Mathematics
In the given figure, ∠OAB = 30° and ∠OCB = 57°, find ∠BOC and ∠AOC.
![In the given figure, ∠OAB = 30° and ∠OCB = 57°, find ∠BOC and ∠AOC. Chapterwise Revision, Concise Mathematics Solutions ICSE Class 10.](https://cdn1.knowledgeboat.com/img/cm10/q78-chapterwise-revision-concise-maths-solutions-icse-class-10-889x930.png)
Circles
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Answer
In △AOB,
⇒ OA = OB [Radius of same circle]
⇒ ∠OBA = ∠BAO = 30° [Angles opposite to equal sides are equal]
Also,
⇒ ∠OBA + ∠BAO + ∠AOB = 180° [By angles sum property of triangle]
⇒ 30° + 30° + ∠AOB = 180°
⇒ ∠AOB = 180° - 60° = 120°
In △OCB,
OC = OB [Radius of same circle]
⇒ ∠OBC = ∠OCB = 57° [Angles opposite to equal sides are equal]
Also,
⇒ ∠OCB + ∠OBC + ∠BOC = 180° [By angles sum property of triangle]
⇒ 57° + 57° + ∠BOC = 180°
⇒ ∠BOC = 180° - 114° = 66°
From figure,
⇒ ∠AOB = ∠AOC + ∠BOC
⇒ 120° = ∠AOC + 66°
⇒ ∠AOC = 120° - 66° = 54°.
Hence, ∠AOC = 54° and ∠BOC = 66°.
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