Question

in fig. angle ACB is:-
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Answer 1

Answer:

AO=OB (radius of a circle are equal)

therefore ΔAOB is isosceles

∠OAB =∠OBA (base angles of an isosceles triangle are equal)

∠OAB = 60°

∠AOB =180-(60+60)              [ sum of angles in a triangle is 180]

          = 180-120

         = 60°

∠ACB = ∠AOB÷2   [The angle subtended by an arc of a circle at its center          is twice the angle it subtends anywhere on the circle's circumference]

∠ACB = 60÷2

          = 30°

Answer 2

Answer:

∆ABO is an isoceles triangle

,Because OA=OB

Hence , angle ABO = angle BAO =60°

So , WE KNOW THAT SUM OF ALL ANGLES OF A TRIANGLE IS 180°

Therefore , angle ABO + angle BAO + angle AOB = 180 °

=> 60° + 60° + angle AOB =180 °

=> angle AOB = 60°

By theorem, angle formed by a chord in radius is double than the angle in circle

so, angleAOB =2angle ACB

=> angle ACB = angle AOB/2

=>angle ACB = 60/2

=> angle ACB = 30 ° } ANS

HOPE THIS WILL HELP U