Two chords AB and CD of a circle with centre O, intersect each other at P. If ∠AOD = 100° and ∠BOC = 70°, then the value of ∠APC is

According to question
Given:

∠AOD = 100°, ∠BOC = 70°
∴ ∠ACD = ∠ACP = $$\frac{{{{100}^ \circ }}}{2}$$ = 50°
∴ The angle subtended at the centre is twice to that of angle subtended at the circumference by the same arc
∠BOC = 70°
∴ ∠BDC = ∠BAC = $$\frac{{{{70}^ \circ }}}{2}$$ = 35°
In ΔAPC
∠PAC + ∠ACP + ∠APC = 180°
∠APC = 180° - 50° - 35°
∠APC = 95°