मराठी

In the given figure, O is the centre of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. ∠AOC is equal to:

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प्रश्न

In the given figure, O is the centre of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. ∠AOC is equal to:

The image displays a circle with center $O$, points $A$, $B$, and $C$ on the circumference, line segments connecting the center $O$ to points $A$ and $C$, chords $AB$ and $BC$, and interior angle measures of $30^\circ$ at vertex $A$ and $40^\circ$ at vertex $C$. It illustrates circle geometry problems involving radii properties, isosceles triangles, central angles, and inscribed angles within a circle.

पर्याय

  • 70°

  • 80°

  • 150°

  • 140°

MCQ
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उत्तर

140°

Explanation:

Join AC.

In △AOC,

Since,

OA = OC (Radius of same circle)

∴ ∠OAC = ∠OCA = x (let)

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ x + x + ∠AOC = 180°

⇒ ∠AOC = 180° − 2x

In △AOC,

By angle sum property of triangle,

⇒ ∠BAC + ∠ACB + ∠CBA = 180°

⇒ (30° + x) + (40° + x) + ∠CBA = 180°

⇒ ∠CBA + 70° + 2x = 180°

⇒ ∠CBA = 180° − 70° − 2x

⇒ ∠CBA = 110° − 2x

We know that,

The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠AOC = 2∠CBA

⇒ 180° − 2x = 2(110° − 2x)

⇒ 180° − 2x = 220° − 4x

⇒ 4x − 2x = 220° − 180°

⇒ 2x = 40°

⇒ x = `(40°)/2`​ = 20°

⇒ ∠AOC = 180° − 2x

⇒ ∠AOC = 180° − 2(20°)

= 180° − 40°

= 140°

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पाठ 17: Circles - EXERCISE 17 (B) [पृष्ठ २६५]

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सेलिना Concise Mathematics [English] Class 10 ICSE
पाठ 17 Circles
EXERCISE 17 (B) | Q 1. (c) | पृष्ठ २६५
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