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 :

70°

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°