High School Maths · Year 2 · Unit 11 · Lesson 1 of 5

The Sine Rule: Find a Side When There Is No Right Angle

About 15 minutes · labelling sides a, b, c opposite angles A, B, C · the sine rule · choosing a side and its opposite angle · finding the third angle first · degrees mode · two-triangle problems · answers to 3 significant figures · calculator allowed · exit quiz

Hook

Two lifeguard towers A and B stand 150 m apart on a straight beach. A swimmer S is out at sea. Angle SAB is 62° and angle SBA is 71°.

How far is the swimmer from tower A? There is no right angle, so SOH CAH TOA cannot be used.

About 194 m. The angle at the swimmer is 180° − 62° − 71° = 47°. AS is opposite the 71° angle, and the 150 m is opposite the 47° angle. The sine rule says each side divided by the sine of its opposite angle gives the same number, so AS = 150 × sin 71° ÷ sin 47° = 193.9… m. In Year 1 you used sine, cosine and tangent in right-angled triangles. Now you can solve any triangle.

Learn

Master the essentials

A small letter names the side opposite the capital-letter angle.

Labelling

Name each side with the small letter of the angle opposite it.

In triangle PQR, side p = QR is opposite angle P.

The sine rule

In any triangle, each side divided by the sine of its opposite angle gives the same value.

asin A = bsin B = csin C

When to use it

You know a side AND the angle opposite it (a pair), plus one more angle.

Know a, A and B: find b.

Finding a side

Use two parts of the rule, with the unknown side on top. Multiply both sides by its sine.

b = 8 × sin 65° ÷ sin 40° = 11.3

Third angle first

The angles of a triangle add up to 180°. Find a missing angle before you start.

A = 52°, B = 71°, so C = 57°

Opposite pairs only

Each side goes with the angle opposite it, never an angle at one of its ends.

Side c goes with angle C, which is across from it.

Degrees mode

Check that your calculator shows D or DEG. In radians mode every sine is wrong.

sin 30° = 0.5, but radians mode gives −0.988

Sense check

The longest side is opposite the largest angle; the shortest side is opposite the smallest angle.

Angles 40°, 65°, 75°: the longest side is opposite 75°.

Worked examples

See it done

1. In triangle ABC, angle A = 48° and angle B = 77°. Side BC = 9.4 cm is opposite angle A. Find side AC, which is opposite angle B. Give your answer correct to 3 significant figures.

  1. Side BC = a = 9.4 cm goes with angle A = 48°. Side AC = b goes with angle B = 77°.
  2. b ÷ sin 77° = 9.4 ÷ sin 48°.
  3. Multiply both sides by sin 77°: b = 9.4 × sin 77° ÷ sin 48°.
  4. b = 12.32… = 12.3 cm (3 s.f.).
  5. Check: b is opposite the larger angle, so it should be longer than 9.4 cm ✓.

12.3 cm

2. In triangle PQR, angle P = 38°, angle Q = 105° and PQ = 14 cm. Find QR. Give your answer correct to 3 significant figures.

  1. PQ is opposite angle R, so find angle R first: R = 180° − 38° − 105° = 37°.
  2. QR is opposite angle P = 38°.
  3. QR ÷ sin 38° = 14 ÷ sin 37°.
  4. QR = 14 × sin 38° ÷ sin 37° = 14.32….
  5. QR = 14.3 cm (3 s.f.).

14.3 cm

3. Two survey posts X and Y are 85 m apart. A tree T stands across a field. Angle TXY = 54° and angle TYX = 68°. How far is the tree from post X? Give your answer correct to 3 significant figures.

  1. The angle at the tree: T = 180° − 54° − 68° = 58°. The 85 m side XY is opposite angle T.
  2. XT is opposite angle Y = 68°.
  3. XT = 85 × sin 68° ÷ sin 58° = 92.93….
  4. XT = 92.9 m (3 s.f.).

92.9 m

Common mistake: Pairing a side with the wrong angle. In the sine rule, each side goes with the angle OPPOSITE it, never with an angle at one of its ends. Before you write anything, draw an arrow from each side across the triangle to its opposite angle. If the opposite angle is not given, find it first from the angle sum of 180°.

PractiseCalculator allowed

10 questions with instant feedback

Choose the correct answer. A calculator is allowed: check it is in degrees mode. Match each side with the angle opposite it, then put the unknown side on top. Give lengths correct to 3 significant figures. Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.

Score: 0 / 10Answered: 0 / 10

Warm-up (The basic skill: one or two steps.)

1.Which equation is the sine rule for triangle KLM?

Correct.

Solution

  1. Side KM does not touch L, so KM is opposite angle L.
  2. Side KL does not touch M, so KL is opposite angle M.
  3. Sine rule: KM ÷ sin L = KL ÷ sin M.

Not quite. Each side must go with the angle opposite it. KM touches K: its opposite angle is L.

Solution

  1. Side KM does not touch L, so KM is opposite angle L.
  2. Side KL does not touch M, so KL is opposite angle M.
  3. Sine rule: KM ÷ sin L = KL ÷ sin M.

Answer: A. KM ÷ sin L = KL ÷ sin M.

Not quite. KL touches angle K. The angle opposite KL is M.

Solution

  1. Side KM does not touch L, so KM is opposite angle L.
  2. Side KL does not touch M, so KL is opposite angle M.
  3. Sine rule: KM ÷ sin L = KL ÷ sin M.

Answer: A. KM ÷ sin L = KL ÷ sin M.

Not quite. The rule divides each side by the sine of its opposite angle. It does not multiply.

Solution

  1. Side KM does not touch L, so KM is opposite angle L.
  2. Side KL does not touch M, so KL is opposite angle M.
  3. Sine rule: KM ÷ sin L = KL ÷ sin M.

Answer: A. KM ÷ sin L = KL ÷ sin M.

2.In triangle ABC, angle A = 35° and angle B = 80°. Side BC = 7.2 cm is opposite angle A. Find side AC, which is opposite angle B. Give your answer correct to 3 significant figures.

Not quite. This puts the sines the wrong way up. The unknown side goes with sin 80°, on top.

Solution

  1. AC goes with angle B = 80°; BC = 7.2 cm goes with angle A = 35°.
  2. AC = 7.2 × sin 80° ÷ sin 35°.
  3. = 12.36… = 12.4 cm (3 s.f.).

Answer: D. 12.4 cm.

Not quite. This uses cosines. The sine rule uses sines.

Solution

  1. AC goes with angle B = 80°; BC = 7.2 cm goes with angle A = 35°.
  2. AC = 7.2 × sin 80° ÷ sin 35°.
  3. = 12.36… = 12.4 cm (3 s.f.).

Answer: D. 12.4 cm.

Not quite. This is what a calculator in radians mode gives. Switch to degrees.

Solution

  1. AC goes with angle B = 80°; BC = 7.2 cm goes with angle A = 35°.
  2. AC = 7.2 × sin 80° ÷ sin 35°.
  3. = 12.36… = 12.4 cm (3 s.f.).

Answer: D. 12.4 cm.

Correct.

Solution

  1. AC goes with angle B = 80°; BC = 7.2 cm goes with angle A = 35°.
  2. AC = 7.2 × sin 80° ÷ sin 35°.
  3. = 12.36… = 12.4 cm (3 s.f.).

3.In triangle ABC, angle B = 57° and angle C = 46°. Side AC = 12.5 cm is opposite angle B. Find side BC, which is opposite angle A. Give your answer correct to 3 significant figures.

Not quite. This uses angle C = 46°. BC is opposite angle A, so find angle A first.

Solution

  1. Angle A = 180° − 57° − 46° = 77°.
  2. BC goes with angle A = 77°; AC = 12.5 cm goes with angle B = 57°.
  3. BC = 12.5 × sin 77° ÷ sin 57° = 14.52… = 14.5 cm.

Answer: B. 14.5 cm.

Correct.

Solution

  1. Angle A = 180° − 57° − 46° = 77°.
  2. BC goes with angle A = 77°; AC = 12.5 cm goes with angle B = 57°.
  3. BC = 12.5 × sin 77° ÷ sin 57° = 14.52… = 14.5 cm.

Not quite. This puts the sines the wrong way up. The unknown side goes with sin A, on top.

Solution

  1. Angle A = 180° − 57° − 46° = 77°.
  2. BC goes with angle A = 77°; AC = 12.5 cm goes with angle B = 57°.
  3. BC = 12.5 × sin 77° ÷ sin 57° = 14.52… = 14.5 cm.

Answer: B. 14.5 cm.

Not quite. This takes A = 180° − 57° = 123° and forgets angle C. A = 180° − 57° − 46°.

Solution

  1. Angle A = 180° − 57° − 46° = 77°.
  2. BC goes with angle A = 77°; AC = 12.5 cm goes with angle B = 57°.
  3. BC = 12.5 × sin 77° ÷ sin 57° = 14.52… = 14.5 cm.

Answer: B. 14.5 cm.

Exam standard (Several steps, like real exam questions.)

4.Two coastguard stations A and B are 24 km apart. A ship S is out at sea. Angle SAB = 52° and angle SBA = 61°. How far is the ship from station B? Give your answer correct to 3 significant figures.

Not quite. This is the distance from A. BS is opposite the 52° angle at A.

Solution

  1. Angle S = 180° − 52° − 61° = 67°. The 24 km side AB is opposite angle S.
  2. BS is opposite angle A = 52°.
  3. BS = 24 × sin 52° ÷ sin 67° = 20.54… = 20.5 km.

Answer: B. 20.5 km.

Correct.

Solution

  1. Angle S = 180° − 52° − 61° = 67°. The 24 km side AB is opposite angle S.
  2. BS is opposite angle A = 52°.
  3. BS = 24 × sin 52° ÷ sin 67° = 20.54… = 20.5 km.

Not quite. There is no right angle, so SOH CAH TOA does not work here. Use the sine rule.

Solution

  1. Angle S = 180° − 52° − 61° = 67°. The 24 km side AB is opposite angle S.
  2. BS is opposite angle A = 52°.
  3. BS = 24 × sin 52° ÷ sin 67° = 20.54… = 20.5 km.

Answer: B. 20.5 km.

Not quite. This puts the sines the wrong way up. BS goes with sin 52°, on top.

Solution

  1. Angle S = 180° − 52° − 61° = 67°. The 24 km side AB is opposite angle S.
  2. BS is opposite angle A = 52°.
  3. BS = 24 × sin 52° ÷ sin 67° = 20.54… = 20.5 km.

Answer: B. 20.5 km.

5.The table gives two angles and one side of three triangles. For triangle Q, find side b, which is opposite angle B. Give your answer correct to 3 significant figures.

TriangleAngle AAngle BSide a
P42°73°6.5 cm
Q39°84°11.2 cm
R58°66°9.7 cm

In each triangle, side a is opposite angle A and side b is opposite angle B.

Correct.

Solution

  1. Triangle Q: A = 39°, B = 84°, a = 11.2 cm.
  2. b = 11.2 × sin 84° ÷ sin 39°.
  3. = 17.69… = 17.7 cm (3 s.f.).

Not quite. This uses the row for triangle P. Read the row for triangle Q.

Solution

  1. Triangle Q: A = 39°, B = 84°, a = 11.2 cm.
  2. b = 11.2 × sin 84° ÷ sin 39°.
  3. = 17.69… = 17.7 cm (3 s.f.).

Answer: A. 17.7 cm.

Not quite. This puts the sines the wrong way up. Side b goes with sin B, on top.

Solution

  1. Triangle Q: A = 39°, B = 84°, a = 11.2 cm.
  2. b = 11.2 × sin 84° ÷ sin 39°.
  3. = 17.69… = 17.7 cm (3 s.f.).

Answer: A. 17.7 cm.

Not quite. This uses cosines. The sine rule uses sines.

Solution

  1. Triangle Q: A = 39°, B = 84°, a = 11.2 cm.
  2. b = 11.2 × sin 84° ÷ sin 39°.
  3. = 17.69… = 17.7 cm (3 s.f.).

Answer: A. 17.7 cm.

6.In triangle DEF, angle D = 30° and angle E = 112°. Side EF = 6 cm is opposite angle D. Which working shows that DF = 11.1 cm, correct to 3 significant figures?

Correct.

Solution

  1. DF does not touch E, so DF is opposite angle E = 112°.
  2. EF = 6 cm is opposite angle D = 30°.
  3. DF = 6 × sin 112° ÷ sin 30° = 11.12… = 11.1 cm ✓.

Not quite. The sines are the wrong way up, and the result is not 11.1.

Solution

  1. DF does not touch E, so DF is opposite angle E = 112°.
  2. EF = 6 cm is opposite angle D = 30°.
  3. DF = 6 × sin 112° ÷ sin 30° = 11.12… = 11.1 cm ✓.

Answer: A. DF = 6 × sin 112° ÷ sin 30° = 11.12….

Not quite. DF is opposite angle E = 112°, not angle F = 38°.

Solution

  1. DF does not touch E, so DF is opposite angle E = 112°.
  2. EF = 6 cm is opposite angle D = 30°.
  3. DF = 6 × sin 112° ÷ sin 30° = 11.12… = 11.1 cm ✓.

Answer: A. DF = 6 × sin 112° ÷ sin 30° = 11.12….

Not quite. This forgets to divide by sin 30°. Each side is divided by the sine of its opposite angle.

Solution

  1. DF does not touch E, so DF is opposite angle E = 112°.
  2. EF = 6 cm is opposite angle D = 30°.
  3. DF = 6 × sin 112° ÷ sin 30° = 11.12… = 11.1 cm ✓.

Answer: A. DF = 6 × sin 112° ÷ sin 30° = 11.12….

7.Quadrilateral ABCD has a diagonal BD. In triangle BCD, BC = 7.5 cm, angle BDC = 40° and angle BCD = 95°. In triangle ABD, angle BAD = 62° and angle ABD = 51°. Find AD. Give your answer correct to 3 significant figures.

Not quite. This is BD, the first step. Now use triangle ABD to find AD.

Solution

  1. Triangle BCD: BD is opposite angle C = 95°, and BC = 7.5 cm is opposite angle BDC = 40°.
  2. BD = 7.5 × sin 95° ÷ sin 40° = 11.62… cm.
  3. Triangle ABD: AD is opposite angle ABD = 51°, and BD is opposite angle BAD = 62°.
  4. AD = 11.62… × sin 51° ÷ sin 62° = 10.23… = 10.2 cm.

Answer: D. 10.2 cm.

Not quite. This swaps the angles. AD is opposite angle ABD = 51°, and BD is opposite angle BAD = 62°.

Solution

  1. Triangle BCD: BD is opposite angle C = 95°, and BC = 7.5 cm is opposite angle BDC = 40°.
  2. BD = 7.5 × sin 95° ÷ sin 40° = 11.62… cm.
  3. Triangle ABD: AD is opposite angle ABD = 51°, and BD is opposite angle BAD = 62°.
  4. AD = 11.62… × sin 51° ÷ sin 62° = 10.23… = 10.2 cm.

Answer: D. 10.2 cm.

Not quite. This is AB, which is opposite angle ADB = 67°. AD is opposite angle ABD = 51°.

Solution

  1. Triangle BCD: BD is opposite angle C = 95°, and BC = 7.5 cm is opposite angle BDC = 40°.
  2. BD = 7.5 × sin 95° ÷ sin 40° = 11.62… cm.
  3. Triangle ABD: AD is opposite angle ABD = 51°, and BD is opposite angle BAD = 62°.
  4. AD = 11.62… × sin 51° ÷ sin 62° = 10.23… = 10.2 cm.

Answer: D. 10.2 cm.

Correct.

Solution

  1. Triangle BCD: BD is opposite angle C = 95°, and BC = 7.5 cm is opposite angle BDC = 40°.
  2. BD = 7.5 × sin 95° ÷ sin 40° = 11.62… cm.
  3. Triangle ABD: AD is opposite angle ABD = 51°, and BD is opposite angle BAD = 62°.
  4. AD = 11.62… × sin 51° ÷ sin 62° = 10.23… = 10.2 cm.

Challenge (Harder problems that mix ideas or need reasoning.)

8.A triangle has angles of 48°, 59° and 73°. Its perimeter is 36 cm. Find the length of its longest side, correct to 3 significant figures.

Not quite. The sides are in proportion to the SINES of the angles, not to the angles: 36 × 73 ÷ 180 is wrong.

Solution

  1. By the sine rule, each side = k × sin(its opposite angle), for the same number k.
  2. k × (sin 48° + sin 59° + sin 73°) = 36, so k = 14.08….
  3. Longest side (opposite 73°) = k × sin 73° = 13.46… = 13.5 cm.

Answer: B. 13.5 cm.

Correct.

Solution

  1. By the sine rule, each side = k × sin(its opposite angle), for the same number k.
  2. k × (sin 48° + sin 59° + sin 73°) = 36, so k = 14.08….
  3. Longest side (opposite 73°) = k × sin 73° = 13.46… = 13.5 cm.

Not quite. This is the shortest side, opposite the smallest angle. The longest side is opposite 73°.

Solution

  1. By the sine rule, each side = k × sin(its opposite angle), for the same number k.
  2. k × (sin 48° + sin 59° + sin 73°) = 36, so k = 14.08….
  3. Longest side (opposite 73°) = k × sin 73° = 13.46… = 13.5 cm.

Answer: B. 13.5 cm.

Not quite. This is the common value a ÷ sin A of the sine rule. Multiply it by sin 73° to get the side.

Solution

  1. By the sine rule, each side = k × sin(its opposite angle), for the same number k.
  2. k × (sin 48° + sin 59° + sin 73°) = 36, so k = 14.08….
  3. Longest side (opposite 73°) = k × sin 73° = 13.46… = 13.5 cm.

Answer: B. 13.5 cm.

9.Two people A and B stand 600 m apart on level ground. A balloon is above the ground between them. The angle of elevation of the balloon is 35° from A and 48° from B. Find the height of the balloon, correct to 3 significant figures.

Not quite. This is the distance from A to the balloon. The height is a second step: multiply by sin 35°.

Solution

  1. The angle at the balloon is 180° − 35° − 48° = 97°, opposite the 600 m side.
  2. Distance from A: AC = 600 × sin 48° ÷ sin 97° = 449.2… m.
  3. Height = AC × sin 35° = 257.6… = 258 m.

Answer: C. 258 m.

Not quite. This uses the distance from B with the angle at A. From B, the height is BC × sin 48°.

Solution

  1. The angle at the balloon is 180° − 35° − 48° = 97°, opposite the 600 m side.
  2. Distance from A: AC = 600 × sin 48° ÷ sin 97° = 449.2… m.
  3. Height = AC × sin 35° = 257.6… = 258 m.

Answer: C. 258 m.

Correct.

Solution

  1. The angle at the balloon is 180° − 35° − 48° = 97°, opposite the 600 m side.
  2. Distance from A: AC = 600 × sin 48° ÷ sin 97° = 449.2… m.
  3. Height = AC × sin 35° = 257.6… = 258 m.

Not quite. This is the horizontal distance from A to the point under the balloon. The height is opposite 35°: use sin.

Solution

  1. The angle at the balloon is 180° − 35° − 48° = 97°, opposite the 600 m side.
  2. Distance from A: AC = 600 × sin 48° ÷ sin 97° = 449.2… m.
  3. Height = AC × sin 35° = 257.6… = 258 m.

Answer: C. 258 m.

10.In triangle ABC, angle A = 36° and angle B = 59°. Side b (opposite angle B) is 4.5 cm longer than side a (opposite angle A). Find side a, correct to 3 significant figures.

Not quite. This is side b. The question asks for side a, which is 4.5 cm shorter.

Solution

  1. Sine rule: b = a × sin 59° ÷ sin 36°.
  2. b − a = 4.5, so a × (sin 59° ÷ sin 36° − 1) = 4.5.
  3. a = 4.5 ÷ 0.4583… = 9.818… = 9.82 cm.

Answer: C. 9.82 cm.

Not quite. This uses the ratio of the angles, 59 ÷ 36. The sine rule needs the ratio of the sines.

Solution

  1. Sine rule: b = a × sin 59° ÷ sin 36°.
  2. b − a = 4.5, so a × (sin 59° ÷ sin 36° − 1) = 4.5.
  3. a = 4.5 ÷ 0.4583… = 9.818… = 9.82 cm.

Answer: C. 9.82 cm.

Correct.

Solution

  1. Sine rule: b = a × sin 59° ÷ sin 36°.
  2. b − a = 4.5, so a × (sin 59° ÷ sin 36° − 1) = 4.5.
  3. a = 4.5 ÷ 0.4583… = 9.818… = 9.82 cm.

Not quite. This treats 4.5 cm as side b. It is the difference b − a.

Solution

  1. Sine rule: b = a × sin 59° ÷ sin 36°.
  2. b − a = 4.5, so a × (sin 59° ÷ sin 36° − 1) = 4.5.
  3. a = 4.5 ÷ 0.4583… = 9.818… = 9.82 cm.

Answer: C. 9.82 cm.

Apply

How wide is the river?

Points A and B are 120 m apart on one straight bank of a river. A tree T stands on the far bank. Angle TAB = 64° and angle TBA = 51°. A calculator is allowed.

1. Find angle ATB.

2. Find the distance AT, correct to 3 significant figures.

3. Find the width of the river, correct to 3 significant figures.

  1. Angle ATB = 180° − 64° − 51° = 65°.
  2. AT is opposite angle B = 51°, and AB = 120 m is opposite angle T = 65°. AT = 120 × sin 51° ÷ sin 65° = 102.8… = 103 m.
  3. The width is the height of T above the line AB. In the right-angled triangle ATW: width = AT × sin 64° = 92.48… = 92.5 m.

Next: the Lesson 1 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.

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