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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 · Calculator · 15 questions

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(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) 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.
(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.
(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) 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) 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.
(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.
(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.
(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) 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.
(b) No — should divide by −2bc, not +2bc; sign is wrong. — Method: rearrange a² = b² + c² − 2bc cos A step by step and compare with the student's version. Working: subtracting b² + c² from both sides gives a² − b² − c² = −2bc cos A, then dividing both sides by −2bc gives cos A = (a² − b² − c²) / (−2bc), which is the same as cos A = (b² + c² − a²) / (2bc). The student divided by +2bc instead of −2bc, so their expression is the negative of the correct one — the verdict is No. Saying the algebra is fine because 'either sign order gives a valid result' ignores that only one of the two signed expressions matches the original equation. Saying the denominator should be bc rather than 2bc is a different, unrelated error — the coefficient 2bc in the original formula is correct and must stay.
(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.
(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.
(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.

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