Two chords AB and CD of a circle with center O intersect each other at P. If ∠APC = 85° and ∠AOD = 120°, then what ∠BOC is :
Answer & explanation
Correct answer: option 2
Angle made by arc at anywhere on circle is half to the angle made by same arc on center
Arc AC,
Let ∠AOC = x° , then ∠ABC = \(\frac{x}{2}\) (using above property)
Arc BD,
Let ∠BOD = y°, then ∠BCD = \(\frac{y}{2}\)°
In ΔBPC
∠APC = ∠PBC + ∠PCB [exterior angle]
85° = \(\frac{x}{2}\) + \(\frac{y}{2}\) ⇒ x + y = 170°
Now, ∠AOD + ∠DOB + ∠AOC + ∠BOC = 360°
120° + x + y + ∠BOC = 360°
∠BOC = 360° - 120° - 170°
∠BOC = 70°