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Cambridge IGCSE0607

Angles

International Mathematics 0607 Chapter Notes

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Angles
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1. Angles at a point, on a line and vertically opposite

Angles round a point add up to 360°. Angles on a straight line add up to 180°. When two straight lines cross, the angles facing each other are vertically opposite angles and they are equal.

When you give a reason, write the whole property: angles on a straight line sum to 180°, not just 'they add up'.

angles at a point = 360^(°)

angles on a straight line = 180^(°)

Examiner insight

“A common answer was ‘opposite angles’, only a few stated ‘vertically opposite angles are equal’.” “Many candidates just stated that the angles should add up to 180 without checking that their three angles did not add up to 180.”

Worked example 14 marks

Angle a = 56^(°).

(i) The ratio a : b : c = 4 : 7 : 2. Work out the size of angle b and the size of angle c.

(ii) Give a geometrical reason why angles a, b and c cannot be joined to form a straight line.

  1. 1

    (i) one part = 56 ÷ 4 = 14^(°) [1]

  2. 2

    b = 7 × 14 = 98^(°) [1]

  3. 3

    c = 2 × 14 = 28^(°) [1]

  4. 4

    (ii) the sum of the three angles is 56 + 98 + 28 = 182^(°), which is more than 180^(°) [1]

2. Triangles and quadrilaterals

The angle sum of a triangle is 180°. The angle sum of a quadrilateral is 360°.

Angle ABC is the angle at B, between the lines BA and BC. The middle letter is the corner. Use only what the question tells you: do not assume a triangle is isosceles unless it says so.

Examiner insight

“In this particular question no marks could be awarded for anything in the diagram as candidates had to identify angles represented in three letter form.” “A few took the triangle to be isosceles and candidates should be reminded that they must only use information given in the question.”

Worked example 13 marks

In quadrilateral ABCD, angle DAB = 85^(°), angle ABC = 110^(°) and angle BCD = 70^(°).

Find angle CDA. Give a geometric reason for your answer.

  1. 1

    85 + 110 + 70 = 265 [1]

  2. 2

    angle CDA = 360 - 265 = 95^(°) [1]

  3. 3

    reason: the angle sum of a quadrilateral is 360° [1]

3. Angles in parallel lines

A line crossing two parallel lines makes angles at both crossing points. Corresponding angles are equal: they are in the same position at each crossing point. Alternate angles are equal: they are between the parallel lines, on opposite sides of the crossing line.

Co-interior angles sum to 180°: they are between the parallel lines, on the same side of the crossing line. Do not treat them as equal, and do not add alternate angles to make 180°.

co-interior angles: a + b = 180^(°)

Examiner insight

“For a, 180 – 94 = 86 (where candidates were using alternate angles adding to 180) or 52 (assuming lines to be parallel which were not) were common wrong answers.”

Worked example 14 marks

Lines PQ and RS are parallel. Line AB meets PQ at A and RS at B. P and R are on the same side of AB, and Q and S are on the other side. Angle PAB = 65^(°).

(a) Find angle ABS. Give a reason.

(b) Find angle ABR. Give a reason.

  1. 1

    (a) angle ABS = 65^(°) [1]

  2. 2

    reason: alternate angles are equal [1]

  3. 3

    (b) angle ABR = 180 - 65 = 115^(°) [1]

  4. 4

    reason: co-interior angles sum to 180° [1]

4. Angle sum of a polygon

The sum of the interior angles of a polygon with n sides is (n - 2) × 180^(°). For a five-sided shape the angle sum is (5 - 2) × 180 = 540^(°), not 360°.

The sum of the exterior angles of a polygon is 360°. The interior and exterior angles at the same corner lie on a straight line, so they add up to 180°.

sum of interior angles = (n-2) × 180^(°)

sum of exterior angles = 360^(°)

Examiner insight

“Although many candidates were successful, the common mistake was the inability of candidates to know (or to be able to find) the sum of the interior angles of a pentagon.”

Worked example 12 marks

Find the sum of the interior angles of an octagon.

  1. 1

    an octagon has 8 sides, so (8 - 2) × 180 [1]

  2. 2

    = 1080^(°) [1]

5. Regular polygons

In a regular polygon all sides and all angles are equal. Each exterior angle is 360° divided by n. Each interior angle is 180° minus the exterior angle.

To find n, divide 360° by the exterior angle.

exterior angle = (360^(°))/n

interior angle = 180^(°) - exterior angle

n = (360^(°))/(exterior angle)

Examiner insight

“It is easier to work with the exterior angle of a polygon, 180 – 168 and then divide 360 by 12.”

Worked example 12 marks

The interior angle of a regular polygon is 168°.

Find the number of sides of the polygon.

  1. 1

    exterior angle = 180 - 168 = 12^(°) [1]

  2. 2

    n = 360 ÷ 12 = 30 [1]

6. Irregular polygons

In an irregular polygon the angles are not all equal, so you cannot divide the angle sum by n. Use the angle sum (n - 2) × 180^(°), then subtract all the angles you know, including any right angle.

At each corner the exterior angle is still 180° minus the interior angle.

Examiner insight

“A significant number of candidates decided that the angle sum was 360° and then subtracted all the angles except for the right angle so their answer was 30°.”

Worked example 13 marks

A pentagon has interior angles 100^(°), 110^(°), 120^(°), 95^(°) and x^(°). Find x.

  1. 1

    angle sum = (5 - 2) × 180 = 540^(°) [1]

  2. 2

    100 + 110 + 120 + 95 = 425 [1]

  3. 3

    x = 540 - 425 = 115 [1]

Recap

  • Angles at a point sum to 360°. Angles on a straight line sum to 180°.
  • Vertically opposite angles are equal.
  • Angle sum of a triangle is 180°. Angle sum of a quadrilateral is 360°.
  • Parallel lines: corresponding and alternate angles are equal. Co-interior angles sum to 180°.
  • Polygon with n sides: interior angle sum is (n − 2) × 180°. Exterior angles sum to 360°.
  • Regular polygon: exterior angle = 360° ÷ n. Interior angle = 180° − exterior angle.
  • Extended: for an irregular polygon, use the angle sum and subtract the known angles.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. State the sum of the angles at a point.1 mark
  2. Two straight lines cross. One of the four angles is 38°. Find the other three angles.3 marks
  3. Do not use a calculator. Three angles meet at a point. Two of them are 127° and 98°. Find the third angle.2 marks
  4. Do not use a calculator. In triangle ABC, angle BAC = 64^(°) and angle ABC = angle ACB. Find angle ABC. Give a reason.3 marks
  5. Do not use a calculator. The angles of a quadrilateral are x^(°), 2x^(°), 3x^(°) and 4x^(°). Find the largest angle.3 marks
  6. Lines PQ and RS are parallel. Line XY meets PQ at X and RS at Y. P and R are on the same side of XY. Angle PXY = 118^(°). Find angle RYX and give a reason.2 marks
  7. Do not use a calculator. Lines PQ and RS are parallel. Line XY meets PQ at X and RS at Y. P and R are on the same side of XY. Angle PXY = (3x + 10)^(°) and angle XYS = (2x + 30)^(°). Find x and angle PXY, and give a reason.4 marks
  8. A regular polygon has 15 sides. Calculate the sum of its interior angles and the size of one interior angle.3 marks
  9. Do not use a calculator. Three floor tiles meet at a point and leave no gap. One tile is a square, one is a regular hexagon and the third is another regular polygon. Find the number of sides of the third tile.4 marks
  10. A garden bed has six straight sides. Two of its interior angles are right angles. The other four interior angles are in the ratio 4 : 5 : 5 : 6. The exterior angle at the smallest of these four angles is equal to the exterior angle of a regular polygon. Find the number of sides of the regular polygon.6 marks

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