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

The Sine Rule for Angles: Why Your Calculator Hides an Answer

About 15 minutes · the sine rule upside down for angles · sin⁻¹ to find the angle · sin x = sin(180° − x) · when the angle is obtuse · checking the angle sum · why sin can never be more than 1 · two possible triangles (challenge) · calculator allowed · exit quiz

Hook

sin x = 0.8

Your calculator says x = 53.1°. Is that the only angle in a triangle with a sine of 0.8?

No. 126.9° works too: sin 126.9° = 0.8 as well, because sin x = sin(180° − x) and 180° − 53.1° = 126.9°. The calculator only ever gives the angle below 90°. When you use the sine rule to find an angle, you must decide whether the angle is acute or obtuse.

Learn

Master the essentials

The same two sides and angle fit two triangles. Angle y = 180° − x, and sin x = sin(180° − x).

Rule for angles

Turn the sine rule upside down, so the unknown angle is on top.

sin Aa = sin Bb

Find the sine first

Multiply both sides by the side opposite the unknown angle.

sin B = 6 × sin 70° ÷ 9 = 0.6264…

Then use sin⁻¹

The angle is sin⁻¹ of the value. Do not stop at the sine.

B = sin⁻¹(0.6264…) = 38.8°

Two angles, one sine

An angle and its supplement have the same sine.

sin 40° = sin 140° = 0.6427…

The calculator gives one

sin⁻¹ always gives an angle from 0° to 90°. The other answer is 180° minus it.

sin⁻¹(0.35) = 20.5° or 159.5°

Acute or obtuse?

If the unknown angle is opposite a shorter side than the known pair, it must be acute. If the question says it is obtuse, use 180° − x.

A side shorter than the known side faces a smaller angle.

Angle sum check

The angles must add up to 180°. An obtuse answer must leave room for the other two.

A = 70°, B = 38.8°: C = 71.2°

Never more than 1

sin x is never more than 1. If you get more than 1, the triangle cannot exist (or the sides are the wrong way up).

sin B = 1.3: no such triangle

Worked examples

See it done

1. In triangle ABC, side BC = 11 cm is opposite angle A = 72°, and side AC = 8 cm is opposite angle B. Find angle B, correct to 1 decimal place.

  1. The known pair is a = 11 cm with A = 72°. The unknown angle B goes with b = 8 cm.
  2. sin B ÷ 8 = sin 72° ÷ 11.
  3. sin B = 8 × sin 72° ÷ 11 = 0.6916….
  4. B = sin⁻¹(0.6916…) = 43.8°.
  5. b = 8 cm is shorter than a = 11 cm, so B is smaller than 72°: the acute answer is right ✓.

43.8°

2. In triangle PQR, angle R = 32°. Side PQ = 9 cm is opposite angle R, and side PR = 15 cm is opposite angle Q. Angle Q is obtuse. Find angle Q, correct to 1 decimal place.

  1. The known pair is PQ = 9 cm with angle R = 32°. Angle Q goes with PR = 15 cm.
  2. sin Q = 15 × sin 32° ÷ 9 = 0.8831….
  3. sin⁻¹(0.8831…) = 62.0°. This is the acute answer.
  4. Q is obtuse, so Q = 180° − 62.0° = 118.0°.
  5. Check: 32° + 118.0° is less than 180° ✓.

118.0°

3. In triangle XYZ, angle Z = 58°. Side XY = 13 cm is opposite angle Z, and side YZ = 10 cm is opposite angle X. Find side XZ, correct to 3 significant figures.

  1. XZ is opposite angle Y, which is not known. Find angle X first, then angle Y.
  2. sin X = 10 × sin 58° ÷ 13, so X = 40.7° (acute: 10 cm is shorter than 13 cm).
  3. Y = 180° − 58° − 40.71…° = 81.28…°.
  4. XZ = 13 × sin 81.28…° ÷ sin 58° = 15.15… = 15.2 cm.

15.2 cm

Common mistake: Stopping at the sine, or missing the obtuse answer. sin B = 0.69 is not an angle: press sin⁻¹ to get B. And the calculator gives only the acute angle. If the angle is obtuse, the answer is 180° minus the calculator value.

PractiseCalculator allowed

10 questions with instant feedback

Choose the correct answer. A calculator is allowed: check it is in degrees mode. Put the unknown angle on top, find its sine, then use sin⁻¹. Decide whether the angle is acute or obtuse. Give angles correct to 1 decimal place. 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.In triangle ABC, side BC = 9 cm is opposite angle A = 64°, and side AC = 7 cm is opposite angle B. Find angle B, correct to 1 decimal place.

Not quite. This is sin B, not the angle. Use sin⁻¹ to find B.

Solution

  1. sin B ÷ 7 = sin 64° ÷ 9.
  2. sin B = 7 × sin 64° ÷ 9 = 0.6990….
  3. B = sin⁻¹(0.6990…) = 44.4°.

Answer: D. 44.4°.

Not quite. 7 cm is shorter than 9 cm, so B is smaller than 64°. It cannot be obtuse.

Solution

  1. sin B ÷ 7 = sin 64° ÷ 9.
  2. sin B = 7 × sin 64° ÷ 9 = 0.6990….
  3. B = sin⁻¹(0.6990…) = 44.4°.

Answer: D. 44.4°.

Not quite. This uses sin B = 7 ÷ 9 and forgets the sin 64°.

Solution

  1. sin B ÷ 7 = sin 64° ÷ 9.
  2. sin B = 7 × sin 64° ÷ 9 = 0.6990….
  3. B = sin⁻¹(0.6990…) = 44.4°.

Answer: D. 44.4°.

Correct.

Solution

  1. sin B ÷ 7 = sin 64° ÷ 9.
  2. sin B = 7 × sin 64° ÷ 9 = 0.6990….
  3. B = sin⁻¹(0.6990…) = 44.4°.

2.Which angle has the same sine as 37°?

Correct.

Solution

  1. sin x = sin(180° − x).
  2. 180° − 37° = 143°.
  3. So sin 143° = sin 37°.

Not quite. sin 37° = cos 53°, not sin 53°. The supplement of 37° is 180° − 37°.

Solution

  1. sin x = sin(180° − x).
  2. 180° − 37° = 143°.
  3. So sin 143° = sin 37°.

Answer: A. 143°.

Not quite. sin 217° is negative. Use 180° − 37°, not 180° + 37°.

Solution

  1. sin x = sin(180° − x).
  2. 180° − 37° = 143°.
  3. So sin 143° = sin 37°.

Answer: A. 143°.

Not quite. sin 323° is negative. Use 180° − 37°, not 360° − 37°.

Solution

  1. sin x = sin(180° − x).
  2. 180° − 37° = 143°.
  3. So sin 143° = sin 37°.

Answer: A. 143°.

3.In triangle DEF, angle F = 76°. Side DE = 12 cm is opposite angle F, and side EF = 8.5 cm is opposite angle D. Find angle D, correct to 1 decimal place.

Correct.

Solution

  1. The known pair is DE = 12 cm with F = 76°.
  2. sin D = 8.5 × sin 76° ÷ 12 = 0.6872….
  3. D = sin⁻¹(0.6872…) = 43.4°.

Not quite. This is sin D. Use sin⁻¹ to find the angle.

Solution

  1. The known pair is DE = 12 cm with F = 76°.
  2. sin D = 8.5 × sin 76° ÷ 12 = 0.6872….
  3. D = sin⁻¹(0.6872…) = 43.4°.

Answer: A. 43.4°.

Not quite. This uses cos⁻¹. The sine rule needs sin⁻¹.

Solution

  1. The known pair is DE = 12 cm with F = 76°.
  2. sin D = 8.5 × sin 76° ÷ 12 = 0.6872….
  3. D = sin⁻¹(0.6872…) = 43.4°.

Answer: A. 43.4°.

Not quite. 8.5 cm is shorter than 12 cm, so D is smaller than 76°: it is acute.

Solution

  1. The known pair is DE = 12 cm with F = 76°.
  2. sin D = 8.5 × sin 76° ÷ 12 = 0.6872….
  3. D = sin⁻¹(0.6872…) = 43.4°.

Answer: A. 43.4°.

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

4.In triangle ABC, angle A = 29°. Side BC = 8.1 cm is opposite angle A, and side AC = 14.2 cm is opposite angle B. Angle B is obtuse. Find angle B, correct to 1 decimal place.

Not quite. This is the acute angle from the calculator. B is obtuse, so take it from 180°.

Solution

  1. sin B = 14.2 × sin 29° ÷ 8.1 = 0.8499….
  2. sin⁻¹(0.8499…) = 58.2°, the acute answer.
  3. B is obtuse: B = 180° − 58.2° = 121.8°.

Answer: C. 121.8°.

Not quite. This is sin B. Use sin⁻¹, then 180° minus the result.

Solution

  1. sin B = 14.2 × sin 29° ÷ 8.1 = 0.8499….
  2. sin⁻¹(0.8499…) = 58.2°, the acute answer.
  3. B is obtuse: B = 180° − 58.2° = 121.8°.

Answer: C. 121.8°.

Correct.

Solution

  1. sin B = 14.2 × sin 29° ÷ 8.1 = 0.8499….
  2. sin⁻¹(0.8499…) = 58.2°, the acute answer.
  3. B is obtuse: B = 180° − 58.2° = 121.8°.

Not quite. This is angle C if B were acute. The question asks for the obtuse angle B.

Solution

  1. sin B = 14.2 × sin 29° ÷ 8.1 = 0.8499….
  2. sin⁻¹(0.8499…) = 58.2°, the acute answer.
  3. B is obtuse: B = 180° − 58.2° = 121.8°.

Answer: C. 121.8°.

5.In triangle ABC, side BC = 15.3 cm is opposite angle A = 81°, and side AC = 9.8 cm is opposite angle B. Find angle C, correct to 1 decimal place.

Not quite. This is angle B, the first step. Then C = 180° − 81° − B.

Solution

  1. sin B = 9.8 × sin 81° ÷ 15.3 = 0.6326…, so B = 39.2° (acute: 9.8 cm is shorter than 15.3 cm).
  2. C = 180° − 81° − 39.24…°.
  3. C = 59.8°.

Answer: B. 59.8°.

Correct.

Solution

  1. sin B = 9.8 × sin 81° ÷ 15.3 = 0.6326…, so B = 39.2° (acute: 9.8 cm is shorter than 15.3 cm).
  2. C = 180° − 81° − 39.24…°.
  3. C = 59.8°.

Not quite. This takes B from 90°, as if the triangle had a right angle. Use 180° − 81° − B.

Solution

  1. sin B = 9.8 × sin 81° ÷ 15.3 = 0.6326…, so B = 39.2° (acute: 9.8 cm is shorter than 15.3 cm).
  2. C = 180° − 81° − 39.24…°.
  3. C = 59.8°.

Answer: B. 59.8°.

Not quite. This uses sin B = 9.8 ÷ 15.3 and forgets the sin 81°.

Solution

  1. sin B = 9.8 × sin 81° ÷ 15.3 = 0.6326…, so B = 39.2° (acute: 9.8 cm is shorter than 15.3 cm).
  2. C = 180° − 81° − 39.24…°.
  3. C = 59.8°.

Answer: B. 59.8°.

6.The table gives two sides and one angle of three triangles. For triangle Y, find angle B, which is opposite side b. Give your answer correct to 1 decimal place.

TriangleSide aSide bAngle A
X10.4 cm7.7 cm66°
Y13.5 cm9.2 cm58°
Z8.8 cm6.1 cm102°

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

Not quite. This uses the row for triangle X. Read the row for triangle Y.

Solution

  1. Triangle Y: a = 13.5 cm, b = 9.2 cm, A = 58°.
  2. sin B = 9.2 × sin 58° ÷ 13.5 = 0.5779….
  3. B = sin⁻¹(0.5779…) = 35.3°.

Answer: B. 35.3°.

Correct.

Solution

  1. Triangle Y: a = 13.5 cm, b = 9.2 cm, A = 58°.
  2. sin B = 9.2 × sin 58° ÷ 13.5 = 0.5779….
  3. B = sin⁻¹(0.5779…) = 35.3°.

Not quite. This is sin B. Use sin⁻¹ to find the angle.

Solution

  1. Triangle Y: a = 13.5 cm, b = 9.2 cm, A = 58°.
  2. sin B = 9.2 × sin 58° ÷ 13.5 = 0.5779….
  3. B = sin⁻¹(0.5779…) = 35.3°.

Answer: B. 35.3°.

Not quite. b = 9.2 cm is shorter than a = 13.5 cm, so B is smaller than 58°: it is acute.

Solution

  1. Triangle Y: a = 13.5 cm, b = 9.2 cm, A = 58°.
  2. sin B = 9.2 × sin 58° ÷ 13.5 = 0.5779….
  3. B = sin⁻¹(0.5779…) = 35.3°.

Answer: B. 35.3°.

7.A triangular garden PQR has PR = 7.4 m, PQ = 5.6 m and angle PQR = 68°. Find angle QPR, correct to 1 decimal place.

Not quite. This is angle PRQ, the first step. Then angle QPR = 180° − 68° − PRQ.

Solution

  1. PR = 7.4 m is opposite angle Q = 68°. PQ = 5.6 m is opposite angle R.
  2. sin R = 5.6 × sin 68° ÷ 7.4 = 0.7016…, so R = 44.6°.
  3. Angle QPR = 180° − 68° − 44.55…° = 67.4°.

Answer: B. 67.4°.

Correct.

Solution

  1. PR = 7.4 m is opposite angle Q = 68°. PQ = 5.6 m is opposite angle R.
  2. sin R = 5.6 × sin 68° ÷ 7.4 = 0.7016…, so R = 44.6°.
  3. Angle QPR = 180° − 68° − 44.55…° = 67.4°.

Not quite. This uses sin R = 5.6 ÷ 7.4 and forgets the sin 68°.

Solution

  1. PR = 7.4 m is opposite angle Q = 68°. PQ = 5.6 m is opposite angle R.
  2. sin R = 5.6 × sin 68° ÷ 7.4 = 0.7016…, so R = 44.6°.
  3. Angle QPR = 180° − 68° − 44.55…° = 67.4°.

Answer: B. 67.4°.

Not quite. This uses cos⁻¹ for angle PRQ. The sine rule needs sin⁻¹.

Solution

  1. PR = 7.4 m is opposite angle Q = 68°. PQ = 5.6 m is opposite angle R.
  2. sin R = 5.6 × sin 68° ÷ 7.4 = 0.7016…, so R = 44.6°.
  3. Angle QPR = 180° − 68° − 44.55…° = 67.4°.

Answer: B. 67.4°.

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

8.In triangle ABC, AB = 10 cm, BC = 7 cm and angle BAC = 40°. Two different triangles fit these facts. Find the two possible sizes of angle ACB, correct to 1 decimal place.

Not quite. This misses the obtuse answer. BC is shorter than AB, so angle C can also be 180° minus the calculator value.

Solution

  1. AB = 10 cm is opposite angle C; BC = 7 cm is opposite angle A = 40°.
  2. sin C = 10 × sin 40° ÷ 7 = 0.9182…, so C = 66.7° or 180° − 66.7° = 113.3°.
  3. Both work: 40° + 113.3° is less than 180°, so there are two triangles.

Answer: C. 66.7° or 113.3°.

Not quite. No angle in a triangle is more than 180°. The second answer is 180° − x, not 180° + x.

Solution

  1. AB = 10 cm is opposite angle C; BC = 7 cm is opposite angle A = 40°.
  2. sin C = 10 × sin 40° ÷ 7 = 0.9182…, so C = 66.7° or 180° − 66.7° = 113.3°.
  3. Both work: 40° + 113.3° is less than 180°, so there are two triangles.

Answer: C. 66.7° or 113.3°.

Correct.

Solution

  1. AB = 10 cm is opposite angle C; BC = 7 cm is opposite angle A = 40°.
  2. sin C = 10 × sin 40° ÷ 7 = 0.9182…, so C = 66.7° or 180° − 66.7° = 113.3°.
  3. Both work: 40° + 113.3° is less than 180°, so there are two triangles.

Not quite. This takes the calculator angle from 90°. Keep it as it is: the two answers are x and 180° − x.

Solution

  1. AB = 10 cm is opposite angle C; BC = 7 cm is opposite angle A = 40°.
  2. sin C = 10 × sin 40° ÷ 7 = 0.9182…, so C = 66.7° or 180° − 66.7° = 113.3°.
  3. Both work: 40° + 113.3° is less than 180°, so there are two triangles.

Answer: C. 66.7° or 113.3°.

9.Ava says she has drawn a triangle ABC with a = 5 cm, b = 11 cm and angle A = 33°. Which working shows that her triangle cannot exist?

Correct.

Solution

  1. sin B = b × sin A ÷ a = 11 × sin 33° ÷ 5.
  2. = 1.198…, which is more than 1.
  3. No angle has a sine bigger than 1, so the triangle cannot exist.

Not quite. This puts the sides the wrong way up. Side b = 11 cm goes with sin B.

Solution

  1. sin B = b × sin A ÷ a = 11 × sin 33° ÷ 5.
  2. = 1.198…, which is more than 1.
  3. No angle has a sine bigger than 1, so the triangle cannot exist.

Answer: A. sin B = 11 × sin 33° ÷ 5 = 1.20, more than 1.

Not quite. The third side is not given, so this shows nothing.

Solution

  1. sin B = b × sin A ÷ a = 11 × sin 33° ÷ 5.
  2. = 1.198…, which is more than 1.
  3. No angle has a sine bigger than 1, so the triangle cannot exist.

Answer: A. sin B = 11 × sin 33° ÷ 5 = 1.20, more than 1.

Not quite. This is true, but the problem is with angle B, not angle A.

Solution

  1. sin B = b × sin A ÷ a = 11 × sin 33° ÷ 5.
  2. = 1.198…, which is more than 1.
  3. No angle has a sine bigger than 1, so the triangle cannot exist.

Answer: A. sin B = 11 × sin 33° ÷ 5 = 1.20, more than 1.

10.In triangle ABC, side a = 10 cm is opposite angle A = 40°. Side b is opposite angle B. How long must side b be for angle B to be exactly 90°? Give your answer correct to 3 significant figures.

Not quite. This multiplies by sin 40°. From sin B = b sin A ÷ a = 1, b = a ÷ sin A.

Solution

  1. For B = 90°, sin B = 1.
  2. Sine rule: b ÷ sin 90° = 10 ÷ sin 40°, so b = 10 ÷ sin 40°.
  3. b = 15.55… = 15.6 cm.

Answer: D. 15.6 cm.

Not quite. This uses cos 40°. The sine rule uses sin 40°.

Solution

  1. For B = 90°, sin B = 1.
  2. Sine rule: b ÷ sin 90° = 10 ÷ sin 40°, so b = 10 ÷ sin 40°.
  3. b = 15.55… = 15.6 cm.

Answer: D. 15.6 cm.

Not quite. This uses tan 40°. When B = 90°, b is the hypotenuse, so use sin.

Solution

  1. For B = 90°, sin B = 1.
  2. Sine rule: b ÷ sin 90° = 10 ÷ sin 40°, so b = 10 ÷ sin 40°.
  3. b = 15.55… = 15.6 cm.

Answer: D. 15.6 cm.

Correct.

Solution

  1. For B = 90°, sin B = 1.
  2. Sine rule: b ÷ sin 90° = 10 ÷ sin 40°, so b = 10 ÷ sin 40°.
  3. b = 15.55… = 15.6 cm.

Apply

Checking a roof truss

A roof truss is a triangle ABC. The base AB = 8.4 m, the rafter AC = 5.1 m and angle ABC = 32°. The builder says angle ACB is obtuse. A calculator is allowed.

1. Find angle ACB, correct to 1 decimal place.

2. Find angle BAC.

3. Find the length of the rafter BC, correct to 3 significant figures.

  1. AB = 8.4 m is opposite angle C, and AC = 5.1 m is opposite angle B = 32°. sin C = 8.4 × sin 32° ÷ 5.1 = 0.8728….
  2. sin⁻¹(0.8728…) = 60.8°, but C is obtuse: C = 180° − 60.8° = 119.2°.
  3. Angle BAC = 180° − 32° − 119.2…° = 28.8°.
  4. BC is opposite angle A: BC = 5.1 × sin 28.78…° ÷ sin 32° = 4.63 m.

Next: the Lesson 2 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.

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