Year 11 Standard · Unit 2 Right-angled triangles and circle geometry · Lesson 7 of 8

Angles at the centre and at the circumference, standing on one chord

Fix two points on a circle and move a third around it. This lesson proves what happens to the angle it makes, and why a diameter is special.

Geometry and TrigonometryNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 7: a triangle has angles 38° and 90°. Find the third angle.
    Answer52°, because 180 − 38 − 90 = 52.
  2. From Year 7: in an isosceles triangle the apex angle is 70°. What are the base angles?
    Answer55° each, because (180 − 70) ÷ 2 = 55.
  3. From Year 9: the exterior angle of a triangle is 120° and one interior opposite angle is 45°. Find the other interior opposite angle.
    Answer75°, because the exterior angle is the sum of the two opposite interior angles.
Words used in this lesson
subtend
to make an angle at a point from two other points. For example: a chord subtends an angle at the centre.
circumference
the edge of a circle, or the distance round it. For example: the rim of a plate.
segment
the region of a circle between a chord and its arc. For example: the cap cut off by a straight cut.
semicircle
half of a circle, cut off by a diameter. For example: half a plate.
major arc
the longer arc between two points on a circle. For example: the long way round.
corollary
a result that follows quickly from one already proved. For example: the semicircle result.

Moving a point round a circle

Two posts, A and B, stand on the edge of a round pond. A surveyor walks round the rest of the edge and at each step measures the angle APB between her lines of sight to the two posts. A second surveyor stands at the centre O and measures the angle AOB.

Predict first: as the first surveyor walks, does her angle APB get bigger, get smaller or stay the same? And how does it compare with the centre angle AOB? Write a guess and a reason, then use the first figure.

80
3
80°angle AOB at the centre
40.0°angle APB at the circumference

Centre angle AOB = 80°, P at position 3: angle APB = 40.0°.

Show me the steps
  1. Centre angle 80°, P at position 3: angle APB = 40.0°, half of the centre angle.
  2. Move P to position 8, the far end of the arc: angle APB is still 40.0°. Angles in the same segment are equal.
  3. Make the centre angle 180°: AB is a diameter, and angle APB = 90.0° at every position. The angle in a semicircle is a right angle.
  4. Make the centre angle 40°: angle APB = 20.0°. Whatever the size of AOB, the angle at the circumference is half of it.

The angle at the centre is twice the angle at the circumference on the same arc. Proof: OP, OA and OB are radii, so triangles OPA and OPB are isosceles. Call the base angles x (at P and A) and y (at P and B). Extend PO to Q. The exterior angle AOQ of triangle OPA is x + x = 2x, and BOQ is 2y. So angle AOB = 2x + 2y = 2(x + y) = 2 × angle APB. That is the case where O lies inside the angle APB. If O lies outside it, A and B are on the same side of the line PQ, and the same two exterior angles give angle AOB = 2x − 2y = 2(x − y), while angle APB = x − y, naming the triangles so that x is the larger base angle. So again angle AOB = 2 × angle APB. If P is on the minor arc, O lies inside the angle APB and the first argument applies unchanged: AOQ + QOB = 2x + 2y is then the reflex angle AOB. So the theorem holds wherever P is, including for a reflex centre angle.

Two consequences. Angles at the circumference on the same arc are equal, because each is half the same centre angle. And if AB is a diameter the centre angle is a straight line, 180°, so the angle in a semicircle is 90°.

35
15
50.0°angle APB
100.0°angle AOB

x = 35°, y = 15°: angle APB = x + y = 50.0° and angle AOB = 2x + 2y = 100.0°, twice the angle at P.

Show me the steps
  1. x = 35°, y = 15°: the angle APB = 35° + 15° = 50.0°, and AOB = 70° + 30° = 100.0°.
  2. Make x = 10° and y = 10°: APB = 20.0° and AOB = 40.0°. The centre angle is always the double.
  3. Make x = 40° and y = 40°: APB = 80.0° and AOB = 160.0°. The relation holds at every setting, because it comes from the proof.
  4. Why 2x: the exterior angle of a triangle is the sum of the two opposite interior angles, and those two are equal (x and x) because OP = OA.

Check that both angles stand on the same arc. The centre angle and the circumference angle must be made by the same two points A and B. And the doubling goes one way: a centre angle of 64° gives a circumference angle of 32°, never 128°. Say “at the centre is double” before you write a number.

Worked examples

1. Prove that the angle at the centre is twice the angle at the circumference. (The centre lies inside the angle APB.)

  1. OP = OA (radii), so triangle OPA is isosceles and angle OAP = angle OPA = x.
  2. Extend PO to Q. Angle AOQ is an exterior angle of triangle OPA, so it equals x + x = 2x.
  3. In the same way OP = OB gives angle BOQ = 2y.
  4. Angle AOB = 2x + 2y = 2(x + y) = 2 × angle APB.

2. The angle AOB at the centre is 116°. Find the angle APB at P on the major arc, and the angle AQB for another point Q on the same arc.

  1. The angle at the circumference is half the angle at the centre: 116 ÷ 2 = 58°.
  2. Q is in the same segment as P, so angle AQB is also 58°.

3. AB is a diameter, P is on the circle and angle PAB = 24°. Find angle PBA.

  1. The angle in a semicircle is 90°, so angle APB = 90°.
  2. Angles in the triangle APB: 24 + 90 + PBA = 180, so PBA = 66°.

Your go. The first step is given. O is the centre. The angle ACB is 34°, where C is on the major arc. Find the angle OAB.

  1. Angle AOB = 2 × 34 = 68°.
  2. Triangle OAB has OA = OB, so it is isosceles. Use the angle sum to find the base angle OAB.
Check the answer

(180 − 68) ÷ 2 = 56°. Check: 56 + 56 + 68 = 180.

Check before you go on

1. The angle AOB at the centre of a circle is 84°. What is the angle at the circumference on the major arc, subtended by the same chord AB?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. A chord AB subtends an angle of 130° at the centre O of a circle. What angle does it subtend at a point on the major arc, in degrees?

3. AB is a diameter of a circle and P is a point on the circle. The angle PAB is 35°. What is the angle PBA, in degrees?

Level 2 · Building

4. O is the centre of a circle, and A, B, C are on the circle with C on the major arc. The angle ACB is 3x° and the angle AOB is (x + 100)°. Find x.

5. AB is a diameter, and C is on the circle. The angle CAB is 3x° and the angle CBA is (2x + 15)°. What is the angle CAB, in degrees?

6. P and Q are two points on the major arc of a circle through A and B, and O is the centre. The angle APB is 38°. What is the angle AOB, in degrees?

Level 3 · Stretch

7. O is the centre of a circle and AB is a chord. The angle OAB is 20°. P is on the major arc. What is the angle APB, in degrees?

8. AB is a diameter of a circle of radius 5 cm. C is on the circle and AC = 6 cm. How long is BC, in cm?

Something to think about

The angle at the centre AOB is 200° (a reflex angle on the major side). What is the angle at the circumference on the minor arc? Use the theorem and then say what the angle in the other segment must be.

One way to think about it

100°. The key card's proof covers this case: P on the minor arc has O inside the angle APB, so AOQ + QOB is the reflex angle 200° and APB = 200° ÷ 2 = 100°. In the major segment the angle is half of the non-reflex angle 360° − 200° = 160°, which is 80°. Notice that 100° + 80° = 180°: lesson 8 proves that this always happens, using this reflex case.

Exit ticket

9. O is the centre of a circle and A, B and C are on the circle. The angle ABC is 31°, where B is on the major arc. What is the angle AOC?