# Arc and Angles

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**Arc and angles Subtended by them**

**Some results on angles subtended by arcs of a circle either at the center of the circle or at a point on its circumference.**

1) The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.

∠POQ = 2 ∠PRQ or ∠ PRQ = ½ ∠ POQ

2) Angles in the same segment of a circle are equal.

∠PRQ = ∠PSQ ( both the angles are subtended in the same arc PQ )

3) The angle in a semicircle is a right angle.

∠PRQ = 90

^{0}(PQ is a diameter )

**Some solved examples on arc and angles**

1) In the given figure, A,B,C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130

^{0}and ∠ECD = 20

^{0}. Find ∠BAC.

**Solution :**∠BEC = 130

^{0}and ∠ECD = 20

^{0}

∠BEC + ∠CED = 180

^{0}( Linear pair angles)

130 + ∠CED = 180

∴ ∠CED = 50

^{0}

∠EDC = 180 – (50 +20) [ In ΔCDE, sum of all the angles in a Δ = 180

^{0})

∠EDC = 110

^{0}

But ∠EDC = ∠CDB ( angles in the same segment are equal )

∴ ∠CDB = 110

^{0}

2) Two circles are drawn with sides AB and AC of a triangle ABC as diameters. The circles intersect at a point D. Prove that D lies on BC.

**Solution :**Join AD.

As AC is a diameter of circle C1

∴ ∠ADC = 90

^{0}[ Angle in a semicircle is a right angle (90) ]

As AB is a diameter of circle C2

∴ ∠ADB = 90

^{0}[ Angle in a semicircle is a right angle (90) ]

∠ ADC + ∠ADB = 90 + 90 = 180

^{0}

⇒ BDC is a straight line ⇒ D lies on BC.

**Circles**

• Circles

• Parts of Circle

• Arc and Chords

• Equal Chords of a Circle

• Arc and Angles

• Cyclic Quadrilaterals

• Tangent to Circle

• Circles

• Parts of Circle

• Arc and Chords

• Equal Chords of a Circle

• Arc and Angles

• Cyclic Quadrilaterals

• Tangent to Circle

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