Angle properties of circles | Year 11

01 / LEARN

The idea, step by step

Identify the arc first

A central angle has its vertex at the centre. An inscribed angle has its vertex on the circle and its sides along chords. For the same intercepted arc, the central angle is twice the inscribed angle. It is the arc not containing the inscribed vertex that matters.

Angles in the same segment subtend the same chord from the same side and are equal. If two vertices lie on opposite sides of a chord, their inscribed angles are supplementary instead. Draw the chord and mark which arc belongs to each angle.

Use the diameter as a special case

A diameter subtends a straight central angle of $180^{\circ}$ . Therefore any triangle with the diameter as one side and its third vertex on the circle has a right angle at that third vertex. This is the angle-in-a-semicircle theorem.

These facts can be justified by joining the centre to the vertices: the resulting triangles are isosceles because radii are equal. In a proof, name the circle theorem as well as giving the numerical calculation. A diagram alone is not a proof.

A common trap

Halving whichever centre angle is drawn without checking its arc.

Instead: Use the central angle, possibly reflex, on the arc intercepted by the inscribed angle.

02 / WORKED EXAMPLES

See the thinking

Example 1

A shared minor arc

The minor arc AB has central angle $132^{\circ}$ . P is on the major arc AB. Find $∠ A P B$ .

P is not on the minor arc AB, so its angle intercepts that minor arc.
An inscribed angle is half the central angle on that arc.
$∠ A P B = 132^{\circ} / 2 = 66^{\circ}$ .

66^{\circ}

Example 2

A diameter gives a length

AB is a diameter and P is on the circle. If $A P = 10$ cm and $B P = 24$ cm, find AB.

The angle in a semicircle gives $∠ A P B = 90^{\circ}$ .
Pythagoras gives $A B^{2} = 10^{2} + 24^{2} = 676$ .
Take the positive root because AB is a length.

A B = 26

cm.

Check your understanding

The minor central angle AOB is $100^{\circ}$ and P is on the major arc AB. Find angle APB.

Show answer

$50^{\circ}$ .

P intercepts the minor arc, so take half of $100^{\circ}$ .

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

An inscribed angle subtends the same arc as a central angle of $146^{\circ}$ . Find it.
2 marks
An inscribed angle is $48^{\circ}$ . Find the central angle on its intercepted arc.
2 marks
P and Q lie on the same segment of chord AB. If $∠ A P B = 57^{\circ}$ , find $∠ A Q B$ .
2 marks
AB is a diameter. In triangle APB, angle PAB is $31^{\circ}$ . Find angle PBA.
2 marks
The major arc AB has central reflex angle $248^{\circ}$ . P lies on the minor arc AB. Find angle APB.
3 marks
Explain why two angles subtending chord AB from opposite segments sum to $180^{\circ}$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

An inscribed angle subtends the same arc as a central angle of $146^{\circ}$ . Find it.
2 marks
$146 / 2 = 73$ .
$73^{\circ}$ .
An inscribed angle is $48^{\circ}$ . Find the central angle on its intercepted arc.
2 marks
The central angle is twice the inscribed angle.
$96^{\circ}$ .
P and Q lie on the same segment of chord AB. If $∠ A P B = 57^{\circ}$ , find $∠ A Q B$ .
2 marks
Angles in the same segment are equal.
$57^{\circ}$ .
AB is a diameter. In triangle APB, angle PAB is $31^{\circ}$ . Find angle PBA.
2 marks
Angle APB is $90^{\circ}$ ; use $180 - 90 - 31 = 59$ .
$59^{\circ}$ .
The major arc AB has central reflex angle $248^{\circ}$ . P lies on the minor arc AB. Find angle APB.
3 marks
P intercepts the major arc, so the correct calculation is $248 / 2 = 124$ .
$124^{\circ}$ .
Explain why two angles subtending chord AB from opposite segments sum to $180^{\circ}$ .
3 marks
Each inscribed angle is half its intercepted arc measure.
The two arcs make the complete circle, so the angle sum is $360^{\circ} / 2 = 180^{\circ}$ .
They intercept arcs whose measures total $360^{\circ}$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: geometric reasoning and real-number/proof material, regrouped into a school chapter; all exercises are original.

Official curriculum source ↗

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