GCSE Maths: The angle at the centre (Higher)
Part of the GCSE Maths circle theorems chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
The angle at the centre
Look at an arc AB of a circle. Two kinds of angle can be drawn standing on that arc, which means their arms end at A and B. One has its corner at the centre O. The other has its corner at a point C on the circumference (the rest of the circle). The theorem says the angle at the centre is exactly twice the angle at the circumference.
The reason to write is “the angle at the centre is twice the angle at the circumference”. To use it, always do this first: name the arc both angles stand on. If you pick the wrong arc, you will double or halve the wrong angle. Then double to go from the circumference to the centre, and halve to go from the centre to the circumference.
Why is it true? Join C to O and carry on to the point D on the other side of the circle. OA, OB and OC are radii, so triangle OAC is isosceles with base angles p. Its angle at O is 180° − 2p. Angles on the straight line CD add to 180°, so angle AOD = 2p. In the same way triangle OBC has base angles q and angle BOD = 2q. So angle AOB = 2p + 2q = 2(p + q) = 2 × angle ACB. The full proof is worked example 26.3.5.
Special cases. (1) If the centre O lies on the line AB (a diameter) the angle at the centre is 180°, which gives the angle in a semicircle: 90°. (2) If C is on the minor arc AB, then the angle at the circumference stands on the major arc, so it matches the reflex angle at the centre (the one bigger than 180°). Find the reflex angle with 360° − the angle shown, then halve it. Worked example 26.3.3 shows this.
Key ideas
- The angle at the centre is twice the angle at the circumference, when both stand on the same arc.
- Reason: “the angle at the centre is twice the angle at the circumference”.
- Circumference to centre: double. Centre to circumference: halve.
- If the point is on the minor arc, use the reflex angle at the centre: 360° minus the angle shown.
- Radii OA and OB make an isosceles triangle, so the centre theorem often combines with base angles.
Watch out
Both angles must stand on the same arc. Do not double an angle that is not made from the same two points A and B. And remember: the angle at the circumference is the smaller one (unless the point is on the minor arc).Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.
Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.
- Spot the two angles. Angle at the centre and angle at the circumference are both on the same arc AB. AOB is at the centre and ACB is at the circumference.Centre: angle AOB = ?
Circumference: angle ACB = 42° - Choose the rule. We are going from the circumference to the centre, so we use the circle theorem for the angle at the centre.angle at the centre = 2 × angle at the circumference
- Double it. The angle at the centre is twice the angle at the circumference.Angle AOB = 2 × 42° = 84°
Check. Half of 84° is 42°. ✓
Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.
Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.
Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.
- Find angle AOB. Angle at the centre and angle at the circumference are both on the same arc AB.Angle AOB = 2 × 50° = 100° (reason: the angle at the centre is twice the angle at the circumference)
- Spot the isosceles triangle. OA and OB are radii, so they are equal. Triangle OAB is isosceles and its base angles are equal.Angle OAB = angle OBA (the base angles of an isosceles triangle are equal)
- Use the angle sum in triangle OAB. The angles add up to 180°. Take off the angle at O and share the rest equally.2 × angle OAB = 180° − 100° = 80°
Angle OAB = 80° ÷ 2 = 40°
Check. 40° + 40° + 100° = 180°. ✓ Also 40 = 90 − 50. ✓
Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.
Key words. The minor arc is the shorter arc between two points and the major arc is the longer one. A reflex angle is bigger than 180°.
Method. If the point is on the minor arc, the angle at the circumference stands on the major arc, so we use the reflex angle at the centre.
- Decide which arc the angle stands on. C is on the minor arc AB, so angle ACB looks at the chord AB from the minor-arc side. It stands on the other arc, the major arc.angle ACB stands on the major arc AB
- Find the reflex angle at the centre. Angle ACB stands on the major arc AB (the long way round, the arc that does not contain C). The matching angle at the centre is the reflex angle AOB.Reflex angle AOB = 360° − 140° = 220° (angles around a point add up to 360°)
- Halve it. The angle at the centre is twice the angle at the circumference.Angle ACB = 220° ÷ 2 = 110° (the angle at the centre is twice the angle at the circumference)
Check. Using the cyclic quadrilateral idea: 110° + 70° = 180°, where 70° is the angle at the circumference from the major arc. ✓
Watch out. An angle at the circumference on the minor arc is obtuse. Do not just halve the angle shown at O; halve the reflex angle.
- Work out the value of x.
- Work out the size of angle ACB.
Method. Make an equation from the circle theorem (angle at the centre = 2 × angle at the circumference), solve it, then substitute back.
- Relate the two angles. Both angles stand on arc AB. The angle at the centre is twice the angle at the circumference.(5x + 12) = 2 × (2x + 12) (the angle at the centre is twice the angle at the circumference)
- Expand the bracket on the right. Multiply both terms inside by 2.2 × (2x + 12) = 4x + 24
- Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.5x + 12 = 4x + 24
- Subtract 4x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.5x + 12 − 4x = 4x + 24 − 4x
x + 12 = 24 - Subtract 12 from both sides. We want the number terms on the other side.x = 24 − 12
x = 12 - Read off x. x = 12
- Work out angle ACB. Put x into the expression for the angle.Angle ACB = 2 × 12 + 12 = 36°
Check. Angle AOB = 5 × 12 + 12 = 72° and 2 × 36° = 72°. ✓
Watch out. Put the whole angle in brackets before doubling: 2(3x + 4) = 6x + 8, not 6x + 4.
Method. Join C to O and carry on to D. Use the isosceles triangles OAC and OBC and angles on a straight line.
- Draw the line COD. It passes through the centre. We can now split the problem into two halves, one for A and one for B.OA = OB = OC (radii)
- Triangle OAC. OA = OC, so the triangle is isosceles and its base angles are equal. Call each one p.angle OAC = angle OCA = p, so angle AOC = 180° − 2p
- Find angle AOD. C, O and D are in a straight line, so angle AOC and angle AOD add up to 180°.angle AOD = 180° − (180° − 2p) = 2p
- Do the same for triangle OBC. OB = OC, so the base angles are equal. Call each one q.angle OBC = angle OCB = q, and in the same way angle BOD = 2q
- Add the two angles at the centre. Angle AOB is made of angle AOD and angle DOB side by side.angle AOB = 2p + 2q = 2(p + q)
- Compare with the angle at C. Angle ACB is made of angle ACO and angle OCB.angle ACB = p + q, so angle AOB = 2 × angle ACB. ✓
Check. Test: if p = 30° and q = 20° then angle ACB = 50° and angle AOB = 2 × 30 + 2 × 20 = 100°, which is twice 50°. ✓
Watch out. The proof shown is for the case where O is inside angle ACB. The other positions of O work the same way.
Exercise 26C The angle at the centre
Non-calculatorKey words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.
Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.
- Spot the two angles. Angle at the centre and angle at the circumference are both on the same arc AB. AOB is at the centre and ACB is at the circumference.Centre: angle AOB = ?
Circumference: angle ACB = 35° - Choose the rule. We are going from the circumference to the centre, so we use the circle theorem for the angle at the centre.angle at the centre = 2 × angle at the circumference
- Double it. The angle at the centre is twice the angle at the circumference.Angle AOB = 2 × 35° = 70°
Check. Half of 70° is 35°. ✓
Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.
Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.
Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.
- Spot the two angles. Angle at the centre and angle at the circumference are both on the same arc AB. AOB is at the centre and ACB is at the circumference.Centre: angle AOB = ?
Circumference: angle ACB = 55° - Choose the rule. We are going from the circumference to the centre, so we use the circle theorem for the angle at the centre.angle at the centre = 2 × angle at the circumference
- Double it. The angle at the centre is twice the angle at the circumference.Angle AOB = 2 × 55° = 110°
Check. Half of 110° is 55°. ✓
Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.
Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.
Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.
- Spot the two angles. Angle at the centre and angle at the circumference are both on the same arc AB. AOB is at the centre and ACB is at the circumference.Centre: angle AOB = ?
Circumference: angle ACB = 26° - Choose the rule. We are going from the circumference to the centre, so we use the circle theorem for the angle at the centre.angle at the centre = 2 × angle at the circumference
- Double it. The angle at the centre is twice the angle at the circumference.Angle AOB = 2 × 26° = 52°
Check. Half of 52° is 26°. ✓
Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.
Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.
Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.
- Spot the two angles. Angle at the centre and angle at the circumference are both on the same arc AB. AOB is at the centre and ACB is at the circumference.Centre: angle AOB = 90°
Circumference: angle ACB = ? - Choose the rule. We are going from the centre to the circumference, so we use the circle theorem the other way round.angle at the circumference = angle at the centre ÷ 2
- Halve it. The angle at the centre is twice the angle at the circumference, so the angle at the circumference is half the angle at the centre.Angle ACB = 90° ÷ 2 = 45°
Check. Double 45° is 90°. ✓
Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.
Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.
Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.
- Spot the two angles. Angle at the centre and angle at the circumference are both on the same arc AB. AOB is at the centre and ACB is at the circumference.Centre: angle AOB = 124°
Circumference: angle ACB = ? - Choose the rule. We are going from the centre to the circumference, so we use the circle theorem the other way round.angle at the circumference = angle at the centre ÷ 2
- Halve it. The angle at the centre is twice the angle at the circumference, so the angle at the circumference is half the angle at the centre.Angle ACB = 124° ÷ 2 = 62°
Check. Double 62° is 124°. ✓
Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.
Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.
Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.
- Spot the two angles. Angle at the centre and angle at the circumference are both on the same arc AB. AOB is at the centre and ACB is at the circumference.Centre: angle AOB = 66°
Circumference: angle ACB = ? - Choose the rule. We are going from the centre to the circumference, so we use the circle theorem the other way round.angle at the circumference = angle at the centre ÷ 2
- Halve it. The angle at the centre is twice the angle at the circumference, so the angle at the circumference is half the angle at the centre.Angle ACB = 66° ÷ 2 = 33°
Check. Double 33° is 66°. ✓
Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.
Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.
Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.
- Find angle AOB. Angle at the centre and angle at the circumference are both on the same arc AB.Angle AOB = 2 × 38° = 76° (reason: the angle at the centre is twice the angle at the circumference)
- Spot the isosceles triangle. OA and OB are radii, so they are equal. Triangle OAB is isosceles and its base angles are equal.Angle OAB = angle OBA (the base angles of an isosceles triangle are equal)
- Use the angle sum in triangle OAB. The angles add up to 180°. Take off the angle at O and share the rest equally.2 × angle OAB = 180° − 76° = 104°
Angle OAB = 104° ÷ 2 = 52°
Check. 52° + 52° + 76° = 180°. ✓ Also 52 = 90 − 38. ✓
Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.
Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.
Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.
- Find angle AOB. Angle at the centre and angle at the circumference are both on the same arc AB.Angle AOB = 2 × 60° = 120° (reason: the angle at the centre is twice the angle at the circumference)
- Spot the isosceles triangle. OA and OB are radii, so they are equal. Triangle OAB is isosceles and its base angles are equal.Angle OAB = angle OBA (the base angles of an isosceles triangle are equal)
- Use the angle sum in triangle OAB. The angles add up to 180°. Take off the angle at O and share the rest equally.2 × angle OAB = 180° − 120° = 60°
Angle OAB = 60° ÷ 2 = 30°
Check. 30° + 30° + 120° = 180°. ✓ Also 30 = 90 − 60. ✓
Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.
Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.
Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.
- Find angle AOB. Angle at the centre and angle at the circumference are both on the same arc AB.Angle AOB = 2 × 32° = 64° (reason: the angle at the centre is twice the angle at the circumference)
- Spot the isosceles triangle. OA and OB are radii, so they are equal. Triangle OAB is isosceles and its base angles are equal.Angle OAB = angle OBA (the base angles of an isosceles triangle are equal)
- Use the angle sum in triangle OAB. The angles add up to 180°. Take off the angle at O and share the rest equally.2 × angle OAB = 180° − 64° = 116°
Angle OAB = 116° ÷ 2 = 58°
Check. 58° + 58° + 64° = 180°. ✓ Also 58 = 90 − 32. ✓
Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.
Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.
Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.
- Spot the isosceles triangle. OA and OB are radii so they are equal. Triangle OAB is isosceles.Angle OBA = angle OAB = 60° (the base angles of an isosceles triangle are equal)
- Find angle AOB. Use the angle sum of triangle OAB.Angle AOB = 180° − 60° − 60° = 60° (the angles in a triangle add up to 180°)
- Halve to get angle ACB. The angle at the circumference is half the angle at the centre.Angle ACB = 60° ÷ 2 = 30° (the angle at the centre is twice the angle at the circumference)
Check. 30 = 90 − 60. ✓
Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.
Key words. An arc is a part of the circumference. An angle stands on an arc if its two arms end at the two ends of that arc.
Method. Find the angle at the centre and the angle at the circumference that stand on the same arc, then use: angle at the centre = 2 × angle at the circumference.
- Spot the isosceles triangle. OA and OB are radii so they are equal. Triangle OAB is isosceles.Angle OBA = angle OAB = 50° (the base angles of an isosceles triangle are equal)
- Find angle AOB. Use the angle sum of triangle OAB.Angle AOB = 180° − 50° − 50° = 80° (the angles in a triangle add up to 180°)
- Halve to get angle ACB. The angle at the circumference is half the angle at the centre.Angle ACB = 80° ÷ 2 = 40° (the angle at the centre is twice the angle at the circumference)
Check. 40 = 90 − 50. ✓
Watch out. Both angles must stand on the same arc (here arc AB). An angle at the centre is always the bigger one.
Key words. The minor arc is the shorter arc between two points and the major arc is the longer one. A reflex angle is bigger than 180°.
Method. If the point is on the minor arc, the angle at the circumference stands on the major arc, so we use the reflex angle at the centre.
- Decide which arc the angle stands on. C is on the minor arc AB, so angle ACB looks at the chord AB from the minor-arc side. It stands on the other arc, the major arc.angle ACB stands on the major arc AB
- Find the reflex angle at the centre. Angle ACB stands on the major arc AB (the long way round, the arc that does not contain C). The matching angle at the centre is the reflex angle AOB.Reflex angle AOB = 360° − 100° = 260° (angles around a point add up to 360°)
- Halve it. The angle at the centre is twice the angle at the circumference.Angle ACB = 260° ÷ 2 = 130° (the angle at the centre is twice the angle at the circumference)
Check. Using the cyclic quadrilateral idea: 130° + 50° = 180°, where 50° is the angle at the circumference from the major arc. ✓
Watch out. An angle at the circumference on the minor arc is obtuse. Do not just halve the angle shown at O; halve the reflex angle.
Key words. The minor arc is the shorter arc between two points and the major arc is the longer one. A reflex angle is bigger than 180°.
Method. If the point is on the minor arc, the angle at the circumference stands on the major arc, so we use the reflex angle at the centre.
- Decide which arc the angle stands on. C is on the minor arc AB, so angle ACB looks at the chord AB from the minor-arc side. It stands on the other arc, the major arc.angle ACB stands on the major arc AB
- Find the reflex angle at the centre. Angle ACB stands on the major arc AB (the long way round, the arc that does not contain C). The matching angle at the centre is the reflex angle AOB.Reflex angle AOB = 360° − 130° = 230° (angles around a point add up to 360°)
- Halve it. The angle at the centre is twice the angle at the circumference.Angle ACB = 230° ÷ 2 = 115° (the angle at the centre is twice the angle at the circumference)
Check. Using the cyclic quadrilateral idea: 115° + 65° = 180°, where 65° is the angle at the circumference from the major arc. ✓
Watch out. An angle at the circumference on the minor arc is obtuse. Do not just halve the angle shown at O; halve the reflex angle.
Key words. The minor arc is the shorter arc between two points and the major arc is the longer one. A reflex angle is bigger than 180°.
Method. If the point is on the minor arc, the angle at the circumference stands on the major arc, so we use the reflex angle at the centre.
- Decide which arc the angle stands on. C is on the minor arc AB, so angle ACB looks at the chord AB from the minor-arc side. It stands on the other arc, the major arc.angle ACB stands on the major arc AB
- Find the reflex angle at the centre. Angle ACB stands on the major arc AB (the long way round, the arc that does not contain C). The matching angle at the centre is the reflex angle AOB.Reflex angle AOB = 360° − 160° = 200° (angles around a point add up to 360°)
- Halve it. The angle at the centre is twice the angle at the circumference.Angle ACB = 200° ÷ 2 = 100° (the angle at the centre is twice the angle at the circumference)
Check. Using the cyclic quadrilateral idea: 100° + 80° = 180°, where 80° is the angle at the circumference from the major arc. ✓
Watch out. An angle at the circumference on the minor arc is obtuse. Do not just halve the angle shown at O; halve the reflex angle.
Method. The centre O is on the line AB, but the circle theorem still works: the angle at the centre is twice the angle at the circumference.
- Spot the two angles. Angle COB is at the centre and angle CAB is at the circumference. Both stand on arc CB.Centre: angle COB = ?
Circumference: angle CAB = 28° - Double it. The angle at the centre is twice the angle at the circumference.Angle COB = 2 × 28° = 56°
Check. Triangle OAC is isosceles (OA = OC), so angle OCA = 28°. The exterior angle COB = 28° + 28° = 56°. ✓
Watch out. Name the arc both angles stand on (here arc CB) before you double or halve.
Method. The centre O is on the line AB, but the circle theorem still works: the angle at the centre is twice the angle at the circumference.
- Spot the two angles. Angle COB is at the centre and angle CAB is at the circumference. Both stand on arc CB.Centre: angle COB = ?
Circumference: angle CAB = 41° - Double it. The angle at the centre is twice the angle at the circumference.Angle COB = 2 × 41° = 82°
Check. Triangle OAC is isosceles (OA = OC), so angle OCA = 41°. The exterior angle COB = 41° + 41° = 82°. ✓
Watch out. Name the arc both angles stand on (here arc CB) before you double or halve.
Method. Make an equation from the circle theorem (angle at the centre = 2 × angle at the circumference), solve it, then substitute back.
- Relate the two angles. Both angles stand on arc AB. The angle at the centre is twice the angle at the circumference.(4x + 32) = 2 × (3x + 7) (the angle at the centre is twice the angle at the circumference)
- Expand the bracket on the right. Multiply both terms inside by 2.2 × (3x + 7) = 6x + 14
- Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.6x + 14 = 4x + 32 (written the other way round so that the x terms stay positive)
- Subtract 4x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.6x + 14 − 4x = 4x + 32 − 4x
2x + 14 = 32 - Subtract 14 from both sides. We want the number terms on the other side.2x = 32 − 14
2x = 18 - Divide both sides by 2. 2x means 2 × x, so dividing by 2 leaves x on its own.x = 18 ÷ 2 = 9
- Work out angle ACB. Put x into the expression for the angle.Angle ACB = 3 × 9 + 7 = 34°
Check. Angle AOB = 4 × 9 + 32 = 68° and 2 × 34° = 68°. ✓
Watch out. Put the whole angle in brackets before doubling: 2(3x + 4) = 6x + 8, not 6x + 4.
Method. Make an equation from the circle theorem (angle at the centre = 2 × angle at the circumference), solve it, then substitute back.
- Relate the two angles. Both angles stand on arc AB. The angle at the centre is twice the angle at the circumference.(4x + 32) = 2 × (3x + 7) (the angle at the centre is twice the angle at the circumference)
- Expand the bracket on the right. Multiply both terms inside by 2.2 × (3x + 7) = 6x + 14
- Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.6x + 14 = 4x + 32 (written the other way round so that the x terms stay positive)
- Subtract 4x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.6x + 14 − 4x = 4x + 32 − 4x
2x + 14 = 32 - Subtract 14 from both sides. We want the number terms on the other side.2x = 32 − 14
2x = 18 - Divide both sides by 2. 2x means 2 × x, so dividing by 2 leaves x on its own.x = 18 ÷ 2 = 9
- Work out angle ACB. Put x into the expression for the angle.Angle ACB = 3 × 9 + 7 = 34°
Check. Angle AOB = 4 × 9 + 32 = 68° and 2 × 34° = 68°. ✓
Watch out. Put the whole angle in brackets before doubling: 2(3x + 4) = 6x + 8, not 6x + 4.
Method. Make an equation from the circle theorem (angle at the centre = 2 × angle at the circumference), solve it, then substitute back.
- Relate the two angles. Both angles stand on arc AB. The angle at the centre is twice the angle at the circumference.(3x + 5) = 2 × (x + 10) (the angle at the centre is twice the angle at the circumference)
- Expand the bracket on the right. Multiply both terms inside by 2.2 × (x + 10) = 2x + 20
- Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.3x + 5 = 2x + 20
- Subtract 2x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.3x + 5 − 2x = 2x + 20 − 2x
x + 5 = 20 - Subtract 5 from both sides. We want the number terms on the other side.x = 20 − 5
x = 15 - Read off x. x = 15
- Work out angle ACB. Put x into the expression for the angle.Angle ACB = 1 × 15 + 10 = 25°
Check. Angle AOB = 3 × 15 + 5 = 50° and 2 × 25° = 50°. ✓
Watch out. Put the whole angle in brackets before doubling: 2(3x + 4) = 6x + 8, not 6x + 4.
Method. Make an equation from the circle theorem (angle at the centre = 2 × angle at the circumference), solve it, then substitute back.
- Relate the two angles. Both angles stand on arc AB. The angle at the centre is twice the angle at the circumference.(3x + 5) = 2 × (x + 10) (the angle at the centre is twice the angle at the circumference)
- Expand the bracket on the right. Multiply both terms inside by 2.2 × (x + 10) = 2x + 20
- Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.3x + 5 = 2x + 20
- Subtract 2x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.3x + 5 − 2x = 2x + 20 − 2x
x + 5 = 20 - Subtract 5 from both sides. We want the number terms on the other side.x = 20 − 5
x = 15 - Read off x. x = 15
- Work out angle ACB. Put x into the expression for the angle.Angle ACB = 1 × 15 + 10 = 25°
Check. Angle AOB = 3 × 15 + 5 = 50° and 2 × 25° = 50°. ✓
Watch out. Put the whole angle in brackets before doubling: 2(3x + 4) = 6x + 8, not 6x + 4.
Method. Find the angle at the centre from the clock, then halve it to get the angle at the circumference.
- Find angle AOB. From 12 to 4 is 4 hours. Each hour is 30° at the centre.4 × 30° = 120°
- Check which arc. The 9 is on the long way round from 12 to 4, so C is on the major arc and the angle at the centre is 120° (not reflex).C is on the major arc AB
- Halve it. The angle at the centre is twice the angle at the circumference.angle ACB = 120° ÷ 2 = 60° (the angle at the centre is twice the angle at the circumference)
Check. The 9 is not between the 12 and the 4, so it is not on the short arc. 2 × 60° = 120°. ✓
Watch out. Check that the point is on the arc that does not lie between A and B.
Method. Check the rule and decide which angle is bigger.
- Recall the rule. The angle at the centre is twice the angle at the circumference. So the centre angle is the bigger one.centre = 2 × circumference
- Spot the mistake. Ben made the circumference angle (100°) twice as big as the centre angle (50°). That is the wrong way round.He doubled when he should have halved.
- Correct it. Halve the angle at the centre.angle at the circumference = 50° ÷ 2 = 25°
Check. 2 × 25° = 50°. ✓
Watch out. The centre angle is the big one, the circumference angle is the small one.
Method. Find angle AOB with the centre theorem, then use the isosceles triangle OAB.
- Angle at the centre. Angle AOB and angle ACB stand on arc AB, and the angle at the centre is twice the angle at the circumference.angle AOB = 2x (angle at the centre is twice the angle at the circumference)
- Spot the isosceles triangle. OA = OB (radii), so the base angles are equal.angle OAB = angle OBA
- Use the angle sum. Subtract 2x from 180°, then halve.angle OAB = (180° − 2x) ÷ 2 = 90° − x ✓
Check. Try x = 30°: angle AOB = 60°, and the base angles are 60° each. 90 − 30 = 60. ✓
Watch out. This result is used in many exam problems. It links the angle at the circumference to the base angles of the radii triangle.
Method. A diameter is a straight line through the centre, so the angle at the centre is a straight angle.
- Find the angle at the centre. A, O and B are on a straight line, so angle AOB is a straight angle.angle AOB = 180°
- Halve it. For any point C on the circle, the angle at the centre is twice the angle at the circumference.angle ACB = 180° ÷ 2 = 90°
- Conclude. So the angle in a semicircle is 90°. ✓
Check. This agrees with the proof in section 26.2 that used isosceles triangles. ✓
Watch out. The two theorems are connected: the semicircle theorem is the special case when the centre angle is 180°.