P5.1Sine and cosine rules, radians, arcs and sectors
Edexcel A level Maths (9MA0) · Pure mathematics › Trigonometry
Practise Sine and cosine rules, radians, arcs and sectors. 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 1Easy4 marks
The diagram shows a sector \(OAB\) of a circle with centre \(O\) and radius 10 cm. The angle \(AOB\) is \(\theta\) radians. The area of the sector is 45 cm². Diagram not drawn to scale
[object Object]
(a) Find the value of \(\theta\).[2]
(b) Find the perimeter of the sector.[2]
Show the answer and mark scheme
(a)Answer: \(\theta = 0.9\)
M1 for \(\frac{1}{2} \times 10^2 \times \theta = 45\)
M1 for \(2 \times 10 + 10 \times \theta\) with their \(\theta\)
A1 for 29
Worked solution: \(P = 2r + r\theta = 20 + 10 \times 0.9 = 29\) cm
Question 2Medium7 marks
In triangle \(ABC\), \(AB = 10\) cm, \(BC = 7\) cm and angle \(BAC = 32^\circ\). A student finds angle \(ACB\) as follows: \(\frac{\sin C}{10} = \frac{\sin 32^\circ}{7}\), so \(\sin C = 0.7570\) and \(C = 49.2^\circ\).
(a) Explain why the student's answer is incomplete, and find the other possible value of angle \(ACB\).[2]
(b) For each possible triangle, find the length of \(AC\). Give your answers to 3 significant figures.[4]
(c) Explain why there would be only one possible triangle if \(BC\) were 12 cm instead of 7 cm.[1]
Show the answer and mark scheme
(a)Answer: \(130.8^\circ\) °
B1 for \(\sin C = 0.7570\) also has the obtuse solution \(180^\circ - 49.2^\circ\), which is possible because \(32^\circ + 130.8^\circ \lt 180^\circ\)
B1 for \(130.8^\circ\)
Worked solution: \(\sin C = 0.7570\) is also satisfied by \(C = 180^\circ - 49.2^\circ = 130.8^\circ\). This gives a valid triangle, since \(32^\circ + 130.8^\circ = 162.8^\circ \lt 180^\circ\). So there are two possible triangles.
(b)Answer: 13.1 cm and 3.91 cm
M1 for angle \(ABC = 180^\circ - 32^\circ - C\): \(98.8^\circ\) or \(17.2^\circ\)
M1 for the sine rule: \(AC = \frac{7\sin B}{\sin 32^\circ}\) (or the cosine rule leading to a quadratic in \(AC\))
A1 for 13.1 (cm)
A1 for 3.91 (cm)
Worked solution: If \(C = 49.2^\circ\), \(B = 98.8^\circ\) and \(AC = \dfrac{7\sin 98.8^\circ}{\sin 32^\circ} = 13.1\) cm. If \(C = 130.8^\circ\), \(B = 17.2^\circ\) and \(AC = \dfrac{7\sin 17.2^\circ}{\sin 32^\circ} = 3.91\) cm.
(c)Answer: \(\sin C = \frac{10\sin 32^\circ}{12} = 0.442\) gives \(26.2^\circ\) or \(153.8^\circ\), but \(32^\circ + 153.8^\circ \gt 180^\circ\), so only the acute angle is possible.
B1 for showing the obtuse value (\(153.8^\circ\)) makes the angle sum exceed \(180^\circ\) (or: \(BC \gt AB\), so \(C\) must be smaller than \(A\), hence acute)
Worked solution: \(\sin C = \dfrac{10\sin 32^\circ}{12} = 0.4416\), giving \(C = 26.2^\circ\) or \(153.8^\circ\). But \(32^\circ + 153.8^\circ = 185.8^\circ \gt 180^\circ\), so only \(C = 26.2^\circ\) is possible.
Question 3Hard9 marks
The diagram shows triangle \(ABC\) with \(AB = 12\) cm, \(AC = 13\) cm and \(BC = 14\) cm. The points \(D\) and \(E\) lie on \(AB\) and \(AC\) respectively, and \(DE\) is an arc of a circle with centre \(A\) and radius 3 cm. The region \(R\), shown shaded, is bounded by \(DB\), \(BC\), \(CE\) and the arc \(DE\). Diagram not drawn to scale
[object Object]
(a) Find the size of angle \(BAC\), in radians, to 4 significant figures.[2]
(b) Find the area of \(R\), giving your answer to 3 significant figures.[4]
(c) Find the perimeter of \(R\), giving your answer to 3 significant figures.[3]
Show the answer and mark scheme
(a)Answer: \(1.186\) rad
M1 for \(\cos A = \frac{12^2 + 13^2 - 14^2}{2 \times 12 \times 13}\)
A1 for awrt \(1.186\)
Worked solution: \(\cos A = \frac{117}{312}\), so \(A = 1.18639\ldots\) radians \(= 1.186\) (4 s.f.)
(b)Answer: \(67.0\) cm²
M1 for area of triangle \(\frac{1}{2} \times 12 \times 13 \times \sin A\) (\(= 72.307\ldots\))
M1 for area of sector \(\frac{1}{2} \times 3^2 \times A\) with \(A\) in radians (\(= 5.338\ldots\))
dM1 for triangle area minus sector area (dependent on both previous M marks)