Two chords AB and CD of a circle with center O intersect at P. If ∠APC = 50°. Then the value of ∠AOC + ∠BOD is ?
Answer & explanation
Correct answer: option 2
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°