AQA GCSE Mathematics Paper 1 (Higher), June 2025: Question 15

4 marks · Medium difficulty · Multi-step Problem

Work out the size of angle x given that A, C and D are points on a circle with diameter AC, ABC is an isosceles triangle with AC = BC, and BCD is a straight line.

Practise this question

Question

A circle contains points A, C, and D on its circumference, where AC is the diameter. Line BCD is a straight line extending past C to point B. Triangle ABC is formed with AC = BC and angle ABC marked as 20 degrees. Angle CAD is labelled as x. The diagram is noted as not drawn accurately.
Question text

15 A, C and D are points on a circle, diameter AC.

ABC is an isosceles triangle with AC = BC

BCD is a straight line.

Not drawn

accurately

Work out the size of angle x.

[4 marks]

x = °

Mark scheme

Show the mark scheme Mark scheme for question 15 giving two alternative methods to find angle x = 50 degrees. Alternative 1 uses triangle ADC: identifies angle CAB = 20, angle ACB = 140 or ACD = 40, angle ADC = 90, and calculates 50. Alternative 2 uses triangle ADB: identifies angle ADC or ADB = 90, angle DAB = 70, angle CAB = 20, leading to x = 50. Total 4 marks.

Q Answer Mark Comments

Alternative method 1: from triangle ADC

CAB = 20 M1

ACB = 180 – 20 – 20 oe

M1dep

or ACB = 140 or ACD = 40

ADC = 90 or ADB = 90 M1

50 A1

Alternative method 2: from triangle ADB

ADC = 90 or ADB = 90 M1

15 DAB = 180 – 90 – 20 oe

M1dep

or DAB = 70

CAB = 20 M1

50 A1

Additional Guidance

Angles may be seen on the diagram throughout

For an incomplete method, for angles not marked in the correct position on

the diagram the correct 3-letter codes must be given, but condone D for

ADC or ADB

ADC = 90 or ADB = 90 may be designated by a square at the angle

How to answer it

Circle Theorems: Angle in a Semicircle & Isosceles Triangles

📌 What this question tests

This multi-step geometry problem assesses your ability to combine fundamental angle rules with circle theorems:

  • Circle Theorem: Recognising that the angle subtended by a diameter at the circumference is a right angle ( 90° ).
  • Isosceles Triangle Properties: Identifying equal base angles opposite equal sides ( AC = BC ).
  • Straight Line & Triangle Rules: Applying angles on a straight line add to 180° and angles inside a triangle sum to 180° .

Question 15: Find the Size of Angle x

4 Marks • Higher Tier

💡 Key Knowledge

  • Angle in a semicircle is 90°: Since AC is a diameter, angle ∠ADC = 90° .
  • Isosceles triangle ABC : Because AC = BC , the angles opposite those sides are equal: ∠CAB = ∠ABC = 20° .
  • Straight line rule: Angles on the straight line BCD sum to 180° .
  • Sum of angles in a triangle: Angles in any triangle sum to 180° .

🧠 Exam Technique

  • Annotate the diagram: Write every calculated angle directly onto the figure. The examiner awards marks if correct angles appear on the diagram.
  • Use 3-letter angle notation: If writing steps below, clearly state the vertex (e.g., ∠ADC rather than just "D") to avoid ambiguity.
  • Check for right angles: Whenever a line is specified as a diameter, immediately look for subtended right angles at the circumference.

📐 Step-by-Step Calculations

Method 1: Using Triangle ADC (Recommended)
  1. Find ∠CAB: Triangle ABC is isosceles with AC = BC .
    Therefore, ∠CAB = ∠ABC = 20° . [M1]
  2. Find ∠ACB and ∠ACD:
    In triangle ABC : ∠ACB = 180° - 20° - 20° = 140° .
    Since BCD is a straight line: ∠ACD = 180° - 140° = 40° . [M1 dep]
  3. Apply Circle Theorem:
    AC is a diameter, so the angle at the circumference is 90° :
    ∠ADC = 90° . [M1]
  4. Calculate angle x (in triangle ADC):
    x = 180° - 90° - 40° = 50° . [A1]
Method 2: Using the Large Triangle ADB
  1. Recognise right angle at D: ∠ADB = 90° (angle in a semicircle). [M1]
  2. Find whole angle ∠DAB: In right-angled triangle ADB :
    ∠DAB = 180° - 90° - 20° = 70° . [M1 dep]
  3. Find ∠CAB: Isosceles triangle ABC gives ∠CAB = 20° . [M1]
  4. Subtract to find x:
    x = ∠DAB - ∠CAB = 70° - 20° = 50° . [A1]

✅ Final Answer

x = 50°

Mark Scheme Breakdown:
• M1: For ∠CAB = 20°
• M1 (dep): For finding ∠ACB = 140° or ∠ACD = 40° (or ∠DAB = 70° )
• M1: For stating or marking ∠ADC = 90° (or ∠ADB = 90° )
• A1: For final answer 50°

❌ Common Errors to Avoid

  • Misidentifying equal sides: Confusing which angles are equal in triangle ABC . Because AC = BC , the equal angles are opposite them ( ∠CAB and ∠ABC ), not ∠ACB .
  • Missing the right angle: Not noticing that line AC is a diameter, which is the key that unlocks ∠ADC = 90° .
  • Assuming cyclic quadrilateral: Trying to use opposite angles sum to 180° on figure ABDC —this is incorrect because vertex B does not lie on the circle!
  • Not labeling working: Writing random calculations without angle labels (e.g., just writing 180 - 40 = 140 ), which risks losing method marks if an arithmetic slip occurs.

Topics

Geometry and measures · 3.4.1 Properties and constructions

Question and mark scheme from the AQA GCSE Mathematics examination, Paper 1 (Higher), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.