Practise 29 exam-style questions on Exact trigonometric values for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
29 questions
1–5 marks each
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20 questions, marked as you go
Sample questions with model answers
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Question 1
2 marks
Work out the exact value of \(\sin 60° \times \cos 30°\).
Substitute the exact values and cancel the \(\sqrt{3}\). This works for any acute angle, because \(\tan x = \dfrac{\sin x}{\cos x}\), so \(\cos x \times \tan x = \sin x\).
Mark scheme
1 mark
M1 for \(\dfrac{\sqrt{3}}{2} \times \dfrac{1}{\sqrt{3}}\)
oe, e.g. \(\dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{3}\)
1 mark
C1 for simplifying to \(\dfrac{1}{2}\) and stating \(\sin 30° = \dfrac{1}{2}\)
answer only scores 0 (show that)
Question 3
3 marks
The diagram shows an equilateral triangle ABC with sides of length 2 cm. M is the midpoint of BC and AM is perpendicular to BC.
Use the triangle to show that \(\sin 60° = \dfrac{\sqrt{3}}{2}\)
Model answer
Each angle of an equilateral triangle is 60°, so angle ABM = 60°. BM = 2 ÷ 2 = 1 cm In the right-angled triangle ABM: \(AM^2 = 2^2 - 1^2 = 3\), so \(AM = \sqrt{3}\) cm \(\sin 60° = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{AM}{AB} = \dfrac{\sqrt{3}}{2}\)
Explanation
Cutting an equilateral triangle of side 2 in half gives a right-angled triangle with angles 30°, 60°, 90° and sides 1, \(\sqrt{3}\), 2. All the exact values for 30° and 60° can be read from this triangle, so you can rebuild them if you forget.
Mark scheme
1 mark
M1 for BM = 1
1 mark
M1 for \(AM^2 = 2^2 - 1^2\), so \(AM = \sqrt{3}\)
1 mark
C1 for \(\sin 60° = \dfrac{AM}{AB} = \dfrac{\sqrt{3}}{2}\), with angle ABM = 60° stated or used
complete argument required
Question 4
4 marks
The diagram shows a regular hexagon with sides of length 4 cm.
Work out the exact area of the hexagon. Give your answer in the form \(a\sqrt{3}\), where \(a\) is an integer.
Model answer
Joining the centre to each vertex makes 6 equilateral triangles with sides 4 cm. Height of one triangle \(= 4 \sin 60° = 4 \times \dfrac{\sqrt{3}}{2} = 2\sqrt{3}\) cm Area of one triangle \(= \dfrac{1}{2} \times 4 \times 2\sqrt{3} = 4\sqrt{3}\) cm² Area of hexagon \(= 6 \times 4\sqrt{3} = 24\sqrt{3}\) cm²
Explanation
The angles at the centre of a regular hexagon are \(360° \div 6 = 60°\), so the six triangles are equilateral. The exact value \(\sin 60° = \dfrac{\sqrt{3}}{2}\) gives the height of each triangle.
Mark scheme
1 mark
P1 for splitting the hexagon into 6 equilateral triangles of side 4 cm
or other complete dissection; correct answer with no working scores full marks
1 mark
P1 for the height of one triangle \(4 \sin 60°\) \((= 2\sqrt{3})\), or \(\dfrac{1}{2} \times 4 \times 4 \times \sin 60°\)
oe, e.g. \(\sqrt{4^2 - 2^2} = \sqrt{12}\)
1 mark
P1 for the area of one triangle \(4\sqrt{3}\)
ft their height
1 mark
A1 for \(24\sqrt{3}\)
cao
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