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Question
In the following figure, ∠OAB = 30º and ∠OCB = 57º. Find ∠BOC and ∠AOC.
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Solution
Given, ∠OAB = 30° and ∠OCB = 57°
In ΔAOB, AO = OB ...[Both are the radius of a circle]
⇒ ∠OBA = ∠BAO = 30° ...[Angles opposite to equal sides are equal]
In ΔAOB,
⇒ ∠AOB + ∠OBA + ∠BAO = 180° ...[By angle sum property of a triangle]
∴ ∠AOB + 30° + 30° = 180°
∴ ∠AOB = 180° – 2(30°)
= 180° – 60°
= 120° ...(i)
Now, in ΔAOB,
OC = OB ...[Both are the radius of a circle]
⇒ ∠OBC = ∠OCB = 57° ...[Angles opposite to equal sides are equal]
In ΔOCB,
∠COB + ∠OCB + ∠CBO = 180° ...[By angle sum property of triangle]
∴ ∠COB = 180° – (∠OCB + ∠OBC)
= 180° – (57° + 57°)
= 180° – 114°
= 66° ...(ii)
From equation (i), ∠AOB = 120°
⇒ ∠AOC + ∠COB = 120°
⇒ ∠AOC + 66° = 120° ...[From equation (ii)]
∴ ∠AOC = 120° – 66° = 54°
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