Printable · GCSE Higher · ages 14-16
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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Answer key: The sine rule and the cosine rule worksheet — GCSE Higher
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- (a) 114.6° — Using the sine rule, sin(ACB) = AB × sin(ABC) ÷ AC = 6 × sin(40°) ÷ 9. Since sin(40°) ≈ 0.64279, this gives sin(ACB) ≈ 3.8567 ÷ 9 ≈ 0.42852, so angle ACB ≈ 25.4° or its supplement, 154.6°. Testing the obtuse candidate: 40° + 154.6° = 194.6°, which already exceeds 180°, so angle BAC would have to be negative — impossible, so 154.6° is rejected. With angle ACB ≈ 25.4°, angle BAC = 180° − 40° − 25.4° = 114.6°. 25.4° is angle ACB, not angle BAC that the question asks for. 14.6° comes from using the invalid 154.6° candidate anyway and then wrongly turning the resulting negative angle sum (−14.6°) positive instead of rejecting it. 65.4° comes from inverting the sine rule ratio — dividing AC × sin(ABC) by AB instead of AB × sin(ABC) by AC — which gives a different, incorrect candidate for angle ACB entirely.
- (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.
- (d) 14.9 cm — By the sine rule, sin(BAC) ÷ BC = sin(ACB) ÷ AB, so sin(BAC) = 7 × sin35° ÷ 10 = 0.4015. This gives angle BAC = 23.7° or angle BAC = 156.3°; since angle BAC is acute, angle BAC = 23.7°. Angle ABC = 180° − 35° − 23.7° = 121.3°. By the sine rule again, AC ÷ sin(ABC) = AB ÷ sin(ACB), so AC = 10 × sin121.3° ÷ sin35° = 14.9 cm. Using angle BAC in the final step instead of angle ABC gives AC = 10 × sin23.7° ÷ sin35° = 7.0 cm — this just recomputes the given side BC. Using angle ACB again in the final step, instead of the newly found angle ABC, gives AC = 10 × sin35° ÷ sin35° = 10.0 cm — this just returns AB unchanged. Making a sign slip when finding angle ABC, computing 180° − 35° + 23.7° = 168.7° instead of subtracting it, and carrying that through gives AC = 10 × sin168.7° ÷ sin35° = 3.4 cm. Find angle BAC, then angle ABC, then apply the sine rule once more, and AC comes to 14.9 cm.
- (c) 11.2 cm — Method: two angles and a side are given, so use the sine rule, a/sin A = b/sin B = c/sin C, taking care that each side is paired with the angle it faces. Working: BC faces angle BAC = 42°, and AC faces angle ABC = 63°, so AC/sin 63° = 8.4/sin 42°. Multiplying up, AC = 8.4 × sin 63° ÷ sin 42° = 7.4845 ÷ 0.6691 = 11.185. Answer: AC = 11.2 cm to 1 decimal place. The distractors: 6.3 cm comes from writing the ratio upside down, 8.4 × sin 42° ÷ sin 63°, which pairs each side with the angle beside it rather than the angle opposite it; 12.1 cm comes from using the third angle, 180° − 42° − 63° = 75°, in the numerator, which gives the length of AB instead of AC; 12.6 cm comes from assuming the sides are in the same ratio as the angles and working out 8.4 × 63 ÷ 42, which is true for arcs of a circle but never for the sides of a triangle.
- (b) 8.4 cm — Two sides and the angle between them are known, so use the cosine rule: BC² = AB² + AC² − 2 × AB × AC × cos(BAC). Substituting, BC² = 81 + 36 − 45.64 = 71.36. Taking the square root: BC = √71.36 = 8.4 cm (1 d.p.). Leaving out the factor of 2 in the formula gives BC² = 81 + 36 − 22.82 = 94.18, so BC = 9.7 cm. Adding the cosine term instead of subtracting it gives BC² = 81 + 36 + 45.64 = 162.64, so BC = 12.8 cm. Using sin65° in place of cos65° gives BC² = 81 + 36 − 97.88 = 19.12, so BC = 4.4 cm. Keep the factor of 2, subtract the cosine term, and the correct length is 8.4 cm.
- (c) 22.3 km — The bearing of E from D is 065°, so the bearing of D from E (the back bearing) is 065° + 180° = 245°. The bearing of F from E is 165°, so the angle at E between ED and EF is 245° − 165° = 80°. Using the cosine rule, DF² = DE² + EF² − 2 × DE × EF × cos(80°) = 14² + 20² − 2 × 14 × 20 × cos(80°) = 196 + 400 − 560 × cos(80°). Since cos(80°) ≈ 0.17365, 560 × cos(80°) ≈ 97.24, so DF² ≈ 596 − 97.24 = 498.76. Taking the square root, DF ≈ 22.333, which rounds to 22.3 km. 26.3 km comes from using 100° as the angle at E — the difference between the two given bearings taken directly (165° − 65°) without converting to the back bearing first. 498.8 km is DF² itself, rounded, with the square root never taken. 23.4 km comes from leaving out the factor of 2 in the cosine rule formula, computing DF² = 196 + 400 − 14 × 20 × cos(80°) ≈ 547.38 instead.
- (a) 89.6° — Angle PQR is at vertex Q, so the side opposite it is PR = 13 cm. Using the cosine rule rearranged for an angle: cos(PQR) = (121 + 49 − 169) ÷ (2 × 11 × 7) = 1 ÷ 154 = 0.0065. Taking the inverse cosine, angle PQR = 89.6° (1 d.p.). Using PR as one of the enclosing sides instead of as the opposite side, finding angle P instead of angle Q, gives (121 + 169 − 49) ÷ (2 × 11 × 13) = 241 ÷ 286 = 0.8427, and angle = 32.6°. Swapping the sign inside the bracket, computing 169 − 121 − 49 instead, gives −1 ÷ 154 = −0.0065, and angle = 90.4°. Using PR in the denominator instead of QR gives (121 + 49 − 169) ÷ (2 × 11 × 13) = 1 ÷ 286 = 0.0035, and angle = 89.8°. Keep PR as the opposite side, QR and PQ as the enclosing sides, and angle PQR comes to 89.6°.
- (c) 36.7 m — Method: use the cosine rule to find the missing side BC, then add all three sides for the perimeter. Working: BC² = 13² + 10² − 2 × 13 × 10 × cos 72°, so BC = 13.7 m, and perimeter = 13 + 10 + 13.7 = 36.7 m. Forgetting the negative sign in the cosine rule (using + instead of −) gives BC = 18.7 m and a perimeter of 41.7 m; leaving AC out of the total and only adding AB and BC gives 26.7 m; and adding AB twice instead of AB and AC gives 39.7 m. Always add all three named sides once the missing one has been found.
- (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.
- (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) 49 cm² — Method: two sides and the angle between them are given, so use the cosine rule a² = b² + c² − 2bc cos A with BC facing the 60° angle. By hand, cos 60° = 0.5. Working: BC² = 8² + 5² − 2 × 8 × 5 × cos 60° = 64 + 25 − 80 × 0.5 = 89 − 40 = 49. Answer: BC² = 49 cm². The distractors: 129 cm² comes from adding the final term instead of subtracting it, 89 + 40; 89 cm² comes from leaving the cosine term out altogether and treating the 60° as though it were a right angle, so that Pythagoras applies; 69 cm² comes from forgetting the factor 2 in 2bc cos A and subtracting only 8 × 5 × 0.5 = 20.
- (a) 120° — Method: three sides are known, so use the cosine rule rearranged as cos A = (b² + c² − a²) ÷ 2bc, with a the side facing the angle wanted. Working: angle ABC lies between AB = 5 cm and BC = 3 cm and faces AC = 7 cm, so cos ABC = (5² + 3² − 7²) ÷ (2 × 5 × 3) = (25 + 9 − 49) ÷ 30 = −15 ÷ 30 = −0.5. The angle between 0° and 180° whose cosine is −0.5 is 180° − 60°. Answer: angle ABC = 120°. The distractors: 60° comes from taking the subtraction the other way round, (49 − 25 − 9) ÷ 30 = 0.5, which loses the minus sign that makes the angle obtuse; 90° comes from the instinct that three known sides always mean Pythagoras, and 5² + 3² = 34 is not 49, so the triangle is not right-angled; 150° comes from knowing the cosine is −0.5 but subtracting 30° from 180°, using the angle whose sine is 0.5 rather than the angle whose cosine is 0.5.
- (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.
- (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.
- (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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