Solve this problem of chapter circle
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i)
in triangle AOB
OA= OB (radii of same circle)
∴ΔAOB is isosceles triangle
<BAO = <ABO = 30 [ in isosceles triangle opposite angles of equal sides are equal]
<BAO + <ABO +<AOB= 180[ sum of the angles in a triangle=180]
30+30 +<AOB = 180
<AOB = 180-60=120
ii)<AOB +<AOC+<BOC =360 [ angle around the point= 360]
120 +140 +<BOC= 360
⇒260 +<BOC = 360
<BOC = 360-260
<BOC = 100
in triangle AOB
OA= OB (radii of same circle)
∴ΔAOB is isosceles triangle
<BAO = <ABO = 30 [ in isosceles triangle opposite angles of equal sides are equal]
<BAO + <ABO +<AOB= 180[ sum of the angles in a triangle=180]
30+30 +<AOB = 180
<AOB = 180-60=120
ii)<AOB +<AOC+<BOC =360 [ angle around the point= 360]
120 +140 +<BOC= 360
⇒260 +<BOC = 360
<BOC = 360-260
<BOC = 100
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