AQA A-Level Mathematics Paper 1, June 2025: Question 16

7 marks · Medium difficulty · Multi-step Problem

Use the sine rule and compound angle identities in triangle ABC with angles π/3 and x to express BC/AB in terms of cot x, and deduce the value of x for a given ratio.

Practise this question

Question

Question text

16 The triangle ABC is shown in the diagram below.

The angle ABC is radians.

The angle BCA is x radians.

3 x

B C

2π

16 (a) Explain why angle BAC = – x

[1 mark]

16 (b) (i) Use the sine rule to show that

BC √3cot x + p

= q

where p and q are integers.

[5 marks]

(24)

16 (b) (ii) Hence state the exact value of x when

BC √3 + 1

AB 2

[1 mark]

Mark scheme

Show the mark scheme

Q Marking instructions AO Marks Typical solution

16(a) Explains that the sum of the The angles in a triangle add up to

angles in a triangle is π π radians.

(radians) and verifies

2π π 2π

BAC = − x 2.4 E1 Angle BAC = π − − x = − x

33 3

2π π – A-LEVEL MATHEMATICS – –

π = − x + + x OE

Subtotal 1

16(b)(i) Forms an equation using the BC AB

sine rule using any two of =

2π sinx

AC BC AB sin − x

= = OE 3

π sin A sin x

sin 3.1a M1

2π 2π 2π

Could use corresponding sin − x = sin cosx−cos sinx

lowercase letters or letters 3 3 3

defined or on the diagram 3 1

throughout. = cosx− − sinx

BC AB

Obtains = 3 1

2π sinx = cosx+ sinx

sin − x 2 2

BC AB 1.1b A1

= 3 1

cosx+ sinx

π sin x BC 2 2

sin π − + x =

3 AB sinx

OE 3cotx+1

Uses compound angle formula =

2π

to expand sin − x

π 3.1a M1

or sin + x

Condone one sign error

Substitutes correct exact values

2π 2π

for sin and cos

π π 1.1a M1

for sin and cos

into their expanded compound

angle

Completes reasoned argument

BC 3cotx+1

to obtain =

AB 2 2.1 R1 23

a 3cotx+1

Accept = – A-LEVEL MATHEMATICS – 7357/1 –

c 2

Subtotal 5

16(b)(ii) π 3cotx+1 π

Deduces from 2.2a R1

42 4

Subtotal 1

24 Question 16 Total 7

How to answer it

Trigonometric Proofs with Sine Rule & Compound Angles

What this question tests

This question assesses geometric reasoning and algebraic manipulation of trigonometric expressions in radians:

Angle sum in a triangle: Expressing a missing angle in radians using angle sum = π.
The Sine Rule: Setting up ratios between side lengths and opposite angles.
Compound angle identities: Expanding expressions of the form sin(A − B).
Exact trigonometric values: Evaluating sin(2π/3) and cos(2π/3) in radical form.
Reciprocal identities: Simplifying cos x / sin x to cot x to achieve a specific target format.

Part (a)

Angle Sum Verification (1 Mark)

✅ Model Answer

The sum of the angles in a triangle is π radians.

Therefore:

Angle BAC = π − (π/3 + x) = π − π/3 − x = 2π/3 − x

❌ Common Errors

Simply writing π − π/3 − x = 2π/3 − x without stating that angles in a triangle sum to π (or 180°).
Mixing degrees and radians (e.g. writing 180 − 60 − x without converting back to π radians).

Mark Scheme [1 Mark]:
E1 (AO 2.4): Explains that the sum of angles in a triangle is π radians AND verifies BAC = 2π/3 − x.

Part (b)(i)

Show that BC / AB = (√3 cot x + p) / q (5 Marks)

📐 Step-by-Step Proof

Step 1: Apply the Sine Rule to triangle ABC.

Opposite to side BC is angle BAC = (2π/3 − x), and opposite to side AB is angle BCA = x.

BC / sin(BAC) = AB / sin(BCA) ⇒ BC / sin(2π/3 − x) = AB / sin x

Step 2: Rearrange for the ratio BC / AB.

BC / AB = sin(2π/3 − x) / sin x

Step 3: Expand the numerator using the compound angle identity:

sin(A − B) = sin A cos B − cos A sin B

sin(2π/3 − x) = sin(2π/3) cos x − cos(2π/3) sin x

Step 4: Substitute exact values:

sin(2π/3) = √3 / 2

cos(2π/3) = −1 / 2

Therefore:

sin(2π/3 − x) = (√3 / 2) cos x − (−1 / 2) sin x = (√3 cos x + sin x) / 2

Step 5: Divide by sin x and express in terms of cot x:

BC / AB = [ (√3 cos x + sin x) / 2 ] / sin x

BC / AB = (√3 (cos x / sin x) + (sin x / sin x)) / 2

Since cot x = cos x / sin x and sin x / sin x = 1 :

BC / AB = (√3 cot x + 1) / 2

Hence p = 1 and q = 2 (both are integers).

💡 Key Knowledge

Sine Rule: a / sin A = b / sin B = c / sin C
Compound Angle: sin(A − B) = sin A cos B − cos A sin B
2nd Quadrant Angles: 2π/3 is in quadrant 2 where sine is positive and cosine is negative:
• sin(2π/3) = sin(π/3) = √3/2
• cos(2π/3) = −cos(π/3) = −1/2
Identity: cot x = cos x / sin x

🧠 Exam Technique & Traps

Watch the double negative: −cos(2π/3)sin x = −(−1/2)sin x = +1/2 sin x. A sign error here destroys the final result.
"Show that" rigor: Every step must be explicit. Do not jump directly from the compound expansion to the final line without clearly dividing each term by sin x.
Alternative expansion: Since sin(2π/3 − x) = sin(π − (π/3 + x)) = sin(π/3 + x), you can expand sin(π/3 + x) instead. Both methods score full marks!

Mark Scheme [5 Marks]:
M1 (AO 3.1a): Forms an equation using the sine rule with correct sides/angles.
A1 (AO 1.1b): Correctly establishes BC / sin(2π/3 − x) = AB / sin x.
M1 (AO 3.1a): Uses compound angle formula to expand sin(2π/3 − x) [condone 1 sign error].
M1 (AO 1.1a): Substitutes correct exact values for sin(2π/3) and cos(2π/3).
R1 (AO 2.1): Completes reasoned argument to reach (√3 cot x + 1)/2 with clear working.

Part (b)(ii)

Deduce the Exact Value of x (1 Mark)

✅ Model Answer

Equating the formula from (b)(i) to the given value:

(√3 cot x + 1) / 2 = (√3 + 1) / 2

Comparing both sides:

√3 cot x = √3 ⇒ cot x = 1 ⇒ tan x = 1

Since x is an acute angle in the triangle:

x = π / 4

❌ Common Errors

Writing the answer in degrees as 45° instead of radians. The question defines the triangle angles in radians, so exact radian form ( π/4 ) is strictly required.
Re-solving the entire triangle from scratch using scratch trigonometry instead of using the "Hence" prompt.

Mark Scheme [1 Mark]:
R1 (AO 2.2a): Deduces π/4 from (√3 cot x + 1)/2 = (√3 + 1)/2.

Topics

Pure Mathematics · E: Trigonometry

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