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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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Answer key: The sine rule and the cosine rule worksheet — GCSE Higher

Maths · GCSE Higher · G22 · The sine rule and the cosine rule · 15 questions

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(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°.
(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.
(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.
(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.
(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.
(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.
(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.
(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.
(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.
(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.
(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.
(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.
(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.
(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.
(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.

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