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E. Geometry and Measurement · Geometry and Measurement: Circles, Triangles, Measurement Systems, Pythagoras and Volume

Circle Properties: Central and Inscribed Angles

E1.2 A1About 70 minutes
Learning goals
  • I can tell a central angle from an inscribed angle and find the arc each one stands on.
  • I can use the property that an inscribed angle is half the central angle on the same arc.
  • I can use the facts that inscribed angles on the same arc are equal and that an angle in a semicircle is 90°.
  • I can use the property that opposite angles of a cyclic quadrilateral add to 180°, and explain why the properties are true.

Words to know

central angle
An angle whose vertex is the centre of a circle and whose arms are radii.
inscribed angle
An angle whose vertex is on the circle and whose arms are chords.
subtend
An arc or chord subtends an angle when the angle's arms go to the ends of that arc.
semicircle
Half of a circle, cut off by a diameter.
cyclic quadrilateral
A quadrilateral whose four vertices all lie on one circle.
supplementary angles
Two angles that add to 180°.

A hook: the best seat

A soccer goal is 7.32 m wide. Players standing at different spots see the goal at different angles: a bigger angle means a bigger target. Is there a set of spots where the angle is exactly the same?

Yes: every point on a circle through the two goalposts (on the same side of the goal) sees the goal at the same angle. That is one of the circle properties you will learn today.

Central and inscribed angles

A central angle has its vertex at the centre O. An inscribed angle has its vertex on the circle. If both angles stand on (subtend) the same arc AB:

inscribed angle = ½ × central angle

Two more facts follow:

  • All inscribed angles on the same arc are equal (they are all half of the same central angle).
  • An inscribed angle on a diameter is 90°, because the central angle is a straight angle, 180°, and half of 180° is 90°. This is called the angle in a semicircle property (often credited to the Greek mathematician Thales).
AB80°40°OPcentral 80°, inscribed 40°AB40°40°40°same arc, same inscribed angle
Left: the central angle AOB is 80°, so the inscribed angle APB on the same arc is 40°. Right: every inscribed angle standing on arc AB is 40°, wherever its vertex is on the larger arc.

Why is the inscribed angle half?

Take the case where one arm of the inscribed angle passes through O. Let P be on the circle, with PB a diameter, and A another point on the circle.

  1. OP = OA (both radii), so triangle OPA is isosceles: ∠OPA = ∠OAP = x.
  2. The exterior angle of a triangle equals the sum of the two opposite interior angles. So the angle at O outside triangle OPA, which is the central angle ∠AOB, equals x + x = 2x.
  3. So the central angle is twice the inscribed angle. Other cases are made by adding or subtracting two of these.

Cyclic quadrilaterals

If all four vertices of a quadrilateral are on a circle, the two opposite angles together stand on the whole circle (360° of central angle). Half of 360° is 180°, so opposite angles add to 180°.

ABCOangle in a semicircle = 90°65°A95°B115°C85°Dopposite angles add to 180°
Left: AB is a diameter, so the angle at C is 90° wherever C is on the circle. Right: in a cyclic quadrilateral, opposite angles are supplementary: 65° + 115° = 180° and 95° + 85° = 180°.
Worked example

Example 1: Central to inscribed

A central angle is 110°. Find an inscribed angle on the same arc.

  1. Inscribed angle = ½ × central angle.
  2. ½ × 110° = 55°.

55°

Worked example

Example 2: Equal inscribed angles

∠APB = 35° is inscribed on arc AB. Find the central angle AOB and another inscribed angle ∠AQB on the same arc.

  1. Central angle = 2 × 35° = 70°.
  2. ∠AQB stands on the same arc, so it equals ∠APB = 35°.

∠AOB = 70°, ∠AQB = 35°

Worked example

Example 3: Triangle in a semicircle

Triangle ABC is drawn in a circle with AB a diameter. ∠A = 32°. Find ∠B.

  1. AB is a diameter, so ∠C = 90°.
  2. Angles in a triangle add to 180°: ∠B = 180° − 90° − 32° = 58°.

∠B = 58°

Worked example

Example 4: Cyclic quadrilateral

ABCD is a cyclic quadrilateral with ∠A = 75° and ∠B = 100°. Find ∠C and ∠D.

  1. ∠C is opposite ∠A: ∠C = 180° − 75° = 105°.
  2. ∠D is opposite ∠B: ∠D = 180° − 100° = 80°.
  3. Check: 75 + 100 + 105 + 80 = 360°. ✓

∠C = 105°, ∠D = 80°

Worked example

Example 5: Using the isosceles triangle

Central angle ∠AOB = 100°. Find ∠OAB.

  1. OA = OB (radii), so triangle OAB is isosceles and ∠OAB = ∠OBA.
  2. ∠OAB = (180° − 100°) ÷ 2 = 40°.

40°

Watch out!

  • The half rule only works when both angles stand on the same arc.
  • The inscribed angle's vertex must be on the circle, not just inside it.
  • If the vertex is on the small arc, the angle stands on the big arc, so it is more than 90°.
Hands-on activity

The paper-corner diameter

You need: a plate or lid, paper, a pencil, a sheet of paper with a square corner, and a protractor if you have one.

  1. Trace the plate to draw a circle.
  2. Push the square corner of a second sheet of paper so the corner just touches the circle and both edges cross the circle.
  3. Mark the two crossing points, then join them. You have drawn a diameter (the angle in a semicircle is 90°, so the reverse is also true).
  4. Do it again from another spot. Where the two diameters cross is the centre.
  5. Bonus: mark two points A and B on the circle. Draw three inscribed angles on arc AB from different points and measure them. They should match.

Technology option: in a free geometry app such as GeoGebra, construct a circle, points A, B and P on it, and measure ∠APB. Drag P around the circle and watch the angle stay the same.

Just for fun

Why did the inscribed angle feel left out?

Wherever it went on the circle, the central angle was always twice as big.

Your turn

1

A central angle is 140°. Find an inscribed angle on the same arc.

2

An inscribed angle is 28°. Find the central angle on the same arc.

3

Two inscribed angles stand on the same arc. One is 47°. What is the other?

4

Triangle ABC has AB as a diameter of its circle. ∠B = 25°. Find ∠A.

5

One angle of a cyclic quadrilateral is 112°. Find the opposite angle.

6

∠AOB = 70° is a central angle. Find ∠OAB.

7

Explain why an angle in a semicircle is 90°.

More practice: warm-up

1

A central angle is 90°. What is an inscribed angle on the same arc?

2

An inscribed angle is 50°. What is the central angle on the same arc?

3

What is an angle in a semicircle?

4

A cyclic quadrilateral has an angle of 80°. What is the opposite angle?

5

Where is the vertex of a central angle?

More practice: on your own

1

Triangle ABC is in a circle with AB a diameter. AC = 6 cm and BC = 8 cm. Find the radius.

2

In cyclic quadrilateral ABCD, ∠A = 3x and ∠C = 2x. Find x, ∠A and ∠C.

3

∠APB = 35° is inscribed on arc AB. Find ∠AOB and ∠OAB.

4

A quadrilateral has angles 80°, 95°, 100° and 85°, in order around the shape. Could its vertices lie on one circle? Explain.

5

A rectangle 9 cm by 12 cm is drawn inside a circle with all four corners on the circle. Find the circle's radius.

Challenge

1

Points A and B are on a circle with ∠AOB = 100°. Point P is on the small arc AB. Find ∠APB.

2

Prove that the angles of any cyclic quadrilateral add to 360° using only the inscribed angle property.

3

A camera must film a 10 m wide stage AB with a 40° view. Describe all the places the camera can stand (in front of the stage) so the stage exactly fills the view.

Check your understanding

Quiz

1.

A central angle is 124°. What is an inscribed angle on the same arc?

2.

Triangle PQR is inscribed in a circle and PQ is a diameter. What is ∠R?

3.

Two inscribed angles stand on the same arc. One is 38°. The other is:

4.

In a cyclic quadrilateral, ∠A = 64°. What is the opposite angle ∠C?

5.

Central angle ∠AOB = 40°. What is ∠OAB?

6.

Which statement is true?

For parents and teachers

Big idea: These circle properties (part of E1.2) let students find unknown angles in designs, logos, wheels and architecture, and explain why using isosceles triangles and the exterior angle property.

Common difficulties:

  • Doubling instead of halving. Ask: "Where is the vertex: at the centre or on the circle?"
  • Using angles that stand on different arcs. Trace the two arms of each angle to their ends on the circle; both angles must land on the same two points.
  • Forgetting the isosceles triangle made by two radii.

How to help: Try the paper-corner trick; it feels like magic. If you have a phone or computer, a free geometry app lets your teen drag points and watch the angles stay constant.