The idea, step by step
The same arc matters
A central angle and an angle on the circumference intercepting the same arc are related by a factor of two. Identify their shared endpoints and check which arc is opposite the circumference vertex.
An angle on the major arc intercepts the minor arc. A vertex on the other arc may instead intercept a reflex central angle. Blindly halving a labelled minor angle can therefore be wrong.
Angles in one segment
Angles at two circumference points on the same side of a chord and intercepting the same arc are equal. If the chord is a diameter, the intercepted semicircle is 180°, so its circumference angle is 90°.
State which theorem applies. A right angle somewhere in a diagram does not prove that an unrelated chord is a diameter. The converse requires a right angle subtended by that particular chord.
A common trap
Always halving the smaller labelled central angle.
Instead: Use the arc opposite the circumference vertex; it may be the major arc.
See the thinking
Central to circumference
A minor central angle AOB is 100°. C lies on the major arc AB. Find ∠ACB.
- C intercepts the same minor arc AB.
- The circumference angle is half the central angle.
- ∠ACB=50°.
Vertex on the other arc
A minor central angle AOB is 140°. D lies on the minor arc AB. Find ∠ADB.
- D intercepts the major arc AB, whose central reflex angle is 360−140=220°.
- ∠ADB=220/2=110°.
Check your understanding
AB is a diameter and C lies elsewhere on the circumference. Find ∠ACB.
Show answer
90°
The intercepted semicircle has central angle 180°.
Your turn
6 questions · 13 marks · Work at your own pace.
Read the questions online
- Minor central angle AOB is 84°. C lies on the major arc AB. Find ∠ACB.2 marks
- ∠ACB=37° subtends minor arc AB. Find the corresponding minor central angle.2 marks
- Minor central angle AOB is 120°. D lies on the minor arc AB. Find ∠ADB.2 marks
- C and D lie on the same arc of chord AB. ∠ACB=46°. Find ∠ADB.2 marks
- AB is a diameter of a circle and C is another point on the circle. ∠CAB=32°. Find ∠ABC.2 marks
- An inscribed angle subtends a chord and is 90°. Explain why the chord is a diameter.3 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- Minor central angle AOB is 84°. C lies on the major arc AB. Find ∠ACB.2 marks
Same minor arc:84/2=42.
42° - ∠ACB=37° subtends minor arc AB. Find the corresponding minor central angle.2 marks
Double 37.
74° - Minor central angle AOB is 120°. D lies on the minor arc AB. Find ∠ADB.2 marks
Opposite arc is major:360−120=240.
240/2=120.
120° - C and D lie on the same arc of chord AB. ∠ACB=46°. Find ∠ADB.2 marks
They are angles in the same segment.
46° - AB is a diameter of a circle and C is another point on the circle. ∠CAB=32°. Find ∠ABC.2 marks
∠ACB=90°.
180−90−32=58.
58° - An inscribed angle subtends a chord and is 90°. Explain why the chord is a diameter.3 marks
The corresponding central angle is 180°.
The endpoints are opposite ends of a diameter.
Its intercepted arc is a semicircle.
Curriculum reference & lesson scope
Australian Curriculum v9 and SA Curriculum
Angle and similarity reasoning; congruent/similar triangles in proofs; chords, radii and tangents; angles at the centre and circumference; circle theorems and their applications; connected deductive arguments; construction/spatial algorithms; testing and justifying solutions.
Official curriculum source ↗