AQA GCSE Maths · 8300/1H · Higher, Non-Calculator

Angles Practice

All 11 angle questions found across the Paper 1 (non-calculator) past papers, recreated with the original diagrams and ordered by marks so they build up in difficulty. Work through in order; click "Show working" to check each one.

11 questions · 1 mark → 4 marks · ~32 marks total

1November 2023, Q2

1 mark

Work out the size of an exterior angle of a regular hexagon.

Show working

Exterior angles of any regular polygon sum to 360°.
360° ÷ 6 = 60°

2November 2018, Q7

2 marks

The sum of the angles in any quadrilateral is 360°
For example, in a rectangle 4 × 90° = 360°

Zak writes,
"5 × 90° = 450° so the sum of the angles in any pentagon must be 450°"

Is he correct? Tick a box: ☐ Yes ☐ No
Show working to support your answer.

Show working

No. Zak's method only happens to work for a rectangle because a rectangle has four right angles — it isn't a general rule. The correct method uses angle sum = (n − 2) × 180°.
For a pentagon: (5 − 2) × 180° = 3 × 180° = 540°, not 450°.

3November 2017, Q14

3 marks

Two congruent regular polygons are joined together.

Not drawn accurately

Work out the number of sides on each polygon.

Show working

Angles round a point sum to 360°. The two (equal, since congruent) interior angles plus the 60° gap make a full turn:
2 × interior angle + 60° = 360° → interior angle = 150°
Regular polygon interior angle formula: (n − 2) × 180° ÷ n = 150°
180n − 360 = 150n → 30n = 360 → n = 12 sides (a regular dodecagon) each.

4June 2023, Q4

3 marks

ABC, BD and BE are straight lines.

Not drawn accurately

angle EBD = 5 × angle ABE
angle DBC = 3 × angle ABE

Work out the size of angle EBD.

Show working

Let angle ABE = y. Since ABC is a straight line, angle ABE + angle EBD + angle DBC = 180°.
y + 5y + 3y = 180° → 9y = 180° → y = 20°
angle EBD = 5 × 20° = 100°

5June 2022, Q10

3 marks

Use a ruler and compasses in this question.
ABCD represents a garden.

Not drawn accurately

A tree is to be planted in the garden. The tree will be in the region that is closer to AB than to BC.

Label the region, R, where the tree could be planted. Show all your construction lines.

Show working

The set of points equidistant from lines AB and BC is the angle bisector of angle ABC.
Construct it properly with compasses: from B, draw an arc crossing both AB and BC; from those two crossing points draw two more equal-radius arcs that intersect; join B to that intersection point.
Region R (closer to AB than BC) is the area inside the garden, on the AB side of that bisector line — shade it and label R.

6November 2017, Q19

3 marks

The diagram shows a triangle and a trapezium.

Not drawn accurately

Prove that a = b

Show working (method only)

This is a proof, not a single numeric answer — work it through live using: angles on a straight line (180°), angles round a point (360°), and the fact that the two arrow-marked sides of the trapezium are parallel (so alternate/corresponding angles are equal) to link angle a back to angle b. I haven't pre-solved this one for you so it's worth doing properly in the session.

7June 2019, Q14

3 marks

Here is a quadrilateral.

Not drawn accurately

a = 90° and a : b = 5 : 3
x : y = 1 : 3

Show that b = x

Show working

b = 90° × 3/5 = 54°
Let x = k, y = 3k. Angles in any simple quadrilateral sum to 360° (this one is concave, but the rule still holds):
90 + b + x + y = 360 → 90 + 54 + k + 3k = 360 → 4k = 216 → k = 54°
So x = 54° = b, as required.

8November 2017, Q11

4 marks

The four candidates in an election were A, B, C and D.
The pie chart shows the proportion of votes for each candidate.

Not drawn accurately

Work out the probability that a person who voted, chosen at random, voted for C.

Show working

A = 90° (right angle shown), B = x, D = 2x, C = 2x + 10°. Angles round a point sum to 360°:
90 + x + 2x + (2x+10) = 360 → 5x = 260 → x = 52°
Angle C = 2(52) + 10 = 114°
P(voted C) = 114/360 = 19/60

9November 2019, Q26

4 marks

Here are a circle and a sector of the circle. They each have radius r.

Not drawn accurately

circumference of circle = perimeter of sector

Work out the size of angle x. Give your answer in terms of π

Show working

Circumference of circle = 2πr
Perimeter of sector = 2r + arc length = 2r + (x/360) × 2πr
Set equal: 2πr = 2r + (x/360) × 2πr
Divide by r: 2π = 2 + (x/360) × 2π
x/360 = (2π − 2)/(2π) = 1 − 1/π
x = 360 − 360/π degrees (≈ 245.5°)

10November 2017, Q24

4 marks

Here is a cyclic quadrilateral.

Not drawn accurately

x : y = 5 : 7

Work out the size of angle w.

Show working

Opposite angles in a cyclic quadrilateral sum to 180°. Here x and y are opposite, and (x+20°) and w are opposite.
x + y = 180°, with x:y = 5:7 → x = 180×5/12 = 75°, y = 180×7/12 = 105°
(x + 20°) + w = 180° → 95° + w = 180° → w = 85°

11June 2023, Q15

4 marks

ABCD is a trapezium. All four sides are different lengths. AB is parallel to CD. The diagonals intersect at X.

Not drawn accurately

For each statement, tick the correct box: True, May be true, or Not true.

Statement	True	May be true	Not true
Triangles AXB and CXD are similar	☐	☐	☐
Triangles AXD and BXC are congruent	☐	☐	☐
Angle ADB = angle BDC	☐	☐	☐
Area of triangle ABC = area of triangle ABD	☐	☐	☐

Show working

AXB & CXD similar — True. AB∥CD gives equal alternate angles (XAB=XCD, XBA=XDC) plus vertically opposite angles at X — AA similarity always holds, whatever the side lengths.

AXD & BXC congruent — Not true. Congruence here would need the trapezium to be isosceles (symmetric legs), but the question rules that out by stating all four sides are different lengths — so it can never happen.

Angle ADB = angle BDC — May be true. This would mean DB bisects angle ADC. Nothing in the given conditions forces this, but nothing rules it out for a specific trapezium either.

Area ABC = Area ABD — True. Both triangles share base AB, and since AB∥DC, C and D are the same perpendicular distance from AB — equal base and height means equal area, always.