Grade 10 Angles in a Circle: Short Notes

10 ශ්‍රේණිය · 3 වාරය · 30 පරිච්ඡේදය · වෘත්තයක කෝණ

46%

of Grade 10 term tests include it (25 of 54)

334

marks in past papers (72 questions)

11 min

full lesson with quick checks

🎯 What you'll learn

Prove that the angle at the centre is twice the angle at the circumference
Use the reflex angle at the centre when the point is on the minor arc
Use the fact that angles in the same segment are equal
Use the fact that the angle in a semicircle is a right angle
Combine these with isosceles triangles and angle sums to solve problems

Two floodlight towers

Two floodlight towers stand on the edge of a round school ground in Galle. Whether you sit at the canteen end or near the scoreboard, as long as you are on the same curved edge, the angle between the two towers looks exactly the same. That isn't luck. It's a circle theorem, and today you'll learn three of them.

Words you need

A chord AB cuts a circle into two arcs: a minor (small) arc and a major (big) arc.The angle AOB at the centre O stands on arc AB: we say arc AB subtends ∠AOB at the centre.An angle APB with P on the circle is an angle at the circumference (on the remaining part of the circle).

Centre angle = 2 × circumference angle

Both angles stand on the same arc AB. The angle at the centre is twice the angle at P.

🧮 Theorem 1

The angle an arc subtends at the centre is twice the angle it subtends at any point on the remaining part of the circle. Going the other way: angle at the circumference = ½ × angle at the centre.

∠AOB = 2 × ∠APB

Why it works

Join PO and extend it to D.• OA = OP (radii), so △OAP is isosceles: ∠OAP = ∠OPA = a.• The exterior angle ∠AOD = a + a = 2a.• In the same way, with ∠OPB = b, ∠BOD = 2b.• Adding: ∠AOB = 2a + 2b = 2(a + b) = 2 × ∠APB.Radii make isosceles triangles: that idea solves many exam questions too.

✏️ Halve it

A, B and P are on a circle with centre O, with P on the major arc. ∠AOB = 130°. Find ∠APB.

1∠AOB and ∠APB both stand on arc AB.
2Angle at the circumference = ½ × angle at the centre.
3So ∠APB = 130° ÷ 2 = 65°.

✅ 65°

✏️ When the point is on the minor arc

∠AOB = 140° and Q is on the minor arc AB, as shown. Find ∠AQB.

1∠AQB stands on the major arc AB, not the minor arc.
2The major arc subtends the reflex angle at O: 360° − 140° = 220°.
3So ∠AQB = 220° ÷ 2 = 110°.

✅ 110°

⚠️ Doubling the wrong way, or the wrong arc

This theorem was misused in O/L papers year after year. Two traps:1. Halving when you should double. The centre angle is always the bigger one.2. Using an angle that stands on a different arc. First find the arc each angle 'looks at'. If the point is on the minor arc, use the reflex angle at O.

Theorem 2: angles in the same segment

P and Q are on the same side of chord AB (the same segment). Both angles are half of the same centre angle, so ∠APB = ∠AQB. Angles in the same segment are equal.

⚠️ Same segment means same side

Students often miss equal angles in the same segment, or match angles that are not in the same segment.Test: both angles must have their arms ending at the same two points (the same chord), and their vertices must be on the same side of that chord. If they're on opposite sides, the angles are not equal in general.

Theorem 3: angle in a semicircle

AB is a diameter, so ∠AOB = 180°. Half of 180° is 90°, so ∠ACB = 90°. The angle in a semicircle is a right angle. Spot a diameter? Look for a right angle, then Pythagoras or the angle sum.

📌 Remember

• Angle at the centre = 2 × angle at the circumference, on the same arc.• If the point is on the minor arc, use the reflex angle at the centre.• Angles in the same segment are equal.• The angle in a semicircle is 90°.• Radii are equal, so triangles with two radii are isosceles. Give a reason for every angle.

🧾 Quick revision sheet

Angle at the centre vs angle at the circumference (same arc)	The angle at the centre is twice the angle at the circumference.
Angles in the same segment	are equal.
Angle in a semicircle	90°, a right angle.
Point on the minor arc: which centre angle do you halve?	The reflex angle at the centre (360° minus the given angle).
A triangle with two radii as sides is…	isosceles, so its base angles are equal.
What does 'arc AB subtends ∠APB' mean?	∠APB is formed by lines from the ends of arc AB to P, so it stands on that arc.

Now test yourself: 4 quick checks and 1 more worked examples

The full lesson checks your answers as you go, and Gedara Sir tracks which parts you've got.

📖 Start the 11-minute lesson free ▶ Practise 5 past-paper questions