Arc  and Angles

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

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