Practicing Success
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: |
60° 55° 70° 65° |
60° |
\(\angle\)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 \({60}^\circ\). |