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) 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.
- (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.
- (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) 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.
- (a) 5 — Method: BC faces the 60° angle, so put the two algebraic sides into the cosine rule and solve the equation that results. By hand, cos 60° = 0.5. Working: 7² = x² + (x + 3)² − 2 × x × (x + 3) × 0.5, so 49 = x² + x² + 6x + 9 − x² − 3x = x² + 3x + 9. That rearranges to x² + 3x − 40 = 0, which factorises as (x + 8)(x − 5) = 0, giving x = −8 or x = 5. A length cannot be negative, so the negative root is rejected. Answer: x = 5. The distractors: 8 comes from expanding (x + 3)² as x² + 9, the error of assuming that squaring a bracket squares each term, which turns the equation into x² − 3x − 40 = 0; −8 is the negative root, kept by a candidate who solves the quadratic but never checks that x has to be a length; 37 comes from applying the cosine rule as though the 60° angle were at C facing AB, which gives x² = (x + 3)² + 49 − 7(x + 3) and a single linear solution.
- (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) 47.2° — Method: two sides and the angle between them are given, so find the third side with the cosine rule and then use the sine rule for the angle. Working: BC² = 12² + 9² − 2 × 12 × 9 × cos 55° = 144 + 81 − 216 × 0.57358 = 225 − 123.89 = 101.11, so BC = 10.0552 cm. Angle ABC faces AC = 9 cm and the 55° angle faces BC = 10.0552 cm, so sin ABC = 9 × sin 55° ÷ 10.0552 = 7.3724 ÷ 10.0552 = 0.73318, and the inverse sine of 0.73318 is 47.154°. Since AC is not the longest side, angle ABC is not the largest angle and the acute value is the one that fits. Answer: angle ABC = 47.2° to 1 decimal place. The distractors: 77.8° comes from pairing the angle at B with AB = 12 cm, the side beside it, instead of AC = 9 cm, the side it faces, which actually produces angle ACB; 23.2° comes from a sign slip in the cosine rule step, 225 + 123.89 = 348.89, giving BC = 18.68 cm before the sine rule is applied; 37.9° comes from dividing by AB = 12 cm rather than by the side facing the 55° angle.
- (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.
- (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°.
- (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.
- (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) 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) 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.
- (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.
- (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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