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Circles - Apply arc-angle relationships including central and inscribed angle theorems

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 centre is double the angle subtended by it at any point on the remaining part of the circle. This means if an arc PQPQ subtends ∠POQ\angle POQ at center OO and ∠PAQ\angle PAQ at point AA on the circle, then ∠POQ=2∠PAQ\angle POQ = 2\angle PAQ.

Diagram showing angle at center is double the angle at the circumference.
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Angles in the same segment of a circle are equal. If points AA and BB are on the same arc, then ∠PAQ=∠PBQ\angle PAQ = \angle PBQ for any points PP and QQ on the other arc.

Diagram showing angles in the same segment are equal.
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The angle in a semicircle is a right angle (90∘90^{\circ}). If ABAB is the diameter of a circle, then any point CC on the circumference will satisfy ∠ACB=90∘\angle ACB = 90^{\circ}.

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In a cyclic quadrilateral, the sum of opposite angles is 180∘180^{\circ}. For quadrilateral ABCDABCD where all vertices lie on a circle, ∠A+∠C=180∘\angle A + \angle C = 180^{\circ} and ∠B+∠D=180∘\angle B + \angle D = 180^{\circ}.

📐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.

Problem 3:

In the given figure, OO is the center of the circle. If ∠OBC=40∘\angle OBC = 40^{\circ}, find the measure of ∠BAC\angle BAC.

Circle with center O, triangle OBC and inscribed angle BAC.

Solution:

  1. In △OBC\triangle OBC, OB=OCOB = OC (Radii of the same circle).
  2. Therefore, ∠OCB=∠OBC=40∘\angle OCB = \angle OBC = 40^{\circ} (Angles opposite to equal sides).
  3. Sum of angles in △OBC\triangle OBC is 180∘180^{\circ}: ∠BOC+∠OBC+∠OCB=180∘\angle BOC + \angle OBC + \angle OCB = 180^{\circ} ∠BOC+40∘+40∘=180∘\angle BOC + 40^{\circ} + 40^{\circ} = 180^{\circ} ∠BOC=180∘−80∘=100∘\angle BOC = 180^{\circ} - 80^{\circ} = 100^{\circ}
  4. By the central angle theorem, ∠BOC=2×∠BAC\angle BOC = 2 \times \angle BAC.
  5. 100∘=2×∠BAC100^{\circ} = 2 \times \angle BAC
  6. ∠BAC=50∘\angle BAC = 50^{\circ}.

Explanation:

The problem uses the properties of isosceles triangles formed by radii and the theorem relating the central angle to the inscribed angle.

Problem 4:

In a circle, PQPQ is a diameter and RR is a point on the circle. If ∠RPQ=35∘\angle RPQ = 35^{\circ}, calculate ∠PQR\angle PQR.

Circle with diameter PQ and point R forming triangle PQR.

Solution:

  1. Since PQPQ is a diameter, ∠PRQ=90∘\angle PRQ = 90^{\circ} (Angle in a semicircle).
  2. In △PQR\triangle PQR, the sum of angles is 180∘180^{\circ}: ∠PRQ+∠RPQ+∠PQR=180∘\angle PRQ + \angle RPQ + \angle PQR = 180^{\circ}
  3. Substitute the known values: 90∘+35∘+∠PQR=180∘90^{\circ} + 35^{\circ} + \angle PQR = 180^{\circ}
  4. 125∘+∠PQR=180∘125^{\circ} + \angle PQR = 180^{\circ}
  5. ∠PQR=180∘−125∘=55∘\angle PQR = 180^{\circ} - 125^{\circ} = 55^{\circ}.

Explanation:

This solution relies on the property that any angle inscribed in a semicircle is a right angle, followed by the triangle angle sum property.