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This page sets out the method for the cosine rule in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the The cosine rule topic page.
When you know two sides and the angle between them and want the side opposite that angle.
When you know all three sides and want any angle.
c² = a² + b² − 2ab cos C, where C is the included angle between sides a and b.
An included angle of 120 degrees is obtuse, so cos 120° = -1/2 and the opposite side is longer.
An included angle of 60 degrees with two equal sides closes an equilateral triangle, so c = a.
Rearrange to cos C = (a² + b² − c²) / (2ab).
A negative cosine means the angle is obtuse. Sides 2, 3 and 4 give an obtuse angle opposite 4.
Pythagoras is a special case of the cosine rule when C = 90° and cos 90° = 0.
Label the triangle before you substitute. Decide whether you are finding a side or an angle. If you know two sides and the angle between them, you are in the SAS case and you want the side opposite that angle. Put the known angle in the cosine, and put the side you want on the left as a. If you know all three sides, you are in the SSS case. Choose the angle you want, call the side opposite it a, and use the rearranged cosine formula.
Do not use the cosine rule for a side and the angle opposite that side when the included angle is still unknown. That layout is a different method. You also do not need the cosine rule for a right-angled triangle when two sides are enough for Pythagoras, though the cosine rule still works because cos 90^° = 0. On a calculator paper, set degrees. Cosine of 60^° is exactly 1/2.
Method: 1. Use the cosine rule for SAS when you want the side opposite the known angle. 2. Use the rearranged form for SSS when you want an angle. 3. Put the unknown side as a, and keep the calculator in degrees.
Write a^2 = b^2 + c^2 - 2bccos A, substitute, and only then square-root. Sides 7 cm and 10 cm enclose 60^°. Cosine of 60^° is 1/2, so a^2 = 7^2 + 10^2 - 2 × 7 × 10 × 1/2 = 49 + 100 - 70 = 79. The third side is √(79), which is 8.89 cm to 2 decimal places. The length is the positive square root. A negative root is not a length.
Watch the product 2bccos A. For 7 and 10 it is 140 × 1/2 = 70, not 140 and not 2 × 7 × 2 × 10. If the angle is acute, you subtract a positive number from b^2 + c^2. If someone adds 2bccos A instead, the side comes out too long. After the square root, compare the new side with the other two. In this example √(79) is about 8.89, which sits between 7 and 10, a sensible third side for a 60^° angle.
Method: 1. Substitute into a^2=b^2+c^2-2bccos A, then square-root. 2. Sides 7cm and 10cm with included 60^° give a^2=49+100-70=79. 3. The side is √(79)=8.89cm to 2 decimal places.
Rearrange to cos A = (b^2 + c^2 - a^2)/2bc. The side opposite the angle goes in the place of a, with a minus sign in front of its square. For sides 8 cm, 11 cm and 13 cm, the angle opposite 13 cm has cos A = (8^2 + 11^2 - 13^2)/(2 × 8 × 11) = (64 + 121 - 169)/176 = 16/176. Inverse cos of 16/176 is 84.8^° to 1 decimal place.
The largest angle is opposite the longest side. Here 13 cm is the longest side, so 84.8^° should be the largest angle, and it is still acute. A quick size check: if a^2 is less than b^2 + c^2, the cosine is positive and the angle is acute. If a^2 is greater than b^2 + c^2, the cosine is negative and the angle is obtuse. If you get a cosine bigger than 1 or less than -1, the sides cannot make a triangle, so check the substitution.
Method: 1. Rearrange to cos A=dfracb^2+c^2-a^22bc. 2. Opposite the 13cm side in a 8, 11, 13 triangle, cos A=16/176. 3. The angle is cos^-1 (16/176)=84.8^° to 1 decimal place.
Sides 6 cm, 7 cm and 11 cm give a negative cosine for the angle opposite 11 cm, because 6^2 + 7^2 = 85 and 11^2 = 121. Then cos A = (36 + 49 - 121)/(2 × 6 × 7) = (-36)/84 = -3/7. Inverse cos of a negative number is obtuse: 115.4^° to 1 decimal place. Do not report the acute angle with the same digits. The minus sign is the difference between about 64.6^° and 115.4^°.
You can find the area from the same angle, using 1/2bcsin A, without a second angle. For sides 9 cm and 14 cm with included angle 52^°, the opposite side is √(9^2 + 14^2 - 2 × 9 × 14 cos 52^°) = 11.0 cm to 1 decimal place, and the area is 1/2 × 9 × 14 × sin 52^° = 49.6 cm^2 to 1 decimal place. Round at the end. If you need the sine of an angle you have just found, use the full calculator value, not a value already chopped to 1 decimal place. Write 11.0 and 19.0 with the zero when the answer is to 1 decimal place, so the rounding is visible.
Method: 1. A negative cosine means the angle is obtuse. 2. Sides 6cm, 7cm and 11cm give cos A=-3/7, so the angle is 115.4^° to 1 decimal place. 3. Then find the area with 1/2bcsin A, rounding only at the end.
In a non-right-angled triangle the choice of rule depends on what you are given. The cosine rule is for two sides and the angle between them (to find the third side), or for all three sides (to find an angle). The sine rule needs a matching pair: an angle and the side opposite it. If you cannot find such a pair in the information, the cosine rule is the only way in.
Write the given facts as a short list under your sketch. For example "two sides and the included angle" makes the cosine rule the obvious choice. A second sign is a question that says "work out the length of the third side" or gives all three sides and asks for an angle. After that the rule is only substitution, so most of the skill is in the labelling. Always label the side opposite angle A as a.
Method: 1. Sketch the triangle and put the given lengths and angles on it. 2. Look for a side with its opposite angle. If there is none, choose the cosine rule. 3. Label the angle you want or the side opposite the given angle as a and A. 4. Write the rule with letters first, then replace each letter with a number.
Tip: If the angle is between the two sides you know, the answer is the side opposite that angle.
Most lost marks come from keying in the formula incorrectly. In a^2 = b^2 + c^2 - 2bccos A the product 2bccos A is subtracted as a whole. Work out b^2 + c^2 and 2bccos A separately, write both down, subtract, and only then take the square root. Doing (b^2 + c^2 - 2bc)cos A is a classic error that gives a much smaller side.
When you find an angle, rearrange the rule to cos A = (b^2 + c^2 - a^2)/2bc and type the numerator and denominator in brackets. A negative result is allowed and means the angle is obtuse, so do not drop the minus sign. The inverse cosine of -tfrac13 is 109.5^°, not 70.5^°. Keep full calculator accuracy and round only at the end to the accuracy asked for.
Method: 1. Write b^2 + c^2 and 2bccos A as two separate numbers. 2. Subtract them, then take the square root to get the side. 3. For an angle, use the rearranged form with brackets on top and bottom. 4. Check that the largest angle is opposite the longest side.
On a non-calculator paper the angle is usually 60^° or 120^°, because cos 60^° = tfrac12 and cos 120^° = -tfrac12. The cosine rule then collapses to whole numbers. With sides 7 and 8 and an included angle of 120^°, a^2 = 49 + 64 - 2 × 7 × 8 × (-tfrac12) = 49 + 64 + 56 = 169, so a = 13. Watch the double negative: the minus in the rule meets the minus in the cosine.
"Show that" questions give you the answer and you must reach it. Write the cosine rule, show the substitution, show the unsimplified arithmetic and then the value. Algebraic versions put x in the sides and ask you to show a quadratic such as x^2 + 3x - 40 = 0. Expand the brackets carefully, collect terms, and solve. Reject any negative value of x, because a length cannot be negative.
Method: 1. Write the cosine rule with the exact value of cos for the angle. 2. Handle the signs carefully, especially for 120^°. 3. Show every line of arithmetic so the printed result is clearly reached. 4. For algebra, expand, collect, rearrange to zero and solve, rejecting negative lengths.
A bearing problem often hides the angle you need. A boat sails from A to B and then from B to C, and you are given the two bearings. Back-bearings differ by 180^°, so the bearing of A from B is the bearing of B from A plus or minus 180^°. Draw a north line at B and work out the interior angle by subtraction, then use the cosine rule with the two legs and the angle between them to find the direct distance AC.
Other two-stage questions ask for a length followed by an area, or an angle followed by a bearing. Use the full unrounded value from the first part in the second part, and quote both results to the accuracy asked for. Keep the diagram in front of you and label every length as you find it. Marks are given for each correct step, so even if one part goes wrong you can still collect marks for method.
Method: 1. Draw north lines and mark the given bearings on the sketch. 2. Use the fact that bearings differ by 180^° in opposite directions to find the interior angle. 3. Apply the cosine rule to the two known legs and the interior angle. 4. Use unrounded values in later parts and give units such as km.
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