Practise 36 exam-style questions on Sine rule and cosine rule for Edexcel GCSE Maths (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.
36 questions
1–5 marks each
Instant marking
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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
The diagram shows triangle ABC. BC = 9.6 cm, angle BAC = 42° and angle ABC = 71°.
Work out the value of x. Give your answer correct to 3 significant figures.
Each side pairs with the angle opposite it: BC = 9.6 cm is opposite the 42° angle, and x (AC) is opposite the 71° angle. With two angles and a side, use the sine rule. A common mistake is to pair 9.6 cm with 71°.
Mark scheme
1 mark
M1 for \(\dfrac{x}{\sin 71^\circ} = \dfrac{9.6}{\sin 42^\circ}\)
oe, e.g. \(x = \dfrac{9.6 \sin 71^\circ}{\sin 42^\circ}\)
1 mark
A1 for 13.6
awrt 13.6 (13.565...); correct answer with no working scores full marks
Question 2
3 marks
The diagram shows triangle DEF. EF = 8.4 cm, angle EDF = 63° and angle DEF = 52°. DF = x cm.
Show that x = 7.4 correct to 1 decimal place.
Model answer
DF is opposite angle E (52°) and EF is opposite angle D (63°). \(\dfrac{x}{\sin 52^\circ} = \dfrac{8.4}{\sin 63^\circ}\) \(x = \dfrac{8.4 \times \sin 52^\circ}{\sin 63^\circ} = 7.4290...\) 7.4290... = 7.4 correct to 1 decimal place.
Explanation
Match each side with its opposite angle: x (DF) is opposite the 52° angle at E, and 8.4 cm (EF) is opposite the 63° angle at D. In a "show that" question you must write a more accurate value (7.429...) before rounding to 7.4.
Mark scheme
1 mark
M1 for \(\dfrac{x}{\sin 52^\circ} = \dfrac{8.4}{\sin 63^\circ}\)
oe
1 mark
M1 for \(x = \dfrac{8.4 \times \sin 52^\circ}{\sin 63^\circ}\)
oe
1 mark
A1 for 7.429... (a value to more than 1 decimal place) which rounds to 7.4
accept 7.42 to 7.43; answer only, or 7.4 with no more accurate value, scores at most M1 M1
Question 3
4 marks
A boat sails 18 km from a port A on a bearing of 050° to a buoy B. It then sails 25 km on a bearing of 130° to a lighthouse C.
The diagram shows the journey.
Work out the distance from A to C. Give your answer correct to 3 significant figures.
Model answer
At B, the bearing back to A is 050° + 180° = 230°. Angle ABC \(= 230 - 130 = 100^\circ\) \(AC^2 = 18^2 + 25^2 - 2 \times 18 \times 25 \times \cos 100^\circ\) \(AC^2 = 324 + 625 + 156.28... = 1105.28...\) \(AC = 33.245...\) AC = 33.2 km (3 s.f.)
Explanation
First find the angle inside the triangle at B, using the north lines: the angle between BA and due south at B is 50° (alternate angles with the 050° bearing at A), and the angle between due south and BC is 180° − 130° = 50°, so angle ABC = 100°. Then use the cosine rule. Because cos 100° is negative, the last term is added. A common mistake is to use 80° (130 − 50) for the angle at B.
Mark scheme
1 mark
P1 for angle ABC = 100°
e.g. 230 − 130, or 50° (alternate angles) + 50° (180 − 130); may be seen on a diagram; correct answer with no working scores full marks
Find the angle at B from the north lines (100°), use the cosine rule for AC, then the sine rule for angle BAC. The bearing of C from A is measured clockwise from north at A, so add angle BAC to the bearing of B (050°). Keep the unrounded AC in your calculator. A common mistake is to give angle BAC (48°) as the bearing.
Mark scheme
1 mark
P1 for angle ABC = 100°
may be seen on a diagram; correct answer with no working scores full marks
