In the given figure, O is the centre of a circle in which OBA = 20° and OCA = 30°. Then, BOC = ?

Given: and


Consider ΔOAB


Here,


OA = OB (radius)


OBA = OAB = 20° (angles opposite to equal sides are equal)


By angle sum property


AOB + OBA + OAB = 180°


AOB + 20° + 20° = 180°


AOB = 180° – 20° – 20°


AOB = 140°


Similarly, in ΔAOC


OA = OC (radius)


OCA = OAC = 30° (angles opposite to equal sides are equal)


By angle sum property


AOC + OCA + OAC = 180°


AOC + 30° + 30° = 180°


AOC = 180° – 30° – 30°


AOC = 120°


Here,


CAB = OAB + OAC = 50°


Here,


2CAB = BOC (The angle subtended by an arc at the center is twice the angle subtended by the same arc on any point on the remaining part of the circle).


2CAB = BOC


2 × 50° = BOC


BOC = 100°.


BOC = 100

19