Plss solve........the given Q.
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In Δ OAB
AB = OA = OB (radii)
Hence ΔOAB is an equilateral triangle
That is each angle of ΔOAB is 60°
∴ ∠AOB = 60°
∠AOB = 2∠ACB [Angle subtended by an arc of a circle at the centre is double the angle subtended by it on any part of the circle]
Hence ∠ACB = 30°
In cyclic quadrilateral ADBC
∠ADB + ∠ACB = 180° (Sum of opposite angles in cyclic quadrilateral is 180°)
⇒ ∠ADB = 180° − 30° = 150°
Therefore, angle subtended by the chord at a point on the major arc and the minor arc are 30° and 150° respectively.
AB = OA = OB (radii)
Hence ΔOAB is an equilateral triangle
That is each angle of ΔOAB is 60°
∴ ∠AOB = 60°
∠AOB = 2∠ACB [Angle subtended by an arc of a circle at the centre is double the angle subtended by it on any part of the circle]
Hence ∠ACB = 30°
In cyclic quadrilateral ADBC
∠ADB + ∠ACB = 180° (Sum of opposite angles in cyclic quadrilateral is 180°)
⇒ ∠ADB = 180° − 30° = 150°
Therefore, angle subtended by the chord at a point on the major arc and the minor arc are 30° and 150° respectively.
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