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Or take a quick quiz: 6 questions on this subtopic in AS Maths (Edexcel), marked as you tap.
This subtopic covers the sine rule, cosine rule and area formula for any triangle. Sine, cosine and tangent are also defined for angles of any size using the unit circle. These skills are used throughout Paper 1 and in mechanics.
Unit circle: the point at angle θ is \((\cos\theta, \sin\theta)\)
Notes
Sine rule, cosine rule and area
diagram
Label the triangle so that side a is opposite angle A, side b is opposite angle B, and so on.
Each side takes the small letter of the angle opposite it.
Sine rule: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\). Use it when you know a side and the angle opposite it.
Cosine rule: \(a^2 = b^2 + c^2 - 2bc\cos A\), or \(\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}\). Use it for two sides and the included angle, or for three sides.
Area \(= \dfrac{1}{2}ab\sin C\), where C is the angle between sides a and b.
Ambiguous case: when you use the sine rule to find an angle, \(\sin B = k\) has two solutions between 0° and 180°, B and 180° − B. The obtuse one is only possible if it, added to the angle you already know, is less than 180°.
Trig ratios for any angle
diagram
Measure θ anticlockwise from the positive x-axis. The point on the unit circle at angle θ is \((\cos\theta, \sin\theta)\), and \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\)
On the unit circle, the x-coordinate of P is cos θ and the y-coordinate is sin θ, for any angle θ.
This defines sin, cos and tan for any angle, including obtuse, reflex and negative angles, and gives their signs in each quadrant.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Find a side and the area of a triangle
4 marksGrade D
Label the sides and angles, and decide which rule fits.
Substitute carefully. The cosine of an obtuse angle is negative, so the \(-2bc\cos A\) term adds on.
Take the square root at the end and round to 3 significant figures.
Example. In triangle PQR, PQ = 6 cm, PR = 9 cm and angle QPR = 110°. (a) Find the length QR. (b) Find the area of the triangle.
Check that each angle leaves room for the third angle.
Example. In triangle ABC, AB = 10 cm, BC = 7 cm and angle BAC = 40°. Find the two possible sizes of angle ACB, to one decimal place.
Show the model answerHide the model answer
\(\dfrac{\sin C}{10} = \dfrac{\sin 40^\circ}{7}\)
M1
\(\sin C = 0.9182\dots\), so C = 66.7°
A1
or C = 180° − 66.7° M1 = 113.3°
A1
(Both are possible, because 40° + 113.3° is less than 180°.)
Don’t lose marks
Missing the obtuse angle in the ambiguous case, or giving it when it does not fit in the triangle.
Working out \((b^2 + c^2 - 2bc)\cos A\) instead of \(b^2 + c^2 - 2bc\cos A\)
Trying to use the sine rule without a known side and opposite angle pair.
Rounding lengths early and then using them again.
More tips
Memory tricks
With three sides, use the cosine rule to find the largest angle first (it may be obtuse); the other two angles must then be acute, so the sine rule is safe.
The longest side is opposite the largest angle: a quick check on any answer.
Exam technique
Give answers to 3 significant figures unless told otherwise, with units (cm, cm2).
For 'show that' results, write every step; the final line must match the given answer exactly.
Draw or label a diagram before choosing a rule.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Use the sine and cosine rulesChoose the rule from what you know: a side and its opposite angle means the sine rule; two sides and the included angle, or three sides, means the cosine rule.
Grade D
Find the area of a triangleUse \(\dfrac{1}{2}ab\sin C\), where C is the angle between the sides a and b.
Grade C
Find both angles in the ambiguous caseIf the sine rule gives \(\sin B = k\), the angle could be B or 180° − B; check whether both fit in the triangle.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
The diagram shows triangle \(ABC\) with \(AB = 9\) cm, \(AC = 4\) cm and angle \(BAC = 29^\circ\). Diagram not drawn to scale
[object Object]
(a) Find the length of \(BC\), giving your answer in cm to 3 significant figures.[2]
(b) Find the size of angle \(ABC\), giving your answer in degrees to 1 decimal place.[2]
(c) Find the area of triangle \(ABC\), giving your answer to 3 significant figures.[2]
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 3Extreme10 marks
A triangle has sides of length \(a\), \(b\) and \(c\), and the angle opposite the side of length \(c\) is \(C\). The area of the triangle is \(\Delta\).
(a) Use \(\Delta = \frac{1}{2}ab\sin C\) and the cosine rule to show that \[16\Delta^2 = 4a^2b^2 - (a^2 + b^2 - c^2)^2.\][4]
(b) Hence show that \(\Delta = \sqrt{s(s - a)(s - b)(s - c)}\), where \(s = \frac{1}{2}(a + b + c)\).[4]
(c) A triangle has sides of length 7 cm, 8 cm and 9 cm. Find the exact area of the triangle, giving your answer in the form \(k\sqrt{5}\) cm².[2]
Show the answer and mark scheme
(a)Answer: Shown
M1 for \(4\Delta^2 = a^2b^2\sin^2 C = a^2b^2(1 - \cos^2 C)\)
M1 for \(\cos C = \frac{a^2 + b^2 - c^2}{2ab}\) substituted