Edexcel GCSE Maths (1MA1), Higher tier · Geometry and measures
Practise Sine rule, cosine rule and ½ab sin C. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy2 marks
In triangle ABC, AB = 16 cm, AC = 8 cm and angle BAC = 30°. Diagram NOT accurately drawn
[object Object]
Work out the area of triangle ABC.[2]
Show the answer and mark scheme
Answer: \(32\) cm²
M1 for \(\frac{1}{2} \times 16 \times 8 \times \sin 30^\circ\) with \(\sin 30^\circ = \frac{1}{2}\)
A1 for 32 cao
Worked solution: Area \(= \frac{1}{2}ab\sin C = \frac{1}{2} \times 16 \times 8 \times \sin 30^\circ = \frac{1}{2} \times 16 \times 8 \times \frac{1}{2} = 32\) cm².
Question 2Medium3 marks
In triangle ABC, AB = 3 cm, AC = 4 cm and angle BAC = 60°. Diagram NOT accurately drawn
[object Object]
The area of triangle ABC is \(k\sqrt{3}\) cm². Work out the value of \(k\).[3]
Show the answer and mark scheme
Answer: \(k = 3\)
M1 for \(\frac{1}{2} \times 3 \times 4 \times \sin 60^\circ\)
M1 for using \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
A1 for \(k = 3\)
Worked solution: Area \(= \frac{1}{2} \times 3 \times 4 \times \frac{\sqrt{3}}{2} = \frac{12\sqrt{3}}{4} = 3\sqrt{3}\) cm², so \(k = 3\).
Question 3Hard3 marks
In triangle \(ABC\), \(AB = 7\) cm, \(AC = 9\) cm and angle \(BAC = 40^\circ\). Kate works out the length of \(BC\). Here is her working. \(BC^2 = 7^2 + 9^2 - 2 \times 7 \times 9 \times \cos 40^\circ\) \(BC^2 = 130 - 126 \times \cos 40^\circ\) \(BC^2 = 4 \times \cos 40^\circ = 3.064\ldots\) \(BC = 1.75\) cm
(a) Explain the mistake Kate made.[1]
(b) Work out the correct length of \(BC\). Give your answer correct to 3 significant figures.[2]
Show the answer and mark scheme
(a)Answer: She subtracted 126 from 130 before multiplying by cos 40°; she should work out \(126 \times \cos 40^\circ\) first.
C1 for explaining that \(126 \times \cos 40^\circ\) must be worked out before subtracting from 130 (she did \((130 - 126) \times \cos 40^\circ\))
Worked solution: Multiplication comes before subtraction: \(130 - 126\cos 40^\circ\) is not \(4\cos 40^\circ\).
(b)Answer: 5.79 cm
M1 for \(130 - 126\cos 40^\circ\) (= 33.47...)
A1 for 5.79
Worked solution: \(BC^2 = 130 - 96.52\ldots = 33.47\ldots\), so \(BC = 5.786\ldots = 5.79\) cm.