Two chords AB and CD of a circle with centre O intersect each other at P. If ∠APC = 95° and ∠AOD = 110°, then ∠BOC is:
Answer & explanation
Correct answer: option 1
\(\angle\)AOD = \({110}^\circ\) and \(\angle\)APC = \({95}^\circ\)
Let \(\angle\)AOC and \(\angle\)BOD be x and y respectively.
\(\angle\)DCB = \(\angle\)BOD/2
= \(\frac{y}{2}\)
\(\angle\)ABC = \(\angle\)AOC/2
= \(\frac{x}{2}\)
\(\angle\)APC is the external angle for \(\Delta \)PBC
So, \(\angle\)APC = \(\angle\)PBC + \(\angle\)PCB
= 95 = \(\frac{x}{2}\) + \(\frac{y}{2}\)
= x + y = 190
\(\angle\)AOD + \(\angle\)AOC + \(\angle\)BOC + \(\angle\)BOD = 360 (Complete angle)
= 110 + x + y + \(\angle\)BOC = 360
= 110 + 190 + \(\angle\)BOC = 360
= \(\angle\)BOC = 60
Therefore, \(\angle\)BOC is \({60}^\circ\).