∠AOC + ∠BOD is ?

80° 100° 120° 150°

100°

Let ∠PBC = x and ∠PCB = y ΔPBC , x + y = 50° (exterior angle of a triangle is equal to sum of opposite two interior angles) ∠AOC = 2 × ∠ABC = 2x ∠BOD = 2 × ∠BCD = 2y  So, ∠AOC + ∠BOD = 2 (x + y)  =  2 (50°) = 100°