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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The sine rule and the cosine rule worksheet — GCSE Higher
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- 1.In triangle ABC, AB = 7.4 cm, AC = 5.9 cm and angle BAC = 68°. Work out the length of BC. Give your answer to 1 decimal place.
- 2.In triangle ABC, AB = 10 cm, BC = 7 cm and angle ACB = 35°. Given that angle BAC is acute, work out the length of AC. Give your answer to 1 decimal place.
- 3.A ship leaves port P and sails 24 km on a bearing of 040° to a buoy Q. It then sails 31 km on a bearing of 115° to a lighthouse R. Work out the direct distance from P to R. Give your answer to 1 decimal place.
- 4.In triangle ABC, AB = 5 cm, BC = 3 cm and AC = 7 cm. Work out the size of angle ABC.
- 5.A tent frame forms a triangle ABC, where AB = 3.5 m is a sloping pole, BC = 2.8 m is the base and AC = 4.2 m is the guy rope. Work out the size of angle ABC, between the pole and the base. Give your answer to 1 decimal place.
- 6.Which of these statements about solving a triangle is correct?
- 7.In triangle ABC, AB = x cm, AC = (x + 2) cm, BC = 15 cm and angle BAC = 100°. Work out the value of x. Give your answer to 1 decimal place.
- 8.In triangle ABC, AB = 8 cm, BC = 11 cm and AC = 6 cm. Work out the size of angle ABC. Give your answer to 1 decimal place.
- 9.In triangle ABC, AB = 7 cm, AC = 6 cm and angle ABC = 40°. Two different lengths of BC are possible. Given that BC is the longer of them, work out the length of BC. Give your answer to 1 decimal place.
- 10.In triangle PQR, PQ = 11 cm, QR = 7 cm and PR = 13 cm. Work out the size of angle PQR. Give your answer to 1 decimal place.
- 11.In triangle ABC, AB = 9 cm, BC = 6 cm and angle BAC = 35°. This description fits two different triangles. Work out the two possible sizes of angle ACB, each to 1 decimal place.
- 12.In triangle ABC, angle ABC = 47°, AC = 12.5 cm and BC = 9.2 cm. Work out the size of angle BAC. Give your answer to 1 decimal place.
- 13.A triangular plot of land has sides AB = 13 m and AC = 10 m, with angle BAC = 72° between them. Work out the perimeter of the plot. Give your answer to 1 decimal place.
- 14.A triangle has one unknown side or angle, marked x. In which one of these situations should the sine rule, rather than the cosine rule, be used to find x?
- 15.In triangle ABC, angle BAC = 42°, angle ABC = 63° and BC = 8.4 cm. Work out the length of AC. Give your answer to 1 decimal place.
Answer key
- (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.
- (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.
- (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.
- (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) 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) 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) 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) 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.
- (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) 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) 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.
- (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.
- (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) 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.
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