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The sine rule and the cosine rule worksheet — GCSE Higher
Fifteen questions on "the sine rule and the cosine rule" — DfE statement G22. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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The sine rule and the cosine rule worksheet — GCSE Higher
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- 1.In triangle ABC, BC = 8 cm, AC = 10 cm and angle BAC = 40°. Given that angle ABC is acute, work out its size. Give your answer to 1 decimal place.
- 2.In triangle ABC, AB = 8 cm, BC = 11 cm and AC = 6 cm. Work out the size of angle ABC. Give your answer to 1 decimal place.
- 3.In triangle ABC, AB = x cm, AC = (x + 2) cm, BC = 15 cm and angle BAC = 100°. Work out the value of x. Give your answer to 1 decimal place.
- 4.A tent frame forms a triangle ABC, where AB = 3.5 m is a sloping pole, BC = 2.8 m is the base and AC = 4.2 m is the guy rope. Work out the size of angle ABC, between the pole and the base. Give your answer to 1 decimal place.
- 5.A ship leaves port P and sails 24 km on a bearing of 040° to a buoy Q. It then sails 31 km on a bearing of 115° to a lighthouse R. Work out the direct distance from P to R. Give your answer to 1 decimal place.
- 6.In triangle ABC, AB = 13 cm, AC = 9 cm and angle ABC = 35°. Given that angle ACB is obtuse, work out its size. Give your answer to 1 decimal place.
- 7.In triangle ABC, angle ABC = 72° and angle ACB = 48°. Side AB = (2x + 1) cm and side AC = (3x − 2) cm. Work out the value of x. Give your answer to 1 decimal place.
- 8.Two hikers walk from a campsite A to a lookout B, a distance of 14 km on a bearing of 038°. From B they walk to a shelter C, a distance of 20 km on a bearing of 142°. Work out the direct distance from A to C. Give your answer to 1 decimal place.
- 9.In triangle ABC, angle ABC = 47°, AC = 12.5 cm and BC = 9.2 cm. Work out the size of angle BAC. Give your answer to 1 decimal place.
- 10.Two coastguard stations A and B are 18 km apart, with B due east of A. A boat at C is on a bearing of 062° from A and on a bearing of 315° from B. Work out the distance of the boat from station B. Give your answer to 1 decimal place.
- 11.In triangle ABC, AB = 9 cm, AC = 6 cm and angle BAC = 110°. Work out the length of BC. Give your answer to 1 decimal place.
- 12.In triangle ABC, AB = 7 cm, AC = 6 cm and angle ABC = 40°. Two different lengths of BC are possible. Given that BC is the longer of them, work out the length of BC. Give your answer to 1 decimal place.
- 13.In triangle ABC, AB = 6.4 cm, BC = 9.1 cm and AC = 12.8 cm. Work out the size of the largest angle in the triangle. Give your answer to 1 decimal place.
- 14.In triangle ABC, AB = 9 cm, BC = 6 cm and angle BAC = 35°. This description fits two different triangles. Work out the two possible sizes of angle ACB, each to 1 decimal place.
- 15.In triangle ABC, AB = 7.4 cm, AC = 5.9 cm and angle BAC = 68°. Work out the length of BC. Give your answer to 1 decimal place.
Answer key
- (a) 53.5° — Method: use the sine rule BC/sin A = AC/sin B, since BC is opposite angle A and AC is opposite angle B. Working: sin(ABC) = AC × sin(BAC) / BC = 10 × sin 40° / 8, which gives angle ABC = 53.5° or its supplement 180 − 53.5 = 126.5°; since angle ABC is acute, the answer is 53.5°. Taking the obtuse supplement instead gives 126.5°; putting the sides the wrong way round in the ratio (sin B = BC × sin A / AC instead of AC × sin A / BC) gives 30.9°; and using sin 50° in place of sin 40° gives 73.2°. Every sine-rule ratio has two possible angle solutions that add to 180° — the word 'acute' tells you which one to keep.
- (a) 32.2° — Method: all three sides are known, so rearrange the cosine rule as cos A = (b² + c² − a²) ÷ 2bc, where a is the side facing the angle you want. Working: angle ABC sits between AB = 8 cm and BC = 11 cm and faces AC = 6 cm, so cos ABC = (8² + 11² − 6²) ÷ (2 × 8 × 11) = (64 + 121 − 36) ÷ 176 = 149 ÷ 176 = 0.84659, and the inverse cosine of 0.84659 is 32.157°. Answer: angle ABC = 32.2° to 1 decimal place. The distractors: 102.6° comes from subtracting the square of the longest side, 11, rather than the square of the side the angle actually faces, giving (64 + 36 − 121) ÷ 96; 57.8° comes from taking the inverse sine of 0.84659 instead of the inverse cosine; 147.8° comes from rearranging with the subtraction the wrong way round, (36 − 64 − 121) ÷ 176 = −0.84659, which turns an acute angle into its supplement.
- (d) 8.8 — Method: substitute into the cosine rule BC² = AB² + AC² − 2 × AB × AC × cos(BAC) and solve the resulting quadratic in x, keeping only the positive root. Working: 15² = x² + (x + 2)² − 2x(x + 2) cos 100°, which expands to a quadratic with two roots, x = 8.8 and x = −10.8; since x is a length, x = 8.8. Expanding (x + 2)² as x² + 4 instead of x² + 4x + 4, missing the middle term, gives x = 9.6; using +2x(x + 2) cos 100° instead of −2x(x + 2) cos 100° (a sign error on the cosine term) gives x = 10.6; and reporting the size of the rejected negative root instead of discarding it gives x = 10.8. Always expand a squared bracket fully before collecting terms.
- (d) 82.8° — Method: angle ABC is opposite the given side AC, so rearrange the cosine rule to cos B = (AB² + BC² − AC²) / (2 × AB × BC). Working: cos B = (3.5² + 2.8² − 4.2²) / (2 × 3.5 × 2.8), so angle ABC = 82.8°. Forgetting the negative sign in the rearrangement gives the supplementary angle 97.2°; leaving out the factor of 2 in the denominator gives 75.5°; and using AB and AC (the pair either side of angle A, not angle B) gives 41.4°, which is angle BAC, not angle ABC. Always check which angle sits opposite the side you left out of the pair you are dividing by.
- (b) 43.8 km — Method: find the angle of the triangle at Q from the two bearings, then use the cosine rule on the two known sides and that included angle. Working: the bearing of P from Q is the back bearing of 040°, which is 040° + 180° = 220°, and the bearing of R from Q is 115°, so angle PQR = 220° − 115° = 105°. Then PR² = 24² + 31² − 2 × 24 × 31 × cos 105° = 576 + 961 − 1488 × (−0.25882) = 1537 + 385.12 = 1922.12, and the square root of 1922.12 is 43.842. Answer: PR = 43.8 km to 1 decimal place. The distractors: 33.9 km comes from subtracting the bearings, 115° − 040° = 75°, and using 75° as the angle at Q, which is the angle the ship turns through, not the angle inside the triangle; 19.9 km comes from using the first bearing itself, 040°, as the angle at Q; 55.0 km comes from adding the two legs, 24 + 31, which would be the distance only if the ship had sailed in a straight line.
- (b) 124.1° — Method: use the sine rule AB/sin(ACB) = AC/sin(ABC), since AB is opposite angle ACB and AC is opposite angle ABC. Working: sin(ACB) = AB × sin(ABC) / AC = 13 × sin 35° / 9 = 0.8285, which gives angle ACB = 55.9° or its supplement 180 − 55.9 = 124.1°; both of these give a valid triangle, and since angle ACB is obtuse the answer is 124.1°. Reporting the acute solution instead, without checking the word 'obtuse' in the question, gives 55.9°; putting the sides the wrong way round in the ratio, sin(ACB) = 9 × sin 35° / 13, gives 23.4°; and subtracting the acute solution and the given 35° from 180° as if angle ACB were the remaining triangle angle gives 89.1°. A given SSA fact always has two possible angle solutions unless the stem states which one is meant.
- (b) 7.4 — Method: AB is opposite angle ACB and AC is opposite angle ABC, so the sine rule gives AB/sin(ACB) = AC/sin(ABC). Working: (2x + 1)/sin 48° = (3x − 2)/sin 72°; cross-multiplying and collecting the x terms gives x = 7.4. Using angle BAC = 180° − 72° − 48° = 60° in place of angle ABC in the ratio gives x = 4.7; ignoring the sine rule altogether and solving 2x + 1 = 3x − 2 as if the two sides were simply equal gives x = 3.0; and pairing each side with the wrong angle — AB with sin 72° and AC with sin 48° — gives x = 1.9. Each side in the sine rule must be paired with the sine of the angle directly opposite it.
- (b) 21.5 km — Method: find angle ABC from the two bearings, then use the cosine rule. Working: the bearing of A from B is 038° + 180° = 218°, so angle ABC = 218° − 142° = 76°. Then AC² = 14² + 20² − 2 × 14 × 20 × cos 76°, so AC = 21.5 km. Leaving out the factor of 2 in the cosine rule gives AC = 23.0 km; using 142° − 38° = 104° as the angle instead of the correct 76° gives AC = 27.0 km; and simply adding the two distances as if the path were a straight line gives 34.0 km. The angle between the two legs must come from the bearings, not from subtracting them directly.
- (c) 32.6° — Method: an angle is wanted from two sides and the angle facing one of them, so use the sine rule in the form sin A/a = sin B/b. Working: BC = 9.2 cm faces angle BAC and AC = 12.5 cm faces the 47° angle, so sin BAC = 9.2 × sin 47° ÷ 12.5 = 6.7285 ÷ 12.5 = 0.5383, and the inverse sine of 0.5383 is 32.566°. Because BC is shorter than AC, angle BAC must be smaller than 47°, so the acute value is the only one that fits. Answer: angle BAC = 32.6° to 1 decimal place. The distractors: 147.4° comes from taking 180° − 32.566°, the obtuse angle with the same sine, without checking it — 147.4° and 47° already add to more than 180°, so no such triangle exists; 83.6° comes from putting the sides the wrong way up, 12.5 × sin 47° ÷ 9.2, which gives a sine of 0.9937; 57.4° comes from pressing the inverse cosine key on 0.5383 instead of the inverse sine key.
- (d) 8.8 km — Method: turn each bearing into an angle of triangle ABC, find the third angle from the angle sum, then use the sine rule. Working: B is due east of A, so AB itself lies on a bearing of 090°, and the angle at A between AB and AC is 090° − 062° = 28°. From B, station A lies due west on a bearing of 270°, and C lies on 315°, so the angle at B is 315° − 270° = 45°. The third angle is 180° − 28° − 45° = 107°. The side BC faces the 28° angle and AB = 18 km faces the 107° angle, so BC = 18 × sin 28° ÷ sin 107° = 8.4505 ÷ 0.95630 = 8.8366. Answer: the boat is 8.8 km from B, to 1 decimal place. The distractors: 13.3 km comes from pairing BC with the 45° angle at B instead of the 28° angle it faces, which gives the distance AC; 12.0 km comes from never working out the third angle and dividing by sin 45° instead of sin 107°; 16.6 km comes from using the bearing 062° itself as the angle at A, instead of the 28° between AC and AB.
- (c) 12.4 cm — Using the cosine rule, BC² = AB² + AC² − 2 × AB × AC × cos(A) = 9² + 6² − 2 × 9 × 6 × cos(110°) = 81 + 36 − 108 × cos(110°). Since cos(110°) ≈ −0.34202, 108 × cos(110°) ≈ −36.94, so BC² ≈ 117 + 36.94 = 153.94. Taking the square root, BC ≈ 12.4072, which rounds to 12.4 cm. 8.9 cm comes from treating cos(110°) as if it were positive (using +0.342 instead of −0.342), which wrongly subtracts instead of adds and gives BC² ≈ 80.06. 153.9 cm is BC² itself, rounded, with the square root never taken. 11.6 cm comes from leaving out the factor of 2 in the formula, computing BC² = 81 + 36 − 9 × 6 × cos(110°) ≈ 135.47 instead.
- (a) 9.3 cm — Method: the 40° angle is not between the two known sides, so call the unknown side x and put it into the cosine rule, which turns into a quadratic equation with two positive roots. Working: AC faces angle ABC, so 6² = 7² + x² − 2 × 7 × x × cos 40°, that is 36 = 49 + x² − 10.7246x, which rearranges to x² − 10.7246x + 13 = 0. The discriminant is 10.7246² − 4 × 13 = 115.02 − 52 = 63.02, whose square root is 7.9384, so x = (10.7246 + 7.9384) ÷ 2 = 9.3315 or x = (10.7246 − 7.9384) ÷ 2 = 1.3931. The longer of the two is wanted. Answer: BC = 9.3 cm to 1 decimal place. The distractors: 1.4 cm is the shorter root, taken by a candidate who solves the quadratic correctly but does not read which of the two lengths is wanted; 4.5 cm comes from treating the 40° as the angle between the two given sides and working out 7² + 6² − 2 × 7 × 6 × cos 40° directly, when 40° lies at B and faces AC; 3.6 cm comes from assuming the triangle is right-angled with AB as the hypotenuse and using the square root of 7² − 6².
- (d) 110.1° — Method: the largest angle in any triangle faces the longest side, so identify that angle first and then use the cosine rule rearranged as cos A = (b² + c² − a²) ÷ 2bc. Working: the longest side is AC = 12.8 cm, which is faced by angle ABC, and the two sides meeting at B are 6.4 cm and 9.1 cm, so cos ABC = (6.4² + 9.1² − 12.8²) ÷ (2 × 6.4 × 9.1) = (40.96 + 82.81 − 163.84) ÷ 116.48 = −40.07 ÷ 116.48 = −0.34401. A negative cosine means an obtuse angle, and the inverse cosine of −0.34401 is 110.12°. Answer: the largest angle is 110.1° to 1 decimal place. The distractors: 28.0° is the angle facing the shortest side, chosen by a candidate who thinks the largest angle sits opposite the smallest side; 69.9° comes from dropping the minus sign and using cos = 0.34401, which turns the obtuse angle into its supplement; 81.4° comes from assuming the angles share out 180° in the same ratio as the sides, 180 × 12.8 ÷ 28.3, which is not how a triangle behaves.
- (d) 59.4° and 120.6° — Method: two sides and an angle that is not between them can describe two triangles, because an acute angle and its supplement have the same sine. Use the sine rule for the angle, then test whether the supplement also fits inside 180°. Working: AB = 9 cm faces angle ACB and BC = 6 cm faces the 35° angle, so sin ACB = 9 × sin 35° ÷ 6 = 5.1622 ÷ 6 = 0.86036. The inverse sine of 0.86036 is 59.358°, and its supplement is 180° − 59.358° = 120.642°. Both survive the angle-sum test, since 35° + 59.4° = 94.4° and 35° + 120.6° = 155.6°, each less than 180°. Answer: angle ACB is 59.4° or 120.6° to 1 decimal place. The distractors: 22.5° and 157.5° come from putting the sides the wrong way up, 6 × sin 35° ÷ 9 = 0.38238; 30.6° and 149.4° come from pressing the inverse cosine key on 0.86036 instead of the inverse sine key; 59.4° and 85.6° are the two unknown angles of the first triangle, 59.358° and 180° − 35° − 59.358°, given by a candidate who has found one triangle and reported its angles rather than the two possible sizes of the same angle.
- (a) 7.5 cm — Method: two sides and the angle between them are given, so use the cosine rule a² = b² + c² − 2bc cos A, with BC as the side facing the 68° angle. Working: BC² = 7.4² + 5.9² − 2 × 7.4 × 5.9 × cos 68° = 54.76 + 34.81 − 87.32 × 0.3746 = 89.57 − 32.71 = 56.86, and the square root of 56.86 is 7.5405. Answer: BC = 7.5 cm to 1 decimal place. The distractors: 11.1 cm comes from adding the last term instead of subtracting it, 89.57 + 32.71 = 122.28, the commonest sign slip on the cosine rule; 56.9 cm is the value of BC² written down as though it were BC, stopping one step before the square root; 2.9 cm comes from pressing sin instead of cos, using 87.32 × sin 68° = 80.96 as the term to subtract.
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