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) 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.
- (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.
- (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°.
- (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.
- (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) 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.
- (d) Two angles and a side: sine rule finds other sides. — Method: match the data you are given to the rule that needs it. Working: the sine rule a/sin A = b/sin B = c/sin C needs a complete angle-side pair to set up its ratio, so the statement that two angles and a side (AAS or ASA) let the sine rule find the other sides is the correct one — the third angle comes from the angle sum, and each unknown side is then opposite a known angle. The statement that two sides and the angle between them (SAS) call for the sine rule is wrong: no angle-side pair is complete, so the cosine rule is what works there. The statement that the sine rule finds any angle from three sides (SSS) is wrong for the same reason in reverse — no angle is known at all, so the cosine rule must find the first one. The statement that the cosine rule finds a missing angle directly from two sides and a non-included angle (SSA) is wrong: the cosine rule reports the angle enclosed by the two sides it uses, so with SSA it is the sine rule that reaches the missing angle, and the ambiguous case is then settled from the wording of the question.
- (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.
- (c) Two angles and one side are known; x is another side — The sine rule needs a matching pair, a side and the angle opposite it, that you already know, so you can set up a ratio with the unknown. When two angles and one side are known, you can find the third angle from the angle sum, giving you an angle opposite the known side and an angle opposite x: the sine rule applies directly. When all three sides are known and x is an angle, there is no side-angle pair available at all, so the cosine rule, rearranged for an angle, is what's needed instead. When two sides and the included angle are known and x is the third side, again there is no matching side-angle pair yet, so the cosine rule finds the third side directly. When two sides and the included angle are known and x is one of the other angles, you still have no side-angle pair to start from — the cosine rule has to be used first, to find the third side, before any angle can be found. Only the two-angles-and-a-side case hands you a ready-made pair, which is exactly what the sine rule needs.
- (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) 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) 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) 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) £982.20 — Method: find the missing side BC with the cosine rule, add it to AB and AC for the perimeter, then multiply by the cost per metre. Working: BC² = 45² + 38² − 2 × 45 × 38 × cos 110° = 4638.71, so BC = 68.108 m, perimeter = 45 + 38 + 68.108 = 151.108 m, and cost = 151.108 × £6.50 = £982.20 — keep the unrounded perimeter, because the rounded 151.1 m would give £982.15. Forgetting the negative sign in the cosine rule gives BC = 48.0 m and a cost of £851.18; costing only the missing side BC and forgetting to include AB and AC gives £442.70; and stopping after finding the perimeter, without multiplying by the cost per metre, gives £151.11. The perimeter is only the halfway point of this question — the cost still has to be worked out.
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