Two chords AB and CD of a circle with centre O intersect each other at P. If $$\angle BOC = 70^\circ$$ and $$\angle AOD = 100^\circ$$, then $$\angle APC$$ is:

$$\angle AOD = 100\degree $$   
$$\angle BOC = 70\degree $$
$$\angle ACD = \angle ACP = \frac{\angle AOD}{2} = \frac{100}{2} = 50\degree$$

($$\because$$ The angle subtended at the centre is twice to that of angle subtended at the circumference by the same arc)

$$\angle BDC = \angle BAC = \frac{\angle BOC}{2} = \frac{70}{2} = 35\degree$$

In $$\triangle APC$$,

$$\angle PAC + \angle ACP + \angle APC = 180$$

$$\angle APC = 180 − 50 − 35$$

$$\angle APC = 95\degree$$