This Year 11 sine rule revision notes is written for pupils aged 15-16 and follows the National Curriculum. In short, it shows you how to solve non-right-angled triangles and find triangle areas. Before you start, it helps to revise transformations and Pythagoras and trigonometry in right-angled triangles, because this lesson builds on them.
Year 11 sine rule revision notes: what you will learn
This is a Higher tier topic. SOH CAH TOA only works in right-angled triangles; for any other triangle you need the sine rule, the cosine rule and the area formula using sine. Label the triangle with capital letters for the angles and lower-case letters for the sides opposite: side \(a\) is opposite angle \(A\).
Starter activity
You cannot walk straight across a river but you want to know how wide it is. You stand at point \(A\) on your side, look at a tree \(T\) on the opposite bank, then walk 50 m along the bank to \(B\) and look at the tree again. You can measure the angles at \(A\) and \(B\) with a protractor. Draw the triangle \(ABT\). Which measurements do you know, and what else would you need to find the width?
The sine rule
Definition (Higher): labelling a triangle
Use capital letters for the angles and lower-case letters for the sides opposite. A matching pair is a side and the angle opposite it, such as \(a\) and \(A\). The included angle is the angle between two known sides.
Theorem (Higher): the sine rule
\[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\] Use it when you know a matching pair (a side and the angle opposite it) and one more piece of information.
- To find a side, put the sides on top: \(a=\dfrac{b\sin A}{\sin B}\).
- To find an angle, put the sines on top: \(\sin A=\dfrac{a\sin B}{b}\).
If you only know two angles, find the third from \(A+B+C=180^\circ\) first.
Worked example 1 (Higher): a side
\(A=50^\circ\), \(B=70^\circ\), \(b=12\) cm. Find \(a\).
\(a=\dfrac{12\sin50^\circ}{\sin70^\circ}=9.78\) cm.
Worked example 2 (Higher): an angle
\(a=9\), \(b=12\), \(A=40^\circ\). Find acute angle \(B\).
\(\sin B=\dfrac{12\sin40^\circ}{9}=0.857\), so \(B=\sin^{-1}0.857=59.0^\circ\). Because sine is positive in two quadrants, there could be a second angle \(180^\circ-59.0^\circ=121.0^\circ\); the question will normally say if the angle is acute or obtuse.
The cosine rule
Theorem (Higher): the cosine rule
\[a^2=b^2+c^2-2bc\cos A\] Use it when you know two sides and the angle between them (to find the third side), or all three sides (to find an angle): \[\cos A=\frac{b^2+c^2-a^2}{2bc}\]
Worked example 3 (Higher): a side
\(b=7\), \(c=9\), \(A=60^\circ\). Find \(a\).
\(a^2=49+81-2\times7\times9\times\cos60^\circ=130-63=67\), so \(a=\sqrt{67}=8.19\).
Worked example 4 (Higher): an angle
Sides 7, 8, 9. Find the angle opposite the side of length 7. For sides 5, 6, 10 find the angle opposite 10.
\(\cos A=\dfrac{8^2+9^2-7^2}{2\times8\times9}=\dfrac{96}{144}=\dfrac23\), so \(A=48.2^\circ\). If the angle is obtuse the cosine is negative: for sides 5, 6, 10, \(\cos C=\dfrac{25+36-100}{60}=-0.65\), so \(C=130.5^\circ\).
Choosing the right rule
Definition (Higher): acute and obtuse angles
An acute angle is less than \(90^\circ\) and an obtuse angle is between \(90^\circ\) and \(180^\circ\). Sine is positive for both, so the sine rule can give two possible angles, but cosine is negative for an obtuse angle, so the cosine rule gives only one.
Method (Higher): choosing and using a rule
- Sketch the triangle and label sides \(a,b,c\) opposite angles \(A,B,C\).
- Mark what you know and what you want to find.
- Pick the sine rule if you have a matching pair, otherwise the cosine rule.
- Write the formula, substitute, and check that the biggest side faces the biggest angle.
Key fact (Higher): which rule?
| You know | Use |
|---|---|
| two angles and any side | sine rule |
| two sides and a non-included angle | sine rule |
| two sides and the included angle | cosine rule (find the third side) |
| three sides | cosine rule (find an angle) |
Area of a triangle
Rule (Higher): area using sine
\[\text{Area}=\tfrac12ab\sin C\] Here \(a\) and \(b\) are two sides and \(C\) is the angle between them.
Worked example 5 (Higher): area and backwards
Find the area when \(a=8\), \(b=11\), \(C=40^\circ\). If the area is 30 with \(a=10\), \(b=8\), find \(C\).
Area \(=\tfrac12\times8\times11\times\sin40^\circ=28.3\) cm\(^2\). Backwards: \(30=40\sin C\), so \(\sin C=0.75\) and \(C=48.6^\circ\) (or \(131.4^\circ\)).
Worked example 6 (Higher): exact values
Find the area of an equilateral triangle of side 6 exactly. In a triangle with \(AB=4\), \(AC=6\), \(A=60^\circ\), find \(BC\).
Area \(=\tfrac12\times6\times6\times\sin60^\circ=18\times\dfrac{\sqrt3}{2}=9\sqrt3\). \(BC^2=16+36-48\times\tfrac12=28\), so \(BC=2\sqrt7\).
Applying the rules to real problems
Worked example 7 (Higher): a ship and a quadrilateral
(a) A ship sails 10 km, then 14 km so that the interior angle is \(110^\circ\). Find the distance back to the start. (b) In quadrilateral \(ABCD\), \(AB=6\), \(BC=8\), angle \(ABC=90^\circ\), \(CD=7\), \(AD=5\) (cm). Find angle \(ADC\).
(a) Cosine rule: \(\sqrt{10^2+14^2-2\times10\times14\times\cos110^\circ}=19.8\) km. (b) Pythagoras in \(ABC\): \(AC=\sqrt{36+64}=10\). In triangle \(ACD\) all three sides are known (5, 7, 10), so \(\cos D=\dfrac{5^2+7^2-10^2}{2\times5\times7}=\dfrac{-26}{70}\), and \(D=111.8^\circ\). The cosine is negative, so the angle must be obtuse, and it faces the longest side \(AC\).
Key fact (Higher): why the formulas work
The cosine rule is Pythagoras' theorem with a correction term: when \(A=90^\circ\), \(\cos A=0\) and the rule becomes \(a^2=b^2+c^2\). If \(A\) is acute, \(a^2\lt b^2+c^2\); if obtuse, \(a^2>b^2+c^2\). The sine rule comes from the height of the triangle: height \(=b\sin A=a\sin B\). The area formula comes from the same height: area \(=\tfrac12\times a\times b\sin C\).
Note: exam technique
- Write the formula first, then substitute: it earns a method mark even if the arithmetic goes wrong.
- Use brackets on your calculator for \(2bc\cos A\), and do not subtract before multiplying.
- When finding an angle with the cosine rule, \(\cos A\) must be between \(-1\) and \(1\). If not, check the numbers.
- Give angles to 1 decimal place and lengths to 3 significant figures unless told otherwise.
Watch out: common mistakes
- Using the cosine rule with an angle that is not between the two known sides.
- Calculating \(2bc\cos A\) wrongly: do \(2\times b\times c\times\cos A\) in one go and subtract at the end.
- Calculator in the wrong mode.
- Forgetting to take the square root at the end of the cosine rule.
- Rounding angles too early.
Worked example 8: Choosing the right rule
In each case say which rule to use: (a) two angles and a side, (b) two sides and the included angle, (c) three sides.
(a) You know a pair of an angle and its opposite side, so use the sine rule. (b) You know two sides and the included angle, so use the cosine rule to find the third side. (c) You know all three sides, so use the cosine rule rearranged for an angle. A fourth case, two sides and a non-included angle, uses the sine rule and may have two solutions.
Worked example 9: Finding the area when not given the included angle
A triangle has sides 9 cm and 12 cm with the angle \(A=35^\circ\) opposite the 9 cm side. Find the area.
First use the sine rule: \(\sin B=\dfrac{12\sin35^\circ}9=0.7648\), so \(B=49.9^\circ\). Then \(C=180-35-49.9=95.1^\circ\). The area uses the sides 9 and 12 and the angle between them, which is \(C\): \(\tfrac12\times9\times12\times\sin95.1^\circ=53.8\) cm\(^2\).
Worked example 10: Rearranging the sine rule
Show that the sine rule can be written \(\dfrac{\sin A}a=\dfrac{\sin B}b\) and use it to find an angle.
Taking the reciprocal of each fraction in \(\dfrac a{\sin A}=\dfrac b{\sin B}\) gives \(\dfrac{\sin A}a=\dfrac{\sin B}b\). Use this form to find an angle, so that the unknown is on the top: with \(a=9\), \(b=12\) and \(A=35^\circ\), \(\sin B=\dfrac{12\sin35^\circ}9\). Use the form with the sides on top to find a side.
Worked example 11: Applying the rules in three dimensions
In a triangle formed on a hillside, \(AB=40\) m, \(AC=55\) m and angle \(BAC=38^\circ\). Find \(BC\).
Use the cosine rule for two sides and the included angle: \(BC^2=40^2+55^2-2\times40\times55\cos38^\circ=1600+3025-3467=1158\). So \(BC=34.0\) m. Even when the triangle lies on a slope, the lengths and angles in the plane of the triangle obey the same rules.
Worked example 12: A bearing problem with two rules
A ship sails from \(P\) to \(Q\), 14 km on bearing \(065^\circ\), then from \(Q\) to \(R\), 10 km on bearing \(140^\circ\). Find \(PR\).
The angle at \(Q\) is \(180-(140-65)=105^\circ\). Cosine rule: \(PR^2=14^2+10^2-2\times14\times10\cos105^\circ=296+72.5=368.5\), so \(PR=19.2\) km. Draw a north line at \(P\) and at \(Q\) and mark the bearing angles to find the interior angle.
Worked example 13: The ambiguous case
In triangle \(ABC\), \(a=6\), \(b=8\) and \(A=40^\circ\). Find both possible values of \(B\).
\(\sin B=\dfrac{8\sin40^\circ}6=0.857\), so \(B=59.0^\circ\) or \(B=180-59.0=121.0^\circ\). Both triangles exist because \(40+121=161\lt 180\). The ambiguous case happens when the side opposite the known angle is shorter than the other known side.
Worked example 14: Area of a triangle from three sides
A triangle has sides 13 cm, 14 cm and 15 cm. Find the area using an angle.
Find the angle opposite 14: \(\cos B=\dfrac{13^2+15^2-14^2}{2\times13\times15}=\dfrac{198}{390}=0.5077\), so \(B=59.5^\circ\). Area \(=\tfrac12\times13\times15\times\sin59.5^\circ=84.0\) cm\(^2\). This is the classic 13, 14, 15 triangle, whose area is exactly 84.
Note: labelling a triangle
In these formulas the side \(a\) is always opposite the angle \(A\), side \(b\) opposite \(B\) and side \(c\) opposite \(C\). Label the triangle before you start. The sine rule needs a matching pair (a side and its opposite angle) and the cosine rule uses an angle between two known sides. Both rules also work for obtuse triangles: the cosine of an obtuse angle is negative, so the formula for the third side then gives a longer side. Always check that the longest side is opposite the largest angle.
Method: choosing and using a rule
- Sketch the triangle and label the known sides and angles with the correct letters.
- Decide: a side-angle pair with one more piece of information means the sine rule; two sides with the included angle, or all three sides, means the cosine rule.
- Write the formula, substitute, and solve. When finding an angle with the sine rule, think about whether it could be obtuse.
- Check the result: the largest angle faces the longest side and the angles add to \(180^\circ\).
Watch out: mistakes with the sine and cosine rules
- In the cosine rule, work out \(2bc\cos A\) before subtracting; do not subtract \(b^2+c^2-2bc\) and then multiply by \(\cos A\).
- Use the sine rule for the angle only when the angle is known to be acute or you have checked the obtuse case.
- For the area formula, the angle must be between the two sides used.
- Keep your calculator in degrees and round only at the end.
Did you know?
The sine rule and the cosine rule were used by navigators and surveyors for centuries before computers existed. In the 1800s, the Great Trigonometrical Survey of India measured a chain of triangles across the whole subcontinent with accurate angle instruments, starting from a carefully measured baseline of just a few miles. Using the sine rule repeatedly, surveyors computed the distances between mountains hundreds of kilometres apart and found the height of Mount Everest.
Otto’s tip
The longest side is always opposite the largest angle. After every answer, check that your biggest side faces your biggest angle. If it does not, look for a slip.
How to use this Year 11 sine rule revision notes
First, read the key points so that you know where the Year 11 sine rule revision notes is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the sine rule, cosine rule and area of a triangle worksheet and check every question with the worksheet answers. When you feel ready, test yourself with the multiple choice questions or the quick quiz. Finally, sit the sine rule, cosine rule and area of a triangle test and mark it with the test answers.
Also, the starters are a good warm-up, so you can keep the key facts fresh. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, this Year 11 sine rule revision notes gives you everything you need for the chapter.
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