Maths › Trigonometry › Triangles and the sine and cosine rules
Triangles and the sine and cosine rules
Right-angled trigonometry stops at right angles. These two rules do not. The sine and cosine rules solve triangles from the usual combinations of given sides and angles, the area can be found from two sides and their included angle, and the sine rule may produce two possible angles, which must be checked geometrically.
IN THIS TOPIC
- Use the sine rule to find sides and angles, pairing each side with the angle facing it.
- Use the cosine rule from two sides and the included angle, or from all three sides.
- Handle the ambiguous case, and find areas from two sides and the included angle.
COMMON MISCONCEPTION
If sin B = 0.83, then B = sin−1 0.83 and that settles it.
The sine rule
Label every side with the lower-case letter of the angle facing it, a opposite A and so on. The sine rule then says the ratio of a side to the sine of its opposite angle stays the same all the way round the triangle,
and it is the tool you reach for whenever the given quantities include an opposite pair, one side together with the angle facing it, plus one further piece of information.
WORKED EXAMPLE
Solving a triangle from two angles and a side
In triangle ABC, A = 40°, B = 75° and a = 8 cm. Find the remaining angle and sides.
Angles first. C = 180° − 40° − 75° = 65°.
The known opposite pair is a and A, so b = 8 sin 75°/sin 40° = 12.0 cm and c = 8 sin 65°/sin 40° = 11.3 cm, both to 3 significant figures.
The largest side has come out opposite the largest angle. That check takes two seconds and it should end every triangle question you answer.
The cosine rule and the area formula
Sometimes the given pieces do not form an opposite pair. Two sides with the angle between them, say, or three sides and no angle at all. The cosine rule covers those cases,
which is Pythagoras carrying a correction term for the tilt of the angle. Alongside it sits the area formula,
where C is the angle squeezed between the sides a and b. Neither formula is in the booklet, so both have to be in your head.
WORKED EXAMPLE
Two sides and the angle between them
A triangle has sides of 5 cm and 7 cm meeting at 60°. Find the third side and the triangle's area.
Cosine rule first. a2 = 52 + 72 − 2 × 5 × 7 × cos 60° = 25 + 49 − 35 = 39, so a = √39 = 6.24 cm.
Area = ½ × 5 × 7 × sin 60° = 15.2 cm2.
Hold a2 = 39 exact until the last line. Rounding to 6.24 and then squaring it back is how avoidable accuracy marks disappear, and √39 is a perfectly good final answer wherever the question has not demanded decimals.
GUIDED PRACTICE
Three sides, no angles
A triangle has sides 4 cm, 6 cm and 8 cm. Find its largest angle, before opening the working.
Show the working
The largest angle faces the longest side, 8. Rearranged, the cosine rule gives cos θ = (42 + 62 − 82)/(2 × 4 × 6) = −12/48 = −0.25.
θ = cos−1(−0.25) = 104.5°.
The negative cosine indicated an obtuse angle before you pressed the inverse button. The sign of the cosine distinguishes acute from obtuse, which the sine cannot do.
The ambiguous case
Sine does not distinguish an angle from its supplement, because sin θ = sin (180° − θ). So when the sine rule gives sin B = 0.83, there are two candidates, B = 56.1° or 123.9°. Whether both are valid is a question about the triangle, not about the algebra. A candidate is valid only if the angles found so far leave a positive value for the third angle.
INDEPENDENT PRACTICE
Both triangles, in full
In triangle ABC, A = 40°, b = 9 cm and a = 7 cm. Find the two possible values of B, and the two possible values of C.
Show the working
Sine rule: sin B = 9 sin 40°/7 = 0.826, so B = 55.7° or its supplement 124.3°.
Now test each. 40° + 55.7° leaves C = 84.3°, and 40° + 124.3° leaves C = 15.7°. Both are positive, so both triangles exist.
B = 55.7° or 124.3°, C = 84.3° or 15.7°. The side a = 7 is shorter than b = 9, and that is what lets it swing down on either side of the perpendicular.
Had a been 9 or more, only the acute B would have been valid. One line of existence-checking after the inverse sine is what earns this mark.
ASSESSMENT FOCUS
- Label the sides and angles in opposite pairs before quoting anything. A mispaired sine rule is wrong from its first line onward.
- Let the given information choose the rule. An opposite pair points to the sine rule; two sides with the included angle, or all three sides, points to the cosine rule.
- After any inverse sine, write the supplement down too and test whether it leaves a positive third angle.
- Keep squared lengths exact while you work, a2 = 39 and not a = 6.244997…, then round once at the end.
- In ½ab sin C the angle must be the one between the two sides you have used. If it is not, find it first.
CHECK YOURSELF
A triangle has sides 5 cm and 7 cm meeting at an angle of 60°. Without a calculator, find the exact length of the third side and the exact area.
Show a hint
cos 60° = ½ and sin 60° = √3/2 are exact values.
Show the answer
Cosine rule: a2 = 25 + 49 − 70 × ½ = 39, so the third side is exactly √39 cm.
Area = ½ × 5 × 7 × √3/2 = 35√3/4 cm2.
Both answers stayed exact because 60° carries exact trig values. The next lesson makes that habit systematic.
Sine rule for opposite pairs, cosine rule for included angles or three sides.
Inverse sine gives two possible angles; the geometry of the triangle determines how many are valid.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the triangles and the sine and cosine rules questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Use the sine rule to find sides and angles, pairing each side with the angle facing it.
- Use the cosine rule from two sides and the included angle, or from all three sides.
- Handle the ambiguous case, and find areas from two sides and the included angle.
Open the full revision checklist to see every objective in the course in one place.