For a triangle with no right angle, choose the sine rule when you have a side and its opposite angle, and the cosine rule when you have two sides and the angle between them, or all three sides. The decision is made from the given information before any button is pressed.
This lesson follows right-triangle trigonometry, which you use whenever a right angle is present. Check the current Cambridge syllabus for your tier, because these rules belong to the Extended content.
What does each rule need?
Label the triangle so that side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C. The two rules then read:
- Sine rule: a/sin A = b/sin B = c/sin C
- Cosine rule: a² = b² + c² − 2bc cos A
The sine rule links each side with its opposite angle. You need one complete pair (a side with its opposite angle) and one more piece of information.
The cosine rule links three sides with one angle. It is the tool when no complete pair exists yet.
A quick decision routine
- Mark what you are given on the diagram: sides and angles.
- Ask: is there a side and its opposite angle, both known? If yes, use the sine rule.
- If not, ask: are two sides and the included angle known, or all three sides? Use the cosine rule.
- Write the formula with your letters before substituting numbers.
If you are given two angles, find the third one first, because angles add to 180°. That creates the pair you need.
Worked example 1: sine rule
In triangle ABC, angle A = 38°, angle B = 75° and BC = 9.2 cm. Find AC.
Step 1, label: BC is opposite A, so a = 9.2. AC is opposite B, so b is the unknown.
Step 2, choose: a and A form a complete pair, and angle B is known. Use the sine rule.
Step 3, substitute: b / sin 75° = 9.2 / sin 38°.
Step 4, rearrange: b = 9.2 × sin 75° ÷ sin 38° = 8.887 ÷ 0.6157 = 14.43.
Answer: AC = 14.4 cm (3 s.f.).
Worked example 2: cosine rule
In triangle PQR, PQ = 7 cm, PR = 10 cm and angle P = 52°. Find QR.
Step 1, choose: there is no side-and-opposite-angle pair, but the angle 52° sits between the two known sides. Use the cosine rule.
Step 2, substitute: QR² = 7² + 10² − 2 × 7 × 10 × cos 52° = 149 − 140 × 0.6157.
Step 3, multiply before subtracting: 140 × 0.6157 = 86.19, so QR² = 62.81.
Step 4, square root: QR = 7.925.
Answer: QR = 7.93 cm (3 s.f.).
The mistake to watch for
A frequent slip is to subtract before multiplying in the cosine rule.
Mistaken working: 7² + 10² − 2 × 7 × 10 = 9, then 9 × cos 52° = 5.54, so QR = 2.35.
The student treated the whole expression as (49 + 100 − 140) × cos 52°.
The term 2bc cos A is one product and is worked out first. Only then is it taken away from b² + c².
A good check is to ask whether the answer is sensible: QR cannot be shorter than the difference of the other two sides, which is 3 cm. A result of 2.35 cm fails that test, and 7.93 cm passes it.
Check yourself
1. A triangle has sides 6 cm, 8 cm and 11 cm. You want an angle. Which rule starts the solution?
Show answer
No side has a known opposite angle, and all three sides are known. Use the cosine rule.
2. In triangle XYZ, angle X = 50°, angle Y = 60° and XZ = 12 cm. Find YZ.
Show answer
XZ is opposite Y, and YZ is opposite X. Sine rule: YZ / sin 50° = 12 / sin 60°. So YZ = 12 × 0.7660 ÷ 0.8660 = 10.61, giving 10.6 cm.
3. Two sides of a triangle measure 5 cm and 8 cm, and the angle between them is 120°. Find the third side.
Show answer
Cosine rule: x² = 25 + 64 − 2 × 5 × 8 × cos 120° = 89 − 80 × (−0.5) = 89 + 40 = 129. So x = 11.36, giving 11.4 cm. The obtuse angle makes the cosine negative, so the third side is longer than either known side.
Where this leads next
Carry the same choice into directions with bearings and north lines, then into two-stage journey diagrams. The triangle and bearings reasoning board makes you state the rule before it calculates, which builds the habit, and the non-calculator working trainer helps with the arithmetic.
Some students can run each formula but hesitate at the choice. That is the kind of decision our teachers watch for in online one-to-one Mathematics tuition.