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Circle geometry · Lesson 1 of 5

The angle at the centre

Why the angle an arc makes at the centre is always twice the angle it makes on the circle, and how to use it.

12 minYou should already know: Angles, triangles & polygons
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  5. 5

Almost every circle question in WAEC, NECO and JAMB comes back to one idea. Learn it well here and the other circle theorems follow from it.

The words you need

  • A chord is a straight line joining two points on the circle, like AB.
  • An arc is a piece of the circle itself. A chord cuts the circle into two arcs: the shorter minor arc and the longer major arc.
  • Join A and B to the centre O and you get ∠AOB\angle AOB, the angle at the centre.
  • Join A and B to any other point P on the circle and you get ∠APB\angle APB, an angle at the circumference.

Both angles open towards the same arc AB. We say they stand on that arc (some textbooks say the arc subtends the angle).

See it for yourself

Drag the gold points and watch the two numbers.

The angle at the centreDrag the gold points
128°64°OABP
64°∠APB, at the circumference× 2 =128°∠AOB, at the centre
Slide P anywhere on this side of AB: ∠APB stays 64°. The angle at the centre, on the same arc AB, is always twice it. Now move A or B, or take P round to the other side.

Try these three things, in order:

  1. Keep A and B still and slide P along the top of the circle. The angle at P doesn’t change at all.
  2. Now move A or B. Both angles change, but compare them: the angle at the centre is always double.
  3. Take P round below the chord. Something different happens to the angle at the centre (we come back to this below).
ABPO2xx
Angle at the centre∠AOB = 2 × ∠APB, on the same arc

Why it’s true

Dragging points only shows that the rule works for the cases you tried. A reason shows it works every time, and WAEC theory questions can ask for it. The reason uses two facts you already know from angles and triangles: the base angles of an isosceles triangle are equal, and an exterior angle of a triangle equals the sum of the two opposite interior angles.

Press Show why it works on the board above and follow along:

  1. Join P to the centre O and carry the line on to D, so PD is a diameter.
  2. OA and OP are both radii, so they are equal. That makes triangle OAP isosceles, and its base angles are equal. Call each one xx.
  3. ∠AOD\angle AOD is an exterior angle of triangle OAP, so it equals the two opposite angles added: x+x=2xx + x = 2x.
  4. In the same way, OB = OP, so triangle OBP is isosceles with base angles yy, and ∠BOD=2y\angle BOD = 2y.
  5. Put the pieces together: ∠APB=x+y\angle APB = x + y and ∠AOB=2x+2y=2(x+y)\angle AOB = 2x + 2y = 2(x + y). The angle at the centre is exactly twice the angle at P.

Using the rule both ways

If you know the angle at the centre, halve it. If you know the angle at the circumference, double it.

When P is on the minor arc

Go back to the board and take P below the chord, onto the minor arc. Now ∠APB\angle APB opens towards the major arc, so the angle at the centre on that arc is the reflex angle AOB (the one bigger than 180∘180^\circ). The rule still works: the reflex angle is twice ∠APB\angle APB.

Notice too that ∠APB\angle APB becomes obtuse. The angle on the major arc and the angle on the minor arc always add up to 180∘180^\circ (half of 360∘360^\circ). You’ll use that in lesson 3.

A past question, step by step

Worked example · WAEC 2023

WAEC 2023 · Paper 1 · Q37

In the diagram, OO is the centre of the circle and ∠SQR=28∘\angle SQR = 28^\circ. Find ∠ORS\angle ORS.

28°OQSR

  1. Find the angle at the centre

    Look for an angle at the circumference and an angle at the centre that stand on the same arc. ∠SQR\angle SQR at Q and ∠SOR\angle SOR at O both open towards arc SR.

    Think first. ∠SQR\angle SQR and ∠SOR\angle SOR stand on the same arc SR. How are they related?

  2. Double it

    ∠SOR=2×28∘=56∘(angle at the centre)\angle SOR = 2 \times 28^\circ = 56^\circ \quad \text{(angle at the centre)}

    Think first. Now look at triangle SOR. What do you notice about OS and OR?

  3. Spot the isosceles triangle

    OS and OR are both radii, so triangle SOR is isosceles and its two base angles, ∠ORS\angle ORS and ∠OSR\angle OSR, are equal.

  4. Share out what's left

    The angles of a triangle add up to 180∘180^\circ, and the two base angles share what is left equally:

    ∠ORS=180∘−56∘2=62∘\angle ORS = \frac{180^\circ - 56^\circ}{2} = 62^\circ

    The answer is D.

Your turn

WAEC 2020 · Paper 1 · Q44

In the diagram, OO is the centre of the circle. If ∠NLM=74∘\angle NLM = 74^\circ, ∠LMN=39∘\angle LMN = 39^\circ and ∠LOM=x\angle LOM = x, find the value of xx.

74°39°xONLM
Worked solution (try it first)
  1. The angles of triangle LMNLMN add up to 180∘180^\circ: ∠LNM=180∘−74∘−39∘\angle LNM = 180^\circ - 74^\circ - 39^\circ
    =67∘= 67^\circ.
  2. ∠LNM\angle LNM is at the circumference on arc LMLM, the same arc as xx at the centre.
  3. The angle at the centre is twice the angle at the circumference: x=2×67∘=134∘x = 2 \times 67^\circ = 134^\circ, option A.

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