Chapter 7 of 8

All 8 chapters
  1. 1Pythagoras' theorem, readunderstood
  2. 2Using Pythagoras, readunderstood
  3. 3Opposite, adjacent and hypotenuse, readunderstood
  4. 4Sine, cosine and tangent, readunderstood
  5. 5Finding angles, readunderstood
  6. 6Exact values, readunderstood
  7. 7The sine rule and the cosine rule, readunderstood
  8. 8Area of any triangle, and bearings, readunderstood

M4.3.7 Chapter 7

The sine rule and the cosine rule

A triangle has no right angle. You know two of its angles and one side. How can you find the other sides?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Use the sine rule to find a side or an angle
  • Use the cosine rule when you know two sides and the angle between

Name a triangle ABC. Use a capital letter for each angle, and the small letter for the side opposite it. So side a faces angle A, side b faces angle B and side c faces angle C.

A triangle ABC. Side a, opposite A, is labelled a. Side b, opposite B, is labelled b. Side c, opposite C, is labelled c. The corners are named A, B and C.ABCabc
Each side is named after the corner it faces.

Draw the height h from C down to the side c. It makes two right angled triangles. In the left one, sin A = h ÷ b, so h = b × sin A. In the right one, sin B = h ÷ a, so h = a × sin B.

A triangle ABC with a dashed height h from C down to the side c, and a small square where the height meets the side. The corners are named A, B and C. The sides are labelled a, b and c.ABCabch
The height h is a side of two right angled triangles.

Both are h, so b × sin A = a × sin B. Divide both sides by sin A × sin B. On the left you are left with b ÷ sin B. On the right you are left with a ÷ sin A. So a ÷ sin A = b ÷ sin B. Cutting from another corner gives the same with c.

This is the sine rule: a ÷ sin A = b ÷ sin B = c ÷ sin C. Our height fell inside the triangle, because angles A and B were both acute. The rule is true for every triangle. A little later we see why it holds for an obtuse angle too.

Use it when you have a matching pair: an angle and the side opposite it. Then one more piece lets you find another pair. Here is a side.

A triangle ABC with the angle 40° at A, the angle 65° at B, the side a labelled 8 cm opposite A, and the side b labelled b opposite B.ABC8 cmb40°65°

Worked example

In triangle ABC, A = 40°, B = 65° and a = 8 cm. Find the side b, to 1 decimal place.

  1. We know a and A, a matching pair, and B. We want b, opposite B. Use the sine rule.
  2. a ÷ sin A = b ÷ sin B, so b = a × sin B ÷ sin A.
  3. b = 8 × sin 65° ÷ sin 40°.
  4. To 4 decimal places, sin 65° = 0.9063 and sin 40° = 0.6428.
  5. To 4 decimal places, 8 × 0.9063 ÷ 0.6428 = 11.2794.
  6. Round to 1 decimal place: b = 11.3 cm.

Answer: To 1 decimal place, b is 11.3 cm.

You can also find an angle. Turn the rule upside down: sin A ÷ a = sin B ÷ b. Then sin B = b × sin A ÷ a, and the inverse sin key gives B.

A triangle ABC with the angle 50° at A, the side a labelled 12 cm opposite A, the side b labelled 9 cm opposite B, and the angle at B marked B.ABC12 cm9 cm50°B

Worked example

In triangle ABC, A = 50°, a = 12 cm and b = 9 cm. Find the angle B, to 1 decimal place.

  1. We know a and A, a matching pair, and b. We want B, opposite b. Use the sine rule upside down.
  2. sin B ÷ b = sin A ÷ a, so sin B = b × sin A ÷ a.
  3. sin B = 9 × sin 50° ÷ 12.
  4. To 4 decimal places, sin B = 9 × 0.7660 ÷ 12 = 0.5745.
  5. Use the inverse sin key on 9 × sin 50° ÷ 12, with all the digits. To 3 decimal places, B = 35.067.
  6. Round to 1 decimal place: B = 35.1°.
  7. Here b is shorter than a, so B is smaller than A and must be acute. 35.1° is the only answer.

Answer: To 1 decimal place, B is 35.1°.

So far every angle was below 90°. A triangle can have an angle above 90°, called an obtuse angle. Then the height from C falls outside the triangle, on the line c extended beyond A.

A triangle ABC with an obtuse angle of 120° at A. The sides are labelled a, b and c. A dashed line continues the side c beyond A. A dashed height h falls from C to meet that line at a right angle. The angle between the dashed line and the side b is marked 60°.120°60°ABCcbah
The height falls outside. The angle outside at A is 180° − 120° = 60°.

The right angled triangle outside has the angle 180° − A at A. In the picture that is 60°. Its height is h = b × sin 60°.

We want h = b × sin A to stay true for the obtuse angle A. So we choose sin A to mean sin(180° − A). It is a definition. We choose it so that the rules in this chapter keep working.

We make the same choice for the cosine: cos A = −cos(180° − A). Here is why. When A is obtuse, the foot of the height lies beyond A. So the far part of the base is c + b cos(180° − A). With our choice that is c − b cos A again, so the cosine rule keeps its form. Check on a calculator. sin 114.6° and sin 65.4° both give 0.9092. cos 100° gives −0.1736 and cos 80° gives 0.1736. A negative cosine for an angle above 90° is not a fault.

Take care. When you find an angle with the sine rule, there can be two answers. Two different angles have the same sine: an acute angle and 180° minus it. The calculator gives only the acute one.

Look at the picture. Both triangles have A = 30°, a = 5.5 cm and b = 10 cm. Swing the 5.5 cm side round C. It touches the base line in two places, B₁ and B₂. Both make real triangles: 30° + 65.4° and 30° + 114.6° are both under 180°.

One picture with a base line through A. The corner C is 10 cm from A, with the angle 30° at A. A dashed arc of 5.5 cm about C meets the base line in two places, B₁ and B₂. The side AC is labelled 10 cm and the side CB₁ is labelled 5.5 cm. A small tick on CB₁ and on CB₂ shows that they are equal. The angle at B₁ is 65.4° and the angle at B₂ is 114.6°, which is obtuse. The corners are named A, C, B₁ and B₂.ACB₁B₂10 cm5.5 cm30°65.4°114.6°
The same three facts, two triangles. The tick marks CB₁ and CB₂ as equal. B is 65.4° at B₁ or 114.6° at B₂.

The two answers for B are 65.4° and 180° − 65.4° = 114.6°. Always test the second angle. Add it to the known angle: 30° + 114.6° = 144.6°, which is below 180°. So the second triangle exists.

There is at most one triangle in these cases. First, the angle you find is opposite the shorter of the two known sides. Then it must be acute. That was the case before: B faced b = 9 cm, shorter than a = 12 cm.

Second, the known angle is 90° or more. Then the side opposite it must be the longest side, or there is no triangle. Third, the known angle and the second angle add to 180° or more. If sin B works out above 1, there is no triangle at all.

The sine rule needs a matching pair. Sometimes you have no pair. You may know two sides and the angle between them, or all three sides. Then use the second rule.

Draw the height h from C onto side c again. It cuts c into two parts. The part next to A is b × cos A. The rest is c − b × cos A. The height is h = b × sin A.

A triangle ABC with a dashed height h from C down to the side c, and a small square where it meets the side. Under the side c, a line labelled b cos A covers the part from A to the foot of the height. A line labelled c − b cos A covers the rest. A third line labelled c covers the whole side. The corners are named A, B and C, and the sides are labelled a and b.ABCbahb cos Ac − b cos Ac
The foot of the height cuts c into b cos A and c − b cos A.

Now use Pythagoras in the right angled triangle on the right: a² = h² + (c − b cos A)². Put in h = b sin A.

First expand the bracket. Multiply each part of (c − b cos A) by each part of the other bracket. Then (c − b cos A)(c − b cos A) = c² − 2bc cos A + b² cos²A.

So a² = b² sin²A + c² − 2bc cos A + b² cos²A. Collect the two b² terms. They give b² × (sin²A + cos²A).

In the left triangle Pythagoras says (b sin A)² + (b cos A)² = b². So sin²A + cos²A = 1, and the sum is b². Therefore a² = b² + c² − 2bc cos A.

This is the cosine rule: a² = b² + c² − 2bc cos A. Our height fell inside, but the rule holds for every triangle. For an obtuse A the cosine is negative, so −2bc cos A adds to the sum. If A = 90°, then cos A = 0 and the rule is just Pythagoras. The extra part −2bc cos A corrects for the angle not being 90°.

A triangle ABC with the angle 50° at A between the side b of 8 cm and the side c of 11 cm. The side a is labelled a and is opposite A.ABCa8 cm11 cm50°

Worked example

In triangle ABC, b = 8 cm, c = 11 cm and A = 50°. Find the side a, to 1 decimal place.

  1. We know two sides, b = 8 and c = 11, and the angle A = 50° between them. We want a. Use the cosine rule.
  2. a² = b² + c² − 2 × b × c × cos A = 8² + 11² − 2 × 8 × 11 × cos 50°.
  3. Work out the squares and the product: a² = 64 + 121 − 176 × cos 50°.
  4. To 4 decimal places, cos 50° = 0.6428.
  5. Then a² = 185 − 176 × 0.6428 = 71.8672.
  6. Take the square root, to 3 decimal places: a = 8.477.
  7. Round to 1 decimal place: a = 8.5 cm.

Answer: To 1 decimal place, a is 8.5 cm.

The cosine rule can also find an angle when you know all three sides. Rearrange it: cos A = (b² + c² − a²) ÷ (2 × b × c). Then use the inverse cos key.

A triangle ABC with the side a of 7 cm, the side b of 8 cm, the side c of 9 cm, and the angle at A marked A.ABC7 cm8 cm9 cmA

Worked example

A triangle has sides a = 7 cm, b = 8 cm and c = 9 cm. Find the angle A, opposite a, to 1 decimal place.

  1. We know all three sides. We want A, opposite a. Use the cosine rule.
  2. Rearrange: cos A = (b² + c² − a²) ÷ (2 × b × c).
  3. cos A = (8² + 9² − 7²) ÷ (2 × 8 × 9).
  4. The top is 64 + 81 − 49 = 96.
  5. The bottom is 2 × 8 × 9 = 144.
  6. To 4 decimal places, cos A = 96 ÷ 144 = 0.6667.
  7. Use the inverse cos key on 96 ÷ 144. To 4 decimal places, A = 48.1897.
  8. Round to 1 decimal place: A = 48.2°.

Answer: To 1 decimal place, A is 48.2°.

See it three ways

With things

Two sticks joined at one end, 6 cm and 8 cm long, opened to a 60° angle. How far apart are the free ends?

As a picture

A triangle ABC with the angle 60° at A between the side b of 6 cm and the side c of 8 cm. The side a is labelled a and is opposite A.ABCa6 cm8 cm60°

In symbols

a² = 6² + 8² − 2 × 6 × 8 × cos 60° = 36 + 64 − 48 = 52. So a = √52, which is 7.2 cm to 1 decimal place.

Try it

Two sticks and a string

You need: two sticks or pencils of 6 cm and 8 cm, a protractor, a ruler.

  1. Lay the two sticks from one point, with a 60° angle between them. Use the protractor.
  2. Measure the distance between the free ends.
  3. Work out 6² + 8² − 2 × 6 × 8 × cos 60°, then take the square root.

What you will see: The sum gives √52, which is 7.2 cm to 1 decimal place. Your measure should be close.

If you cannot try it, this is what happens.
A triangle ABC made by two sticks with the angle 60° at A. The sticks are labelled 6 cm and 8 cm, and the gap between their free ends is labelled 7.2 cm.ABC7.2 cm6 cm8 cm60°

A wrong idea many people have

Use the sine rule or the cosine rule, whichever comes to mind first.

Why it feels right: Both rules have a side and an angle in them, so either seems to fit.

The test that breaks it: Try two sides and the angle between them. The sine rule has no matching pair, so you cannot start. With a pair, the sine rule needs just one more piece.

What is true: Pick the rule from what you know. A matching pair (an angle and its opposite side) means the sine rule. Two sides and the angle between, or three sides, means the cosine rule.

A wrong idea many people have

The sine rule always gives exactly one angle.

Why it feels right: The calculator shows one answer, and one answer looks final.

The test that breaks it: Take A = 30°, a = 5.5 cm and b = 10 cm. sin B = 0.909, so the calculator gives 65.4°. But 114.6° has the same sine, and a triangle with 30° and 114.6° is fine.

What is true: Sine gives two angles, B and 180° − B. The second fits only when the known angle plus 180° − B is under 180°. That needs B opposite the longer of the two known sides. If sin B works out above 1, no triangle has those measures.

Key words

sine rule
a ÷ sin A = b ÷ sin B = c ÷ sin C, true for every triangle.
cosine rule
a² = b² + c² − 2bc cos A, true for every triangle.

Back to your guess

The answer is A: Cut it into right angled triangles, and the work gives two rules that work for any triangle.

Cutting a triangle with a height makes two right angled triangles. That gives the sine rule and the cosine rule, which work in every triangle. This chapter builds both rules from the height, then uses them.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. You know two sides of a triangle and the angle between them. Which rule finds the third side?

    A triangle ABC with the angle 40° at A between the side b of 5 cm and the side c of 7 cm. The side a is labelled a and is opposite A.ABCa5 cm7 cm40°
    Tap your answer.

    1. Not this one. The sine rule needs a matching pair, an angle and the side opposite it. Here the 40° angle is between the known sides, and the side opposite it is the one you want. Go to M4.3.7. Try another.

    2. Yes, that is right. There is no matching pair of an angle and its opposite side. Two sides and the angle between them is the case for the cosine rule.

    3. Not this one. Pythagoras needs a right angle, and this angle is 40°. The cosine rule is Pythagoras with a correction for the angle. Go to M4.3.1. Try another.

    See the answer

    The cosine rule

    There is no matching pair of an angle and its opposite side. Two sides and the angle between them is the case for the cosine rule.

    • If you chose A: The sine rule needs a matching pair, an angle and the side opposite it. Here the 40° angle is between the known sides, and the side opposite it is the one you want. Go to M4.3.7.
    • If you chose C: Pythagoras needs a right angle, and this angle is 40°. The cosine rule is Pythagoras with a correction for the angle. Go to M4.3.1.

  2. You know all three sides of a triangle and want one angle. Which rule do you use?

    Tap your answer.

    1. Yes, that is right. With three sides and no angle there is no matching pair, so the sine rule cannot start. The cosine rule gives cos A from the three sides.

    2. Not this one. The sine rule needs a matching pair of an angle and its opposite side. With three sides and no angle you have none, so start with the cosine rule. Go to M4.3.7. Try another.

    3. Not this one. Pythagoras needs a right angle. A triangle with three given sides may not have one. Go to M4.3.1. Try another.

    See the answer

    The cosine rule, rearranged to give cos A

    With three sides and no angle there is no matching pair, so the sine rule cannot start. The cosine rule gives cos A from the three sides.

    • If you chose B: The sine rule needs a matching pair of an angle and its opposite side. With three sides and no angle you have none, so start with the cosine rule. Go to M4.3.7.
    • If you chose C: Pythagoras needs a right angle. A triangle with three given sides may not have one. Go to M4.3.1.

  3. In triangle ABC, A = 35°, B = 75° and b = 10 cm. Find a, to 1 decimal place.

    A triangle ABC with the angle 35° at A, the angle 75° at B, the side b labelled 10 cm opposite B, and the side a labelled a opposite A.ABCa10 cm35°75°
    Tap your answer.

    1. Not this one. You turned the ratio upside down: 10 × sin 75° ÷ sin 35°. The side you want is opposite 35°, so sin 35° goes on the top. Go to M4.3.7. Try another.

    2. Not this one. You worked out 10 × sin 35° and stopped. You also need to divide by sin 75°, the sine of the angle opposite the side 10. Go to M4.3.7. Try another.

    3. Yes, that is right. a ÷ sin 35° = 10 ÷ sin 75°. So a = 10 × sin 35° ÷ sin 75°. That is 10 × 0.5736 ÷ 0.9659 = 5.938, which is 5.9 cm to 1 decimal place.

    4. Not this one. Your calculator was in radians mode. It worked out the sines of 35 and 75 radians. Switch to degrees mode. Go to M4.3.4. Try another.

    See the answer

    5.9 cm

    a ÷ sin 35° = 10 ÷ sin 75°. So a = 10 × sin 35° ÷ sin 75°. That is 10 × 0.5736 ÷ 0.9659 = 5.938, which is 5.9 cm to 1 decimal place.

    • If you chose A: You turned the ratio upside down: 10 × sin 75° ÷ sin 35°. The side you want is opposite 35°, so sin 35° goes on the top. Go to M4.3.7.
    • If you chose B: You worked out 10 × sin 35° and stopped. You also need to divide by sin 75°, the sine of the angle opposite the side 10. Go to M4.3.7.
    • If you chose D: Your calculator was in radians mode. It worked out the sines of 35 and 75 radians. Switch to degrees mode. Go to M4.3.4.

  4. In triangle ABC, b = 6 cm, c = 9 cm and A = 70°. Find a, to 1 decimal place.

    A triangle ABC with the angle 70° at A between the side b of 6 cm and the side c of 9 cm. The side a is labelled a and is opposite A.ABCa6 cm9 cm70°
    Tap your answer.

    1. Not this one. You worked out (6² + 9² − 2 × 6 × 9) × cos 70°, which is 3.08, and gave that as the side. The cosine multiplies only the 2bc part, and a² still needs a square root. Go to M4.3.7. Try another.

    2. Not this one. You added 2bc cos A. The rule subtracts it: a² = b² + c² − 2bc cos A. For an angle below 90° the side is shorter than √(b² + c²) = √117, about 10.8 cm, and 12.4 is not. Go to M4.3.7. Try another.

    3. Not this one. 80.1 is a², the square of the side. You forgot to take the square root at the end. Go to M4.3.7. Try another.

    4. Yes, that is right. a² = 6² + 9² − 2 × 6 × 9 × cos 70° = 117 − 108 × 0.3420 = 80.062. The square root is 8.948, which is 8.9 cm to 1 decimal place.

    See the answer

    8.9 cm

    a² = 6² + 9² − 2 × 6 × 9 × cos 70° = 117 − 108 × 0.3420 = 80.062. The square root is 8.948, which is 8.9 cm to 1 decimal place.

    • If you chose A: You worked out (6² + 9² − 2 × 6 × 9) × cos 70°, which is 3.08, and gave that as the side. The cosine multiplies only the 2bc part, and a² still needs a square root. Go to M4.3.7.
    • If you chose B: You added 2bc cos A. The rule subtracts it: a² = b² + c² − 2bc cos A. For an angle below 90° the side is shorter than √(b² + c²) = √117, about 10.8 cm, and 12.4 is not. Go to M4.3.7.
    • If you chose C: 80.1 is a², the square of the side. You forgot to take the square root at the end. Go to M4.3.7.

  5. In triangle ABC, A = 40°, a = 15 cm and b = 11 cm. Find the angle B, to 1 decimal place. Here a is longer than b, so B is acute.

    Tap your answer.

    1. Not this one. You put a and b the wrong way round: sin B = 15 × sin 40° ÷ 11. The side b belongs on the top, with sin A, and a goes underneath. Go to M4.3.7. Try another.

    2. Yes, that is right. sin B = b × sin A ÷ a = 11 × sin 40° ÷ 15 = 0.4714, so B = 28.124, which is 28.1° to 1 decimal place.

    3. Not this one. You used the inverse cos key. The sine rule gives sin B, so the inverse sin key is the one to press. Go to M4.3.5. Try another.

    4. Not this one. You stopped at the ratio, sin B = 0.4712. Press the inverse sin key to get the angle. Go to M4.3.5. Try another.

    See the answer

    28.1°

    sin B = b × sin A ÷ a = 11 × sin 40° ÷ 15 = 0.4714, so B = 28.124, which is 28.1° to 1 decimal place.

    • If you chose A: You put a and b the wrong way round: sin B = 15 × sin 40° ÷ 11. The side b belongs on the top, with sin A, and a goes underneath. Go to M4.3.7.
    • If you chose C: You used the inverse cos key. The sine rule gives sin B, so the inverse sin key is the one to press. Go to M4.3.5.
    • If you chose D: You stopped at the ratio, sin B = 0.4712. Press the inverse sin key to get the angle. Go to M4.3.5.

  6. A triangle has A = 40°, a = 8 cm and b = 11 cm. The sine rule gives B = 62.1°. What other value can B have?

    Tap your answer.

    1. Yes, that is right. Two angles have the same sine: an acute angle and 180° minus it. So B can also be 180° − 62.1° = 117.9°. Then C = 180° − 40° − 117.9° = 22.1°, so it is a valid triangle.

    2. Not this one. That is the angle C of the first triangle: 180° − 40° − 62.1°. The other value of B is 180° minus 62.1°. Go to M4.3.7. Try another.

    3. Not this one. You took 62.1° from 90°. The two angles with the same sine add up to 180°, not 90°. Go to M4.3.7. Try another.

    4. Not this one. There is one. Add it to the known angle: 40° + 117.9° = 157.9°, which is below 180°. So a second triangle exists. Go to M4.3.7. Try another.

    See the answer

    117.9°

    Two angles have the same sine: an acute angle and 180° minus it. So B can also be 180° − 62.1° = 117.9°. Then C = 180° − 40° − 117.9° = 22.1°, so it is a valid triangle.

    • If you chose B: That is the angle C of the first triangle: 180° − 40° − 62.1°. The other value of B is 180° minus 62.1°. Go to M4.3.7.
    • If you chose C: You took 62.1° from 90°. The two angles with the same sine add up to 180°, not 90°. Go to M4.3.7.
    • If you chose D: There is one. Add it to the known angle: 40° + 117.9° = 157.9°, which is below 180°. So a second triangle exists. Go to M4.3.7.

  7. From an earlier chapter

    What is cos 60° exactly?

    Tap your answer.

    1. Not this one. That is cos 30°, and sin 60°. For 60° the adjacent side is the short side, 1, and the hypotenuse is 2. Go to M4.3.6. Try another.

    2. Not this one. That is tan 30°. A cosine divides by the hypotenuse, 2. Go to M4.3.6. Try another.

    3. Yes, that is right. In half an equilateral triangle the adjacent side to 60° is 1 and the hypotenuse is 2, so cos 60° = 1/2.

    4. Not this one. cos 60° is not 1. Only cos 0° is 1. At 60° the adjacent side is half the hypotenuse. Go to M4.3.6. Try another.

    See the answer

    1/2

    In half an equilateral triangle the adjacent side to 60° is 1 and the hypotenuse is 2, so cos 60° = 1/2.

    • If you chose A: That is cos 30°, and sin 60°. For 60° the adjacent side is the short side, 1, and the hypotenuse is 2. Go to M4.3.6.
    • If you chose B: That is tan 30°. A cosine divides by the hypotenuse, 2. Go to M4.3.6.
    • If you chose D: cos 60° is not 1. Only cos 0° is 1. At 60° the adjacent side is half the hypotenuse. Go to M4.3.6.

  8. In triangle ABC, A = 50°, B = 70° and a = 12 cm. How many centimetres is b, to 1 decimal place?

    See the answer

    14.7

    b = a × sin B ÷ sin A = 12 × sin 70° ÷ sin 50° = 14.720, which is 14.7 to 1 decimal place.

  9. In triangle ABC, b = 5 cm, c = 9 cm and A = 30°. How many centimetres is a, to 1 decimal place?

    See the answer

    5.3

    a² = 5² + 9² − 2 × 5 × 9 × cos 30° = 28.058, and the square root is 5.297, which is 5.3 to 1 decimal place.

Keep it

Copy this into your exercise book.

Sine rule: a ÷ sin A = b ÷ sin B. Use it with a matching pair. Cosine rule: a² = b² + c² − 2bc cos A. Use it with two sides and the angle between, or three sides.

Draw: A triangle ABC with its height h dashed, and the two rules written under it.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

No internet tomorrow? Save the whole book as one file (776 KB). It opens on any phone.