Two chords AB and CD of a circle with centre O intersect at P. If ∠APC = 40°. Then the value of ∠AOC + ∠BOD is?
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e. 80°
∠APC = 40°
So, ∠APD = ∠BPC = 140°
Now,
∠AOC + ∠BOD
= 2∠ABC + 2∠BCD
= 2(∠ABC + ∠BCD)
= 2(180° - ∠BPC)
= 2(180° - 140°) = 80°