AQA AS Level Mathematics Paper 2, June 2025: Question 12

9 marks · Medium difficulty · Multi-step Problem

Use right-angled trigonometry to find expressions for side lengths in terms of $\cos\theta$ and solve a trigonometric quadratic equation to find $\theta$.

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Question

A right-angled triangle ABC is shown with the right angle at B and base AB equal to 12 cm. Angle CAB is labeled theta. A point D lies on the hypotenuse AC such that line segment BD is perpendicular to AC. Part (a) asks to show that AC = 12/cos(theta). Part (b) asks to write down an expression for AD in terms of cos(theta). Part (c) asks to hence show that CD = (12 - 12cos^2(theta))/cos(theta). Part (d) gives CD = 8*sqrt(3) cm and asks to find theta to the nearest degree, fully justifying the answer.
Question text

12 In the triangle ABC, the length of AB = 12 cm and the angle ABC is a right angle.

The point D lies on AC such that BD is perpendicular to AC as shown in the diagram.

C

D

θ

A B

12cm

12 (a) Show that AC =

cosθ [1 mark]

12 (b) Write down an expression for AD in terms of cosθ

[1 mark]

12 – 12cos2θ

12 (c) Hence show that CD =

cosθ [2 marks]

12 (d) It is given that CD = 8√3 cm

Find the value of θ

(14)

Give your answer to the nearest degree.

Fully justify your answer.

[5 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 12 shows: 12(a) R1 for starting with cos(theta) = 12/AC to obtain AC = 12/cos(theta). 12(b) B1 for AD = 12cos(theta). 12(c) M1 for CD = AC - AD = 12/cos(theta) - 12cos(theta), R1 for completing the algebra to get (12 - 12cos^2(theta))/cos(theta). 12(d) B1 for setting equal to 8*sqrt(3), M1 for rearranging to 12cos^2(theta) + 8*sqrt(3)cos(theta) - 12 = 0, M1 for solving quadratic, A1 for identifying cos(theta) = sqrt(3)/3 and rejecting -sqrt(3), A1 for theta = 55 degrees (AWRT 55). Total 9 marks.

12(a) Demonstrates given result 2.1 R1 12

12 cosθ =

starting with cosθ = AC

AC 12

OE (in terms of adj/hyp) AC =

cosθ

Subtotal 1

Q Marking instructions AO Marks Typical solution

12(b) States 12cosθ 1.1b B1 AD = 12cosθ

CAO

Subtotal 1

12(c) Expresses CD as the difference 3.1a M1 12

between AC and their AD CD = −12cosθ

cosθ

Or 2

Expresses BD as12sinθ and 12 12cos θ

= −

uses tanθ in triangle BCD cosθ cosθ

Completes reasoned argument 2.1 R1 12 −12cos2θ

to obtain given result =

Must see two separate fractions cosθ

with a common denominator

Or

12sin2θ

CD =

cosθ

Or

CDcos θ = 12 −12cos2 θ

before final answer

AG

Subtotal 2

12 Q Marking instructions AO Marks Typical solution

12(d) 2θ 1.1b B1 12 −12cos2θ

12 −12cos

States 8 3 = 8 3 =

cosθ cosθ

8 3cosθ = 12 −12cos2θ

Rearranges to give a three-term 3.1a M1

12cos2θ + 8 3cosθ −12 = 0

quadratic equation in cosθ

(must be = 0) 3

cosθ = or − 3

Solves their quadratic to obtain 1.1a M1

at least one real value for cosθ cosθ = − 3 not valid

∴cosθ =

Obtains cosθ = and explains 2.4 A1

3 θ = 55°

that − 3 is not valid

OE

Obtains AWRT 55 1.1b A1

ISW CSO

Subtotal 5

Question 12 Total 9

How to answer it

Geometric Trigonometry & Hidden Quadratics

📋 What This Question Tests
  • Right-angled triangle trigonometry: Identifying adjacent, opposite, and hypotenuse across different overlapping triangles (ΔABC and ΔADB).
  • Algebraic fraction manipulation: Combining terms over a common denominator to complete a formal geometric proof.
  • Forming and solving quadratic equations: Transforming trigonometric expressions into standard form a·cos²θ + b·cosθ + c = 0 .
  • Mathematical justification & domain validity: Rejecting impossible trigonometric solutions based on the range of cosine ( -1 ≤ cos θ ≤ 1 ).

Part (a) — Show that AC = 12 / cos θ

[1 Mark]

💡 Key Knowledge

Consider the large right-angled triangle ΔABC (right-angled at B):

  • Side adjacent to θ is AB = 12 cm
  • Hypotenuse is AC
  • cos θ = Adjacent / Hypotenuse = AB / AC

📐 Step-by-Step Proof

  1. State the ratio:
    cos θ = 12 / AC
  2. Multiply both sides by AC and divide by cos θ :
    AC = 12 / cos θ

🧠 Exam Technique

In a "Show that..." question with 1 mark, you must write down the initial trigonometric ratio ( cos θ = 12 / AC ) before rearranging. Simply jumping to the final line gains zero marks.

✅ Mark Scheme Breakdown

R1 (2.1): Correctly establishes cos θ = 12 / AC leading directly to AC = 12 / cos θ .

Part (b) — Expression for AD in terms of cos θ

[1 Mark]

💡 Key Knowledge

Now focus on the smaller right-angled triangle ΔADB:

  • Angle at D is 90° (since BD ⊥ AC).
  • The hypotenuse is side AB = 12 cm .
  • The side adjacent to angle θ is AD .

📐 Step-by-Step Deduction

  1. Set up the ratio:
    cos θ = Adjacent / Hypotenuse = AD / 12
  2. Multiply by 12:
    AD = 12 cos θ

❌ Common Errors

  • Mixing up triangles: Confusing the hypotenuse of ΔABC (which is AC) with the hypotenuse of ΔADB (which is AB).
  • Writing AD = 12 / cos θ by falsely assuming AD is a hypotenuse.

✅ Final Answer

AD = 12 cos θ

B1 (1.1b): Correct expression written down clearly.

Part (c) — Show that CD = (12 − 12 cos²θ) / cos θ

[2 Marks]

🧠 Exam Technique: Segment Subtraction

The word "Hence" tells you to use previous results:

CD = AC − AD

Substitute the exact expressions found in parts (a) and (b).

📐 Step-by-Step Proof

  1. State geometric relation:
    CD = AC − AD
  2. Substitute expressions from (a) and (b):
    CD = (12 / cos θ) − 12 cos θ
  3. Put both terms over a common denominator of cos θ :
    CD = (12 / cos θ) − (12 cos²θ / cos θ)
  4. Combine fractions:
    CD = (12 − 12 cos²θ) / cos θ

❌ Common Errors

  • Missing the intermediate fraction step: Examiners specifically require seeing the two separate fractions with a common denominator, or showing clearly that 12 cos θ = (12 cos²θ)/cos θ .
  • Writing 12 − 12 cos θ instead of 12 − 12 cos²θ due to sloppy fraction multiplication.

✅ Mark Scheme Breakdown

M1 (3.1a): Expresses CD as AC − AD using their earlier answers.
R1 (2.1): Fully reasoned algebraic argument showing two separate fractions with common denominator to achieve the given result.

Part (d) — Solve for θ when CD = 8√3 cm

[5 Marks]

💡 Key Knowledge

  • Rearranging trigonometric equations into standard quadratic form: a·x² + b·x + c = 0 where x = cos θ .
  • Validity condition: Real cosine values must satisfy -1 ≤ cos θ ≤ 1 . Furthermore, in a triangle where θ is acute, 0 < cos θ < 1 .

📐 Step-by-Step Calculation

  1. Equate expressions:
    8√3 = (12 − 12 cos²θ) / cos θ
  2. Form quadratic equation:
    8√3 cos θ = 12 − 12 cos²θ
    12 cos²θ + 8√3 cos θ − 12 = 0
    Divide by 4:
    3 cos²θ + 2√3 cos θ − 3 = 0
  3. Solve quadratic for cos θ:
    Using quadratic formula with a = 3 , b = 2√3 , c = -3 :
    cos θ = [−2√3 ± √((2√3)² − 4(3)(−3))] / (2 × 3)
    cos θ = [−2√3 ± √(12 + 36)] / 6 = [−2√3 ± √48] / 6
    Since √48 = 4√3 :
    cos θ = (−2√3 + 4√3) / 6 = 2√3 / 6 = √3 / 3
    OR
    cos θ = (−2√3 − 4√3) / 6 = −6√3 / 6 = −√3
  4. Justify rejection of invalid root:
    −√3 ≈ −1.732 , which is < −1 (outside the range of cosine, and impossible for an acute angle). Therefore, cos θ = −√3 is not valid.
  5. Compute θ:
    cos θ = √3 / 3 ≈ 0.57735
    θ = arccos(√3 / 3) = 54.7356...°
    Rounding to the nearest whole degree: θ = 55°

❌ Common Errors & Mark Traps

  • Losing the justification mark (A1): Simply ignoring the second root without stating why it is discarded. You must explicitly write that cos θ = −√3 is invalid / has no real solution because −√3 < −1 .
  • Incorrect rounding: Forgetting to round to the nearest whole degree (leaving it as 54.7° loses the final mark).
  • Sign errors when rearranging: Watch signs when moving terms to one side: +12 cos²θ + 8√3 cos θ − 12 = 0 .

✅ Mark Scheme Breakdown

B1 (1.1b): Sets up equation: 8√3 = (12 − 12 cos²θ) / cos θ .
M1 (3.1a): Rearranges into a 3-term quadratic equal to 0: 12 cos²θ + 8√3 cos θ − 12 = 0 .
M1 (1.1a): Solves quadratic by formula or factorisation to find at least one value for cos θ .
A1 (2.4): Obtains cos θ = √3 / 3 AND explicitly explains that −√3 is not valid.
A1 (1.1b): Obtains θ = 55° (AWRT 55).

Topics

Pure Mathematics · B: Algebra and functions · E: Trigonometry

Question and mark scheme from the AQA AS Level Mathematics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.