Question

In a given Figure O is the center of the circle Angle OBA =20 and angle OCA= 30 then find angle BAC​

Answer 1

Answer:

(i) We know that OB = OC which is the radius The base angles of an isosceles triangle are equal So we get ∠OBC = ∠OCB = 55o In △ BOC Using the angle sum property ∠BOC + ∠OCB + ∠OBC = 180o By substituting the values ∠BOC + 55o + 55o = 180o On further calculation ∠BOC = 180o – 55o – 55o By subtraction ∠BOC = 180o – 110o So we get ∠BOC = 70o

(ii) We know that OA = OB which is the radius The base angles of an isosceles triangle are equal So we get ∠OBA= ∠OAB = 20o In △ AOB Using the angle sum property ∠AOB + ∠OAB + ∠OBA = 180o By substituting the values ∠AOB + 20o + 20o = 180o On further calculation ∠AOB = 180o – 20o – 20o By subtraction ∠AOB = 180o – 40o So we get ∠AOB = 140o We know that ∠AOC = ∠AOB – ∠BOC By substituting the values ∠AOC = 140o – 70o So we get ∠AOC = 70o

Step-by-step explanation:

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