मराठी

O is centre of the circle, OB = BC and ∠BOC = 20° Statement (1): x = 2 × 20° = 40° Statement (2): ∠BOC = 20° x = ∠OAB + 20° = ∠OBA + 20° = 40° + 20° = 60°

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प्रश्न

O is centre of the circle, OB = BC and ∠BOC = 20°

A circle with centre O, OB equal to BC, and angle BOC marked 20°.

Statement (1): x = 2 × 20° = 40°

Statement (2): ∠BOC = 20°
x = ∠OAB + 20° 
= ∠OBA + 20° = 40° + 20° = 60°

पर्याय

  • Both the statements are true.

  • Both the statements are false.

  • Statement 1 is true, and statement 2 is false.

  • Statement 1 is false, and statement 2 is true.

MCQ
विधान आणि तर्क
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उत्तर

Statement 1 is false, and statement 2 is true.

Explanation: 

Given,

⇒ OB = OC

⇒ ∠BOC = ∠BCO = 20° (Angles opposite to equal sides of a triangle are always equal)

In △ OBC, using angle sum property,

⇒ ∠OBC + ∠BCO + ∠BOC = 180°

⇒ ∠OBC + 20° + 20° = 180°

⇒ ∠OBC + 40° = 180°

⇒ ∠OBC = 180° − 40°

⇒ ∠OBC = 140°

∠OBC and ∠OBA forms linear pairs of angle.

⇒ ∠OBC + ∠OBA = 180°

⇒ 140° + ∠OBA = 180°

⇒ ∠OBA = 180° − 140°

⇒ ∠OBA = 40°

Since OB = OA (Radii of same circle)

⇒ ∠OBA = ∠OAB = 40° (Angles opposite to equal sides of a triangle are always equal)

Using exterior angle property, the exterior angle of a triangle is equal to the sum of the two opposite interior angles.

In triangle OAC,

⇒ ∠EOA = ∠OAC + ∠OCA

⇒ x = ∠OAB + 20°

⇒ x = ∠OBA + 20° = 40° + 20° = 60°

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पाठ 17: Circles - TEST YOURSELF [पृष्ठ २७३]

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सेलिना Concise Mathematics [English] Class 10 ICSE
पाठ 17 Circles
TEST YOURSELF | Q 1. (m) | पृष्ठ २७३
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