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7.13D The Cosine Rule: Fluency and Exam Drill

EDEXCEL 1MA1 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The Cosine Rule: Fluency and Exam Drill

Revision Library original resource, written for Edexcel 1MA1 (Higher tier)

The cosine rule connects the three sides of a triangle with one of its angles. For a triangle with sides a, b and c, and angle A opposite side a:

a^2 = b^2 + c^2 - 2*b*c*cos(A)

Rearranged to find an angle when all three sides are known:

cos(A) = (b^2 + c^2 - a^2) / (2*b*c)

Use the cosine rule whenever you know two sides and the included angle (to find the third side), or all three sides (to find any angle). A calculator is allowed and expected throughout; give lengths correct to 3 significant figures and angles correct to 1 decimal place unless told otherwise. Questions 1 to 14 build fluency with the rule in both directions. Questions 15 to 20 apply it to exam-style problems, including bearings, perimeters, a combined area calculation, a two-triangle quadrilateral problem and an algebraic setup. The final two questions are grade 9 stretch problems requiring an exact surd answer and multi-step reasoning.

In triangle ABC, AB = 8 cm, AC = 10 cm and angle BAC = 55 degrees. Calculate the length of BC. Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangle ABC with AB = 8 cm and AC = 10 cm marked as the two sides from vertex A, and angle BAC = 55 degrees marked between them. BC is the unknown side opposite vertex A.

(Total for Question 1 is 2 marks)

In triangle PQR, PQ = 6 cm, PR = 9 cm and angle QPR = 40 degrees. Calculate the length of QR. Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangle PQR with PQ = 6 cm and PR = 9 cm marked as the two sides from vertex P, and angle QPR = 40 degrees marked between them. QR is the unknown side opposite vertex P.

(Total for Question 2 is 2 marks)

In triangle XYZ, XY = 12 cm, XZ = 15 cm and angle YXZ = 110 degrees. Work out the length of YZ. Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangle XYZ with XY = 12 cm and XZ = 15 cm marked as the two sides from vertex X, and angle YXZ = 110 degrees (obtuse) marked between them. YZ is the unknown side opposite vertex X.

(Total for Question 3 is 2 marks)

A triangle has sides of length 7 cm, 9 cm and 13 cm. Calculate the size of the largest angle in the triangle. Give your answer correct to 1 decimal place.

Figure (to be drawn): A scalene triangle with sides 7 cm, 9 cm and 13 cm marked. The angle opposite the 13 cm side (the longest side) is marked with a question mark, since this is the largest angle.

(Total for Question 4 is 3 marks)

In triangle PQR, PQ = 5 cm, PR = 8 cm and angle QPR = 120 degrees. Work out the length of QR. Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangle PQR with PQ = 5 cm and PR = 8 cm marked as the two sides from vertex P, and angle QPR = 120 degrees (obtuse) marked between them. QR is the unknown side opposite vertex P.

(Total for Question 5 is 2 marks)

Triangle ABC has AB = 10 cm, AC = 10 cm and angle BAC = 60 degrees. Calculate the length of BC, then state (with a reason) what type of triangle ABC is.

Figure (to be drawn): Isosceles triangle ABC with AB = 10 cm and AC = 10 cm marked as the two equal sides from vertex A, and angle BAC = 60 degrees marked between them. BC is the unknown side opposite vertex A.

(Total for Question 6 is 3 marks)

In triangle XYZ, XY = 4.2 cm, XZ = 6.7 cm and angle YXZ = 75 degrees. Work out the length of YZ. Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangle XYZ with XY = 4.2 cm and XZ = 6.7 cm marked as the two sides from vertex X, and angle YXZ = 75 degrees marked between them. YZ is the unknown side opposite vertex X.

(Total for Question 7 is 2 marks)

In triangle PQR, PQ = 9 cm, PR = 11 cm and angle QPR = 15 degrees. Calculate the length of QR. Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangle PQR with PQ = 9 cm and PR = 11 cm marked as the two sides from vertex P, and a small angle QPR = 15 degrees marked between them. QR is the unknown side opposite vertex P.

(Total for Question 8 is 2 marks)

Triangle ABC has AB = 9 cm, AC = 7 cm and BC = 5 cm. Calculate the size of angle BAC. Give your answer correct to 1 decimal place.

Figure (to be drawn): Triangle ABC with all three sides marked: AB = 9 cm, AC = 7 cm and BC = 5 cm. Angle BAC, at vertex A between sides AB and AC, is marked with a question mark.

(Total for Question 9 is 3 marks)

Triangle XYZ has XY = 20 cm, YZ = 25 cm and XZ = 30 cm. Work out the size of angle XYZ. Give your answer correct to 1 decimal place.

Figure (to be drawn): Triangle XYZ with all three sides marked: XY = 20 cm, YZ = 25 cm and XZ = 30 cm. Angle XYZ, at vertex Y between sides XY and YZ, is marked with a question mark.

(Total for Question 10 is 2 marks)

Fatima is fencing a triangular flower bed in her garden. Two of the sides measure 1.5 m and 2.3 m, with an angle of 48 degrees between them. Calculate the length of the third side. Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangular flower bed with two sides of 1.5 m and 2.3 m meeting at a vertex, with an angle of 48 degrees between them. The third side is unknown.

(Total for Question 11 is 2 marks)

Triangle ABC is isosceles with AB = AC = 14 cm and angle BAC = 100 degrees. Work out the length of BC. Give your answer correct to 3 significant figures.

Figure (to be drawn): Isosceles triangle ABC with AB = 14 cm and AC = 14 cm marked as the two equal sides from vertex A, and angle BAC = 100 degrees (obtuse) marked between them. BC is the unknown side opposite vertex A.

(Total for Question 12 is 2 marks)

Triangle ABC has AB = 10 cm, AC = 13 cm and BC = 8 cm. Calculate the size of angle BAC. Give your answer correct to 1 decimal place.

Figure (to be drawn): Triangle ABC with all three sides marked: AB = 10 cm, AC = 13 cm and BC = 8 cm. Angle BAC, at vertex A between sides AB and AC, is marked with a question mark.

(Total for Question 13 is 3 marks)

A roof truss has two struts of length 3.6 m and 5.1 m, meeting at an angle of 130 degrees. Work out the length of the third side of the triangular truss. Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangular roof truss with two struts of length 3.6 m and 5.1 m meeting at a vertex, with an angle of 130 degrees (obtuse) between them. The third side of the truss is unknown.

(Total for Question 14 is 2 marks)

A ship's captain, Priya, sails from port P to point Q, a distance of 42 km, on a bearing of 070 degrees. She then sails from Q to point R, a distance of 35 km, on a bearing of 150 degrees. Calculate the distance PR. Give your answer correct to 3 significant figures.

Figure (to be drawn): Bearings diagram showing P, Q and R. The bearing of Q from P is 070 degrees (distance 42 km). The bearing of R from Q is 150 degrees (distance 35 km). North lines are shown at P and at Q. Angle PQR is the interior angle at Q between QP and QR.

(Total for Question 15 is 4 marks)

Amara is edging a triangular garden plot ABC. AB = 18 m, BC = 24 m and angle ABC = 105 degrees. Calculate the total length of edging needed to go around the whole perimeter of the plot. Give your answer correct to 3 significant figures.

Figure (to be drawn): Triangular garden plot ABC with AB = 18 m and BC = 24 m marked as the two sides from vertex B, and angle ABC = 105 degrees (obtuse) marked between them. AC is the unknown side opposite vertex B.

(Total for Question 16 is 4 marks)

Triangle PQR has PQ = 11 cm, QR = 6 cm and PR = 14 cm.

Figure (to be drawn): Triangle PQR with all three sides marked: PQ = 11 cm, QR = 6 cm and PR = 14 cm. Angle PQR, at vertex Q between sides PQ and QR, is marked with a question mark.

(a)Calculate the size of angle PQR. Give your answer correct to 1 decimal place.(3)

(b)Hence, without further calculation, state whether triangle PQR is acute-angled, right-angled or obtuse-angled. Give a reason for your answer.(1)

(Total for Question 17 is 4 marks)

Liam is making a triangular sail for a small dinghy. Two sides of the sail measure 5.2 m and 3.8 m, with an angle of 72 degrees between them.

Figure (to be drawn): Triangular sail with two sides of 5.2 m and 3.8 m meeting at a vertex, with an angle of 72 degrees between them. The third side is unknown.

(a)Calculate the length of the third side of the sail. Give your answer correct to 3 significant figures.(3)

(b)Calculate the area of the sail. Give your answer correct to 3 significant figures.(2)

(Total for Question 18 is 5 marks)

Quadrilateral ABCD is made up of two triangles, ABD and BCD, joined along the diagonal BD. In triangle ABD, AB = 9 cm, AD = 7 cm and angle BAD = 95 degrees. In triangle BCD, BC = 10 cm and CD = 12 cm.

Figure (to be drawn): Quadrilateral ABCD split by diagonal BD into triangle ABD (AB = 9 cm, AD = 7 cm, angle BAD = 95 degrees) and triangle BCD (BC = 10 cm, CD = 12 cm). BD is shared by both triangles.

(a)Calculate the length of BD. Give your answer correct to 3 significant figures.(3)

(b)Using an unrounded value of BD² from part (a), calculate the size of angle BCD. Give your answer correct to 1 decimal place.(3)

(Total for Question 19 is 6 marks)

Chidi is designing a triangular metal bracket with sides of length x cm, (x + 2) cm and (x + 4) cm. The angle between the two shorter sides is 120 degrees, and this angle is opposite the longest side.

Figure (to be drawn): Triangle with sides x cm, (x+2) cm and (x+4) cm. The angle between the sides of length x cm and (x+2) cm is marked as 120 degrees, opposite the side of length (x+4) cm.

(a)Show that x satisfies the equation x² - x - 6 = 0.(3)

(b)Hence find the value of x, and state the three side lengths of the bracket.(3)

(Total for Question 20 is 6 marks)

Triangle ABC has AB = 4 cm, BC = 6 cm and angle ABC = 60 degrees. Show that AC = 2*√7 cm.

Figure (to be drawn): Triangle ABC with AB = 4 cm and BC = 6 cm marked as the two sides from vertex B, and angle ABC = 60 degrees marked between them. AC is the unknown side opposite vertex B.

(Total for Question 21 is 3 marks)

Quadrilateral ABCD is made up of two triangles, ABD and BCD, joined along the diagonal BD. In triangle ABD, AB = 6 cm, AD = 10 cm and angle BAD = 130 degrees. Triangle BCD is isosceles with BC = CD, and angle BCD = 70 degrees.

Figure (to be drawn): Quadrilateral ABCD split by diagonal BD into triangle ABD (AB = 6 cm, AD = 10 cm, angle BAD = 130 degrees) and isosceles triangle BCD (BC = CD, angle BCD = 70 degrees). BD is shared by both triangles.

(a)Calculate the length of BD. Give your answer correct to 3 significant figures.(3)

(b)Using an unrounded value of BD² from part (a), and the fact that triangle BCD is isosceles with BC = CD, calculate the length of BC. Give your answer correct to 3 significant figures.(4)

(Total for Question 22 is 7 marks)

Mark scheme · 7.13D The Cosine Rule: Fluency and Exam Drill

Question 1

M1 correct substitution, e.g. BC² = 8² + 10² - 2*8*10*cos(55) oe
A1 BC = 8.50 cm (awrt 8.50), cao
Answer: BC = 8.50 cm (3 s.f.)

Question 2

M1 correct substitution, e.g. QR² = 6² + 9² - 2*6*9*cos(40) oe
A1 QR = 5.85 cm (awrt 5.85), cao
Answer: QR = 5.85 cm (3 s.f.)

Question 3

M1 correct substitution, e.g. YZ² = 12² + 15² - 2*12*15*cos(110) oe
A1 YZ = 22.2 cm (awrt 22.2), cao
Answer: YZ = 22.2 cm (3 s.f.)

Question 4

M1 identifies that the largest angle is opposite the longest side (13 cm) and attempts cos C = (a²+b²-c²)/(2ab) with correct values, e.g. cos C = (7²+9²-13²)/(2*7*9)
M1 cos C = -39/126 oe (awrt -0.310)
A1 C = 108.0 degrees (awrt 108)
Answer: Largest angle = 108.0 degrees (1 d.p.)

Question 5

M1 correct substitution, e.g. QR² = 5² + 8² - 2*5*8*cos(120) oe
A1 QR = 11.4 cm (awrt 11.4), cao
Answer: QR = 11.4 cm (3 s.f.)

Question 6

M1 correct substitution, e.g. BC² = 10² + 10² - 2*10*10*cos(60) oe
A1 BC = 10 cm, cao
B1 equilateral (ft their BC), since AB = AC = BC = 10 cm (all three sides equal)
Answer: BC = 10 cm; triangle ABC is equilateral (all three sides equal 10 cm)

Question 7

M1 correct substitution, e.g. YZ² = 4.2² + 6.7² - 2*4.2*6.7*cos(75) oe
A1 YZ = 6.93 cm (awrt 6.93), cao
Answer: YZ = 6.93 cm (3 s.f.)

Question 8

M1 correct substitution, e.g. QR² = 9² + 11² - 2*9*11*cos(15) oe
A1 QR = 3.28 cm (awrt 3.28), cao
Answer: QR = 3.28 cm (3 s.f.)

Question 9

M1 correct rearrangement/substitution, e.g. cos(BAC) = (7²+9²-5²)/(2*7*9)
M1 cos(BAC) = 105/126 oe (awrt 0.833)
A1 angle BAC = 33.6 degrees (awrt 33.5-33.6)
Answer: Angle BAC = 33.6 degrees (1 d.p.)

Question 10

M1 correct substitution, e.g. cos(XYZ) = (20²+25²-30²)/(2*20*25) = 125/1000 = 0.125 oe
A1 angle XYZ = 82.8 degrees (awrt 82.7-82.8)
Answer: Angle XYZ = 82.8 degrees (1 d.p.)

Question 11

M1 correct substitution, e.g. third side² = 1.5² + 2.3² - 2*1.5*2.3*cos(48) oe
A1 1.71 m (awrt 1.71), cao
Answer: Third side = 1.71 m (3 s.f.)

Question 12

M1 correct substitution, e.g. BC² = 14² + 14² - 2*14*14*cos(100) oe
A1 BC = 21.4 cm (awrt 21.4), cao
Answer: BC = 21.4 cm (3 s.f.)

Question 13

M1 correct rearrangement/substitution, e.g. cos(BAC) = (10²+13²-8²)/(2*10*13)
M1 cos(BAC) = 205/260 oe (awrt 0.788)
A1 angle BAC = 38.0 degrees (awrt 38.0)
Answer: Angle BAC = 38.0 degrees (1 d.p.)

Question 14

M1 correct substitution, e.g. third side² = 3.6² + 5.1² - 2*3.6*5.1*cos(130) oe
A1 7.91 m (awrt 7.91), cao
Answer: Third side = 7.91 m (3 s.f.)

Question 15

B1 angle PQR = 100 degrees (bearing of P from Q is 070+180 = 250 degrees; angle PQR = 250-150 = 100 degrees)
M1 correct substitution into the cosine rule, e.g. PR² = 42² + 35² - 2*42*35*cos(100)
A1 PR² = 3500 (awrt 3499-3500)
A1 PR = 59.2 km (awrt 59.1-59.2)
Answer: PR = 59.2 km (3 s.f.)

Question 16

M1 correct substitution, e.g. AC² = 18² + 24² - 2*18*24*cos(105)
A1 AC = 33.5 m (awrt 33.5)
M1 adds all three sides to find the perimeter (ft their AC), e.g. perimeter = 18 + 24 + their AC
A1 perimeter = 75.5 m (awrt 75.5)
Answer: Perimeter = 75.5 m (3 s.f.)

Question 17

(a) M1 correct substitution, e.g. cos(PQR) = (11²+6²-14²)/(2*11*6)
(a) A1 cos(PQR) = -39/132 oe (awrt -0.295)
(a) A1 angle PQR = 107.2 degrees (awrt 107.1-107.2)
(a) Answer: Angle PQR = 107.2 degrees (1 d.p.)
(b) B1 obtuse-angled (ft their angle from part (a)), because angle PQR (awrt 107 degrees) is greater than 90 degrees, oe (equivalently, because cos(PQR) is negative)
(b) Answer: Obtuse-angled (angle PQR > 90 degrees)

Question 18

(a) M1 correct substitution, e.g. side² = 5.2² + 3.8² - 2*5.2*3.8*cos(72)
(a) A1 side² = 29.3 (awrt 29.2-29.3)
(a) A1 side = 5.41 m (awrt 5.41)
(a) Answer: Third side = 5.41 m (3 s.f.)
(b) M1 area = 0.5*5.2*3.8*sin(72) oe
(b) A1 9.40 m² (awrt 9.39-9.40)
(b) Answer: Area = 9.40 m² (3 s.f.)

Question 19

(a) M1 correct substitution, e.g. BD² = 9² + 7² - 2*9*7*cos(95)
(a) A1 BD² = 141 (awrt 140.9-141.0)
(a) A1 BD = 11.9 cm (awrt 11.9)
(a) Answer: BD = 11.9 cm (3 s.f.)
(b) M1 correct substitution using their unrounded BD², e.g. cos(BCD) = (10²+12²-BD²)/(2*10*12)
(b) A1 cos(BCD) = 0.429 oe (awrt 0.429), ft their BD²
(b) A1 angle BCD = 64.6 degrees (awrt 64.5-64.6)
(b) Answer: Angle BCD = 64.6 degrees (1 d.p.)

Question 20

(a) M1 correct cosine rule statement with cos(120) = -1/2 substituted, e.g. (x+4)² = x²+(x+2)² - 2*x*(x+2)*(-1/2)
(a) M1 correct expansion of both sides, e.g. x²+8x+16 = 3x²+6x+4 oe
(a) A1 correct rearrangement to x² - x - 6 = 0 (cso, all algebraic steps shown)
(a) Answer: Shown: x² - x - 6 = 0
(b) M1 correct factorisation (x-3)(x+2) = 0, or correct use of the quadratic formula
(b) A1 x = 3, rejecting x = -2 since a length cannot be negative
(b) B1 side lengths 3 cm, 5 cm and 7 cm (ft their x)
(b) Answer: x = 3; side lengths are 3 cm, 5 cm and 7 cm

Question 21

M1 correct substitution, e.g. AC² = 4² + 6² - 2*4*6*cos(60)
M1 AC² = 28 oe (using cos(60) = 1/2 exactly, e.g. 16+36-24)
A1 AC = √28 = 2*√7 cm (cso, must be given in this exact simplified surd form)
Answer: AC = 2*√7 cm (exact)

Question 22

(a) M1 correct substitution, e.g. BD² = 6² + 10² - 2*6*10*cos(130)
(a) A1 BD² = 213 (awrt 213.0-213.2)
(a) A1 BD = 14.6 cm (awrt 14.6)
(a) Answer: BD = 14.6 cm (3 s.f.)
(b) M1 correct isosceles cosine rule relationship, e.g. BD² = BC²+BC²-2*BC*BC*cos(70) = 2*BC²*(1-cos(70))
(b) M1 correct rearrangement, e.g. BC² = BD² / (2*(1-cos(70))), dependent on previous method mark
(b) A1 BC² = 162 (awrt 161.9-162.0), ft their BD²
(b) A1 BC = 12.7 cm (awrt 12.7)
(b) Answer: BC = 12.7 cm (3 s.f.)

The Cosine Rule: Fluency and Exam Drill - Worksheets, Questions and Revision