← IGCSE / O Level Mathematics

IGCSE / O Level Mathematics · Topic 4 Geometry · Chapter 25 · pp. 318–340

Angle properties

Almost every geometry question is an angle puzzle: you are given a few angles and asked to deduce the rest, giving a reason each time. This chapter collects the reasons. The first half is about straight lines, triangles and polygons, and the second half is the circle theorems.

Learning objectives

  1. Use the angle facts for a point, a straight line and vertically opposite angles.
  2. Find equal angles in parallel lines: corresponding (F), alternate (Z) and co-interior (C).
  3. Use the angle sums of triangles, quadrilaterals and polygons, and the exterior angle sum.
  4. State and use the circle theorems: semicircle, tangent and radius, angle at the centre, same segment, cyclic quadrilateral and alternate segment.
  5. Give a reason for every angle you calculate.

Book exercises: 25.1 to 25.10 and Student assessments 1 and 2 (pp. 318–340). The angle-at-the-centre, same-segment, cyclic-quadrilateral and alternate-segment theorems are Extended content.

Angles at a point and on a line

🔑 Key ideas

  • Angles around a point add up to 360° (one full turn).
  • Angles on a straight line add up to 180° (half a turn).
  • Vertically opposite angles (opposite each other where two lines cross) are equal.
  • Angles are calculated, not measured. Diagrams are not drawn to scale.
FactRuleReason to write
Around a pointsum = 360°“angles at a point add up to 360°”
On a straight linesum = 180°“angles on a straight line add up to 180°”
Two crossing linesopposite angles equal“vertically opposite angles are equal”

📘 Example 1 · Angles of 120° and 170° and x° fill a point. Find x

  1. Angles at a point add up to 360°: 120 + 170 + x = 360.
  2. Subtract: x = 360 − 120 − 170.
  3. x = 70°.

📘 Example 2 · Angles 40°, 88°, a° and 25° lie on a straight line. Find a

  1. Angles on a straight line add up to 180°: 40 + 88 + a + 25 = 180.
  2. Subtract: a = 180 − 40 − 88 − 25.
  3. a = 27°. Check: 40 + 88 + 27 + 25 = 180. ✔

Angles in parallel lines

🔑 Key ideas

  • Corresponding angles are equal: look for an F shape.
  • Alternate angles are equal: look for a Z shape.
  • Co-interior angles add up to 180°: look for a C (or U) shape.
  • Proof of alternate angles: a = b (corresponding) and b = c (vertically opposite), so a = c.
1 2 3 4 5 6 7 8 F: 1 = 5, 2 = 6 corresponding, equal Z: 3 = 6, 4 = 5 alternate, equal C: 3 + 5 = 180° and 4 + 6 = 180°
Two parallel lines (arrows) cut by a transversal. Number the angles 1 to 4 at the top and 5 to 8 below, left to right and top to bottom, and read the pairs off.

📘 Example 3 · Parallel lines. One angle is 65°. Find the angle that is (a) alternate to it, (b) co-interior to it

  1. (a) Alternate angles are equal (Z shape), so the angle is 65°.
  2. (b) Co-interior angles add up to 180° (C shape): 180 − 65.
  3. The angle is 115°.

📘 Example 4 · Parallel lines. Co-interior angles are 3a° and 2a°. Find a

  1. Co-interior angles add up to 180°: 3a + 2a = 180.
  2. 5a = 180, so a = 180 ÷ 5.
  3. a = 36. The angles are 108° and 72°, and 108 + 72 = 180. ✔

💡 Write a reason beside every angle you find. “a = 65° (alternate angles)” earns marks; “a = 65°” alone often does not.

✏️ Try it

  1. Angles of 120°, 95° and x° fill a point. Find x.
  2. Two lines cross and one of the angles is 68°. Find the other three angles.
  3. Parallel lines: corresponding angles are 4x° and (2x + 30)°. Find x.
Answers

1. 360 − 120 − 95 = 145° · 2. Vertically opposite 68°, and 180 − 68 = 112° twice · 3. 4x = 2x + 30, so 2x = 30 and x = 15 (both angles are 60°)

Next: triangles and quadrilaterals →

Angles in a triangle and a quadrilateral

🔑 Key ideas

  • The angles in a triangle add up to 180°.
  • A quadrilateral splits into two triangles by one diagonal, so its angles add up to 2 × 180 = 360°.
  • An isosceles triangle has two equal sides and the two angles opposite them are equal.
  • An equilateral triangle has three equal angles of 60°.
180° 180° one diagonal makes two triangles Triangle 3 sides → 1 triangle → 180° Quadrilateral 4 sides → 2 triangles → 360° Pentagon (next tab) 5 sides → 3 triangles → 540°
Drawing a diagonal from one corner splits a quadrilateral into two triangles, so its four angles total 2 × 180° = 360°.

📘 Example 5 · A triangle has angles 37° and 64°. Find the third angle x

  1. Angles in a triangle add up to 180°: 37 + 64 + x = 180.
  2. x = 180 − 37 − 64.
  3. x = 79°.

📘 Example 6 · An isosceles triangle has an angle of 35° at one base. Find the angle at the top

  1. The base angles of an isosceles triangle are equal, so both base angles are 35°.
  2. Top angle = 180 − 35 − 35.
  3. The top angle is 110°.

📘 Example 7 · A quadrilateral has angles 90°, 80° and 60°. Find the fourth angle p

  1. Angles in a quadrilateral add up to 360°: 90 + 80 + 60 + p = 360.
  2. p = 360 − 90 − 80 − 60.
  3. p = 130°.

📘 Example 8 · A triangle has angles a°, 2a° and 3a°. Find a

  1. Angles add up to 180°: a + 2a + 3a = 180.
  2. 6a = 180, so a = 30.
  3. The angles are 30°, 60° and 90°. Check: 30 + 60 + 90 = 180. ✔

✏️ Try it

  1. An isosceles triangle has a top angle of 40°. Find each base angle.
  2. Three angles of a quadrilateral are 85°, 95° and 110°. Find the fourth.
  3. A triangle has angles x°, 2x° and 3x°. Find x and the largest angle.
Answers

1. (180 − 40) ÷ 2 = 70° each · 2. 360 − 85 − 95 − 110 = 70° · 3. 6x = 180, so x = 30 and the largest angle is 90°

Next: polygons →

Polygons

🔑 Key ideas

  • A regular polygon has all sides equal and all angles equal.
  • Sum of the interior angles of an n-sided polygon: 180(n − 2)°. From one vertex you can draw n − 2 triangles.
  • Sum of the exterior angles of any convex polygon: 360°.
  • Regular polygon: each exterior angle = °, and interior + exterior = 180°.
  • Both sums also hold for irregular polygons.
3 triangles → 3 × 180° = 540° interior angles of a pentagon 72° 72° 72° 72° 72° 5 × 72° = 360° exterior angles of a regular pentagon
Left: n − 2 = 3 triangles fill a pentagon. Right: walking round the edge you turn through the exterior angles, a full 360° in total, so each of the five is 360 ÷ 5 = 72°.
Regular polygonnInterior sumEach interiorEach exterior
Triangle3180°60°120°
Square4360°90°90°
Pentagon5540°108°72°
Hexagon6720°120°60°
Octagon81080°135°45°
Decagon101440°144°36°

📘 Example 9 · A regular pentagon: the interior angle sum and each interior angle

  1. A pentagon has n = 5 sides.
  2. Sum = 180(5 − 2) = 180 × 3 = 540°.
  3. The angles are equal, so each = 540 ÷ 5 = 108°.

📘 Example 10 · (a) The exterior angle of a regular nonagon. (b) The sides of a regular polygon with exterior angle 15°

  1. (a) A nonagon has 9 sides: 360 ÷ 9 = 40°.
  2. (b) Number of sides = 360 ÷ 15.
  3. The polygon has 24 sides.

📘 Example 11 · Each interior angle of a regular polygon is 156°. How many sides?

  1. Interior + exterior = 180°, so exterior = 180 − 156 = 24°.
  2. Number of sides = 360 ÷ 24.
  3. The polygon has 15 sides.
  4. Check: 180 × 13 = 2340, and 2340 ÷ 15 = 156. ✔

📘 Example 12 · Irregular pentagon with angles x°, (x + 20)°, (x + 40)°, (x + 60)° and (x + 80)°. Find x

  1. A pentagon's interior angles add up to 540°.
  2. x + (x + 20) + (x + 40) + (x + 60) + (x + 80) = 540.
  3. 5x + 200 = 540, so 5x = 340.
  4. x = 68. The angles are 68°, 88°, 108°, 128° and 148°.
  5. Check: 68 + 88 + 108 + 128 + 148 = 540. ✔

📘 Example 13 · A quadrilateral has exterior angles x°, 2x°, 3x° and 4x°. Find its interior angles

  1. Exterior angles add up to 360°: x + 2x + 3x + 4x = 360.
  2. 10x = 360, so x = 36.
  3. The exterior angles are 36°, 72°, 108° and 144°.
  4. Interior = 180 − exterior: 144°, 108°, 72° and 36°.
  5. Check: 144 + 108 + 72 + 36 = 360, the sum for a quadrilateral. ✔

🧪 Try it: polygon angle calculator

n = 9
A regular polygon with the number of sides found below.

    ✏️ Try it

    1. Find the sum of the interior angles of a 12-sided polygon.
    2. Find each interior angle of a regular nonagon.
    3. A regular polygon has exterior angles of 15°. How many sides does it have?
    Answers

    1. 180 × 10 = 1800° · 2. 180 × 7 = 1260, and 1260 ÷ 9 = 140° · 3. 360 ÷ 15 = 24 sides

    Next: semicircle and tangent →

    The angle in a semicircle

    🔑 Key ideas

    • If AB is a diameter and C is any other point on the circle, then angle ACB = 90°.
    • Reason to write: “angle in a semicircle is 90°”.
    • So a triangle drawn on a diameter is right-angled, and Pythagoras works on it.

    The angle between a tangent and a radius

    🔑 Key ideas

    • A tangent touches the circle at one point. It meets the radius to that point at 90°.
    • Two tangents from one outside point are equal in length. Triangles OAC and OBC are congruent: right angles at A and B, OA = OB (radii), OC common.
    • Quadrilateral OACB therefore has two right angles, so angle AOB + angle ACB = 180°.
    A B C O AB is a diameter, so ∠ACB = 90° O A B C ∠OAC = ∠OBC = 90°, CA = CB
    Left: a triangle on a diameter has a right angle at C. Right: two tangents from C. The radii to the points of contact are at right angles to the tangents.

    📘 Example 14 · AB is a diameter. Angle CAB = 35°. Find angle CBA

    1. Angle ACB = 90° (angle in a semicircle).
    2. Angles in triangle ABC add up to 180°: 35 + 90 + CBA = 180.
    3. Angle CBA = 180 − 35 − 90 = 55°.

    📘 Example 15 · Tangent at A, B is on the tangent, O is the centre and angle AOB = 55°. Find angle ABO

    1. Angle OAB = 90° (tangent and radius meet at a right angle).
    2. Angles in triangle OAB add up to 180°: 55 + 90 + ABO = 180.
    3. Angle ABO = 180 − 55 − 90 = 35°.

    📘 Example 16 · Tangents CA and CB meet at C with angle ACB = 40°. Find angle AOB

    1. Angles OAC and OBC are both 90° (tangent and radius).
    2. The four angles of quadrilateral OACB add up to 360°.
    3. AOB = 360 − 90 − 90 − 40 = 140°.

    📘 Example 17 · The radius is 5 cm and the tangent length is 12 cm. Find OC, the distance from O to the end of the tangent

    1. The tangent meets the radius at 90°, so triangle OAC is right-angled with hypotenuse OC.
    2. OC² = 5² + 12² = 25 + 144 = 169.
    3. OC = √169 = 13 cm.

    ✏️ Try it

    1. AB is a diameter and C is on the circle with angle CAB = 28°. Find angle CBA.
    2. The tangent at A meets a line from the centre O at B. Angle AOB = 65°. Find angle ABO.
    3. Radius 9 cm, and the point C is 15 cm from O. How long is the tangent from C?
    Answers

    1. 90 − 28 = 62° · 2. 180 − 90 − 65 = 25° · 3. √(15² − 9²) = √(225 − 81) = √144 = 12 cm

    Next: circle theorems →

    Angle at the centre Extended

    🔑 Key ideas

    • The angle at the centre is twice the angle on the circumference, when both stand on the same arc.
    • Angles in the same segment (standing on the same arc) are equal. Each is half the angle at the centre.
    • If the arc is a semicircle, the centre angle is 180° and the circumference angle is 90°: the semicircle theorem is a special case.

    🧪 Try it: move point C round the circle

    O A B C D
    C slides along the major arc. The angles are measured from the points' coordinates, not typed in.

      The circle theorems at a glance Extended

      2x x Centre = 2 × circumference same arc AB p q r s Cyclic quadrilateral p + r = 180°, q + s = 180° θ θ A B C Alternate segment tangent-chord = angle at C
      Three theorems in one picture. Left: the angle at the centre is double. Middle: opposite angles of a cyclic quadrilateral add up to 180°. Right: the angle between a tangent and a chord equals the angle in the alternate segment.
      TheoremReason to write
      Angle at centre = 2 × angle at circumference“angle at the centre is twice the angle at the circumference”
      Angles in the same segment are equal“angles in the same segment are equal”
      Opposite angles of a cyclic quadrilateral add up to 180°“opposite angles of a cyclic quadrilateral are supplementary”
      Exterior angle of a cyclic quadrilateral = interior opposite angle“exterior angle of a cyclic quadrilateral equals the interior opposite angle”
      Tangent-chord angle = angle in the alternate segment“angle in the alternate segment”

      📘 Example 18 · The angle at the centre is 110°. Find the angle at the circumference on the same arc

      1. The angle at the centre is twice the angle at the circumference.
      2. x = 110 ÷ 2.
      3. x = 55°.

      📘 Example 19 · Two angles stand on the same arc. One is 33°. Find the other

      1. Both are on the circumference, in the same segment.
      2. Angles in the same segment are equal.
      3. The other angle is 33°. The angle at the centre on that arc is 66°.

      📘 Example 20 · ABCD is a cyclic quadrilateral with angle A = 82° and angle B = 110°. Find angles C and D

      1. Opposite angles of a cyclic quadrilateral add up to 180°.
      2. Angle C is opposite A: 180 − 82 = 98°.
      3. Angle D is opposite B: 180 − 110 = 70°.
      4. Check: 82 + 110 + 98 + 70 = 360. ✔

      📘 Example 21 · Tangent at A, chord AB, and the angle between them is 50°. C is in the alternate segment. Find angle ACB, angle AOB and angle OAB

      1. Angle ACB = 50° (angle in the alternate segment equals the tangent-chord angle).
      2. Angle AOB = 2 × 50 = 100° (angle at the centre is twice the angle at the circumference).
      3. Triangle OAB is isosceles (OA = OB, radii), so angle OAB = (180 − 100) ÷ 2 = 40°.
      4. Check: the tangent meets the radius at 90°, and 50 + 40 = 90. ✔

      ✏️ Try it

      1. The angle at the centre is 150°. Find the angle at the circumference on the same arc.
      2. One angle of a cyclic quadrilateral is 95°. Find the opposite angle.
      3. The tangent-chord angle is 36°. Find the angle in the alternate segment and the angle at the centre.
      Answers

      1. 150 ÷ 2 = 75° · 2. 180 − 95 = 85° · 3. 36° (alternate segment) and 2 × 36 = 72° at the centre

      Next: summary and quiz →

      Chapter summary

      • Around a point 360°, on a straight line 180°, vertically opposite angles equal.
      • Parallel lines: F corresponding equal, Z alternate equal, C co-interior add to 180°.
      • Triangle 180°, quadrilateral 360°, n-sided polygon 180(n − 2)°.
      • Exterior angles of any convex polygon add up to 360°. Regular: each = 360 ÷ n, and interior = 180 − exterior.
      • Angle in a semicircle = 90°. Tangent is at 90° to the radius. Tangents from one point are equal.
      • Extended: centre angle = 2 × circumference angle; same segment equal; cyclic quadrilateral opposite angles add to 180°; alternate segment theorem.
      • Always write a reason beside each angle you calculate.

      📄 Practice questions

      The book's Exercises 25.1 to 25.10 and Student assessments 1 and 2 are in the printable assignment for this chapter. The book's diagrams are described in words there.

      Mini-quiz

      1. What is the sum of the interior angles of a hexagon?
      2. What is each exterior angle of a regular polygon with 20 sides?
      3. Each exterior angle of a regular polygon is 40°. How many sides does it have?
      4. Two angles of a triangle are 47° and 85°. What is the third angle?
      5. A quadrilateral has angles x, 2x, 3x and 4x. What is the largest angle?
      6. AB is a diameter of a circle and C is on the circumference. What is angle ACB?
      7. An angle of 130° is at the centre of a circle. What is the angle on the circumference on the same arc?
      8. ABCD is a cyclic quadrilateral and angle A = 78°. What is the opposite angle C?
      9. The tangent-chord angle at A is 52°. What is the angle in the alternate segment?
      10. A radius meets a tangent at the point of contact. What is the angle between them?
      11. Each interior angle of a regular polygon is 150°. How many sides does it have?
      12. A line crosses two parallel lines. Angles in a Z shape between the lines are called what?

      Source: Cambridge IGCSE Mathematics Core and Extended (4th edition, Pimentel and Wall), Chapter 25 (Angle properties). Explanations, the Try it answers, the tools and the quiz are Guru Jazzy's own. This page is not made or endorsed by Cambridge or Hodder Education.