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Circles - Angle Subtended by an Arc of a Circle

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The angle subtended by an arc at the center of a circle is double the angle subtended by it at any point on the remaining part of the circle. Visually, if you have an arc PQPQ and the center OO, the angle ∠POQ\angle POQ will be twice the size of ∠PAQ\angle PAQ, where AA is any point on the major arc.

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Angles subtended by the same arc (or in the same segment) of a circle are equal. Imagine a chord ABAB forming a segment; if you draw multiple angles from AA and BB to any points CC and DD on the circumference within that same segment, then ∠ACB=∠ADB\angle ACB = \angle ADB.

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The angle subtended by a semicircle at any point on the circle is a right angle (90∘90^{\circ}). Visually, if a triangle is constructed using the diameter of the circle as its base and the third vertex lies anywhere on the circumference, that triangle will always be a right-angled triangle.

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If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the segment, then the four points lie on a circle (they are concyclic). This is the converse of the 'angles in the same segment' theorem.

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In a cyclic quadrilateral, the sum of either pair of opposite angles is 180∘180^{\circ}. Visually, if a four-sided figure has all its vertices touching the circle's boundary, the angles across from each other will always add up to a straight line's degree measure.

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If the sum of a pair of opposite angles of a quadrilateral is 180∘180^{\circ}, the quadrilateral is cyclic. This is a crucial test to determine if a circle can be circumscribed around a given four-sided shape.

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Congruent arcs of a circle subtend equal angles at the center. Visually, if two different arcs ABAB and CDCD have the same length, the angles ∠AOB\angle AOB and ∠COD\angle COD formed at the center OO must be equal.

📐Formulae

∠at center=2×∠at circumference\angle \text{at center} = 2 \times \angle \text{at circumference}

∠BAC=∠BDC\angle BAC = \angle BDC (Angles in the same segment)

∠in a semicircle=90∘\angle \text{in a semicircle} = 90^{\circ}

In cyclic quadrilateral ABCDABCD: ∠A+∠C=180∘\angle A + \angle C = 180^{\circ} and ∠B+∠D=180∘\angle B + \angle D = 180^{\circ}

Degree measure of arc ABAB: m(AB)=∠AOBm(AB) = \angle AOB

💡Examples

Problem 1:

In a circle with center OO, an arc ABCABC subtends an angle of 110∘110^{\circ} at the center. Find the measure of the angle ∠ADC\angle ADC where DD is a point on the major arc.

Solution:

Step 1: Identify the given information. The angle subtended by arc ABCABC at the center is ∠AOC=110∘\angle AOC = 110^{\circ}.\nStep 2: Apply the theorem that the angle at the center is double the angle at the circumference. Therefore, ∠AOC=2×∠ADC\angle AOC = 2 \times \angle ADC.\nStep 3: Substitute the value: 110∘=2×∠ADC110^{\circ} = 2 \times \angle ADC.\nStep 4: Solve for ∠ADC\angle ADC: ∠ADC=110∘2=55∘\angle ADC = \frac{110^{\circ}}{2} = 55^{\circ}.

Explanation:

We use the central angle theorem which relates the angle at the center to the angle at any point on the remaining part of the circle.

Problem 2:

Points A,B,CA, B, C and DD are four points on a circle. ACAC and BDBD intersect at a point EE such that ∠BEC=130∘\angle BEC = 130^{\circ} and ∠ECD=20∘\angle ECD = 20^{\circ}. Find ∠BAC\angle BAC.

Solution:

Step 1: In △CED\triangle CED, ∠BEC\angle BEC is an exterior angle. Therefore, ∠BEC=∠ECD+∠EDC\angle BEC = \angle ECD + \angle EDC.\nStep 2: Substitute the known values: 130∘=20∘+∠EDC130^{\circ} = 20^{\circ} + \angle EDC.\nStep 3: Calculate ∠EDC=130∘−20∘=110∘\angle EDC = 130^{\circ} - 20^{\circ} = 110^{\circ}.\nStep 4: Recognize that ∠BAC\angle BAC and ∠EDC\angle EDC (which is the same as ∠BDC\angle BDC) are angles in the same segment subtended by the arc BCBC.\nStep 5: Since angles in the same segment are equal, ∠BAC=∠BDC=110∘\angle BAC = \angle BDC = 110^{\circ}.

Explanation:

This problem combines the exterior angle property of a triangle with the theorem that angles in the same segment of a circle are equal.