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7.13 The Cosine Rule

EDEXCEL 1MA1 · Calculator allowed · about 95 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

In triangle ABC, angle A is opposite side a, angle B is opposite side b, and angle C is opposite side c. Which equation correctly applies the cosine rule to find side a?

A) a² = b² + c² - 2bc x cosA
B) a² = b² + c² + 2bc x cosA
C) a = b² + c² - 2bc x cosA
D) a² = b² - c² - 2bc x cosA

(Total for Question 1 is 1 mark)

In triangle PQR, PQ = 8 cm, PR = 6 cm and angle QPR = 70 degrees. Calculate the length of QR. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 2 is 3 marks)

In triangle ABC, AB = 9 cm, BC = 7 cm and CA = 5 cm. Calculate the size of angle C. Give your answer correct to 1 decimal place.

Diagram NOT accurately drawn

(Total for Question 3 is 3 marks)

A triangular garden plot has two sides of length 15 m and 22 m, with an angle of 58 degrees between them.

Diagram NOT accurately drawn

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

(b)A fence is to be built along this third side, costing 8.50 pounds per metre. Using your unrounded answer to part (a), work out the total cost of the fence. Give your answer to the nearest penny.(2)

(Total for Question 4 is 5 marks)

For each triangle described below, state whether the sine rule or the cosine rule should be used to find the unknown length, and give a reason.

(a)Triangle PQR has PQ = 14 cm, QR = 9 cm, angle PQR = 48 degrees, and PR is unknown. State which rule should be used to find PR, and give a reason.(1)

(b)Triangle STU has angle S = 52 degrees, side s = 11 cm (opposite angle S), and angle T = 67 degrees, with side t unknown. State which rule should be used to find t, and give a reason.(1)

(Total for Question 5 is 2 marks)

A triangle has sides of length 11 cm, 13 cm and 18 cm. Show that the triangle is obtuse-angled.

Diagram NOT accurately drawn

(Total for Question 6 is 3 marks)

In triangle ABC, AB = 6 cm, BC = 8 cm and CA = 11 cm.

Diagram NOT accurately drawn

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

(b)Using your answer to part (a), calculate the area of triangle ABC. Give your answer correct to 3 significant figures.(3)

(Total for Question 7 is 6 marks)

In triangle DEF, DE = 10 cm, DF = 14 cm and angle EDF = 50 degrees. Show that the length of EF is 10.8 cm, correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 8 is 3 marks)

In triangle ABC, angle A = 65 degrees, AB = 10 cm and AC = 7 cm.

Diagram NOT accurately drawn

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

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

(Total for Question 9 is 6 marks)

A boat sails from port P on a bearing of 040 degrees for 12 km to point A. It then changes course and sails on a bearing of 120 degrees for 18 km to point B.

Diagram NOT accurately drawn

(Total for Question 10 is 4 marks)

Three towns X, Y and Z are positioned so that XY = 11 km, YZ = 14 km and XZ = 19 km. Calculate the size of angle XYZ. Give your answer correct to 1 decimal place.

Diagram NOT accurately drawn

(Total for Question 11 is 3 marks)

ABCD is a parallelogram with AB = 9 cm, BC = 13 cm and angle ABC = 55 degrees. Calculate the length of the diagonal AC. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 12 is 3 marks)

Jayden is finding the size of angle X in a triangle with sides 7 cm, 9 cm and 12 cm, where X is the angle opposite the 12 cm side. His working is shown below:
cosX = (7² + 9² - 12²) / (2 x 7 x 9)
cosX = (49+81-144)/126
cosX = -14/126
X = cos^-1(-14/126)
X = 83.6 degrees
Jayden's final answer is wrong.

Diagram NOT accurately drawn

(a)Identify the error in Jayden's working.(1)

(b)Work out the correct value of X. Give your answer correct to 1 decimal place.(2)

(Total for Question 13 is 3 marks)

A yacht race course is a triangle with buoys X, Y and Z, where XY = 4.2 km, YZ = 3.6 km and angle XYZ = 115 degrees.

Diagram NOT accurately drawn

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

(b)The race requires sailing around the course (X to Y to Z to X) twice. Using your unrounded answer to part (a), work out the total distance sailed. Give your answer correct to 3 significant figures.(2)

(Total for Question 14 is 5 marks)

The diagram shows quadrilateral ABCD, made up of triangle ABC and triangle ACD joined along the diagonal AC. In triangle ABC, AB = 25 m, BC = 18 m and angle ABC = 112 degrees. In triangle ACD, AD = 22 m and DC = 24 m.

Diagram NOT accurately drawn

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

(b)Using your unrounded answer to part (a), calculate the size of angle ADC. Give your answer correct to 1 decimal place.(3)

(Total for Question 15 is 6 marks)

In triangle ABC, AB = x cm, BC = (x + 6) cm, angle ABC = 60 degrees and AC = 14 cm.

Diagram NOT accurately drawn

(a)Show that x² + 6x - 160 = 0.(4)

(b)Hence find the value of x, giving a reason for rejecting any invalid solution.(3)

(Total for Question 16 is 7 marks)

VABC is a tetrahedron on a horizontal base ABC, with VB perpendicular to the plane ABC. In the base, AB = 6 cm, BC = 8 cm and angle ABC = 100 degrees. The edge VB = 24 cm.

Diagram NOT accurately drawn

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

(b)Calculate the size of angle AVC. Give your answer correct to 1 decimal place.(4)

(Total for Question 17 is 7 marks)

Triangle ABC has angle A less than 90 degrees. The perpendicular from C meets AB at point D, with CD = h and AD = k.

Diagram NOT accurately drawn

(Total for Question 18 is 4 marks)

Mark scheme · 7.13 The Cosine Rule

Question 1

B1 A oe
Answer: A) a² = b² + c² - 2bc x cosA

Question 2

M1 QR² = 8² + 6² - 2 x 8 x 6 x cos70 oe
M1 QR² = 67.2 awrt (dep)
A1 8.20 cm awrt
Answer: 8.20 cm (3 sf)

Question 3

M1 cosC = (7² + 5² - 9²) / (2 x 7 x 5) oe
M1 cosC = -0.1 awrt (dep)
A1 95.7 degrees awrt
Answer: 95.7 degrees (1 dp)

Question 4

(a) M1 x² = 15² + 22² - 2 x 15 x 22 x cos58 oe
(a) M1 x² = 359 awrt (dep)
(a) A1 19.0 m awrt
(a) Answer: 19.0 m (3 sf)
(b) M1 their unrounded length x 8.50 ft
(b) A1 156.11 pounds to 161.11 pounds awrt (accept 161.11 pounds cao using full accuracy)
(b) Answer: 161.11 pounds

Question 5

(a) B1 cosine rule, because two sides and the included angle (SAS) are known oe
(a) Answer: Cosine rule, because two sides and the angle between them are known (SAS).
(b) B1 sine rule, because a complete angle-opposite-side pair (S and s) is known along with another angle oe
(b) Answer: Sine rule, because a full angle-side pair (S and s) is known along with another angle (T).

Question 6

M1 considers the angle opposite the longest side (18 cm) using cosC = (11²+13²-18²)/(2x11x13) oe
A1 cosC = -34/286 = -0.119 awrt (or compares 18²=324 with 11²+13²=290)
A1 since cosC is negative (or 324 > 290), the angle opposite the 18 cm side is obtuse, so the triangle is obtuse-angled cso
Answer: Obtuse-angled (angle opposite 18 cm is approximately 96.8 degrees)

Question 7

(a) M1 cosB = (6²+8²-11²)/(2x6x8) oe
(a) M1 cosB = -0.219 awrt (dep)
(a) A1 102.6 degrees awrt
(a) Answer: 102.6 degrees (1 dp)
(b) M1 Area = 0.5 x 6 x 8 x sin(their B) oe ft
(b) M1 sin(102.6) = 0.976 awrt used (dep)
(b) A1 23.4 cm² awrt (ft their B)
(b) Answer: 23.4 cm² (3 sf)

Question 8

M1 EF² = 10² + 14² - 2 x 10 x 14 x cos50 oe
M1 EF² = 116 awrt (dep)
A1 EF = 10.8 cm cso (answer given, all working shown)
Answer: 10.8 cm (answer given)

Question 9

(a) M1 BC² = 10² + 7² - 2 x 10 x 7 x cos65 oe
(a) M1 BC² = 89.8 awrt (dep)
(a) A1 9.48 cm awrt
(a) Answer: 9.48 cm (3 sf)
(b) M1 sin(ABC)/7 = sin65/their BC oe ft
(b) M1 sin(ABC) = 7 x sin65 / their BC (dep, ft)
(b) A1 42.0 degrees awrt (ft their BC)
(b) Answer: 42.0 degrees (1 dp)

Question 10

B1 angle PAB = 100 degrees (bearing back to P is 040+180=220, and 220-120=100)
M1 PB² = 12² + 18² - 2 x 12 x 18 x cos100 oe
M1 PB² = 543 awrt (dep)
A1 23.3 km awrt
Answer: 23.3 km (3 sf)

Question 11

M1 cos(XYZ) = (11² + 14² - 19²) / (2 x 11 x 14) oe
M1 cos(XYZ) = -0.143 awrt (dep)
A1 98.2 degrees awrt
Answer: 98.2 degrees (1 dp)

Question 12

M1 AC² = 9² + 13² - 2 x 9 x 13 x cos55 oe
M1 AC² = 116 awrt (dep)
A1 10.8 cm awrt
Answer: 10.8 cm (3 sf)

Question 13

(a) B1 he evaluated cos^-1 of the positive value 0.1111 instead of the negative value -0.1111, so found the acute angle instead of the obtuse angle oe
(a) Answer: He found cos^-1(14/126) instead of cos^-1(-14/126), giving the acute angle instead of the required obtuse angle.
(b) M1 X = cos^-1(-14/126) oe
(b) A1 96.4 degrees awrt
(b) Answer: 96.4 degrees (1 dp)

Question 14

(a) M1 XZ² = 4.2² + 3.6² - 2 x 4.2 x 3.6 x cos115 oe
(a) M1 XZ² = 43.4 awrt (dep)
(a) A1 6.59 km awrt
(a) Answer: 6.59 km (3 sf)
(b) M1 2 x (4.2 + 3.6 + their XZ) ft
(b) A1 28.8 km awrt (ft their XZ)
(b) Answer: 28.8 km (3 sf)

Question 15

(a) M1 AC² = 25² + 18² - 2 x 25 x 18 x cos112 oe
(a) M1 AC² = 1290 awrt (dep)
(a) A1 35.9 m awrt
(a) Answer: 35.9 m (3 sf)
(b) M1 cos(ADC) = (22² + 24² - their AC²) / (2 x 22 x 24) oe ft
(b) M1 cos(ADC) = -0.214 awrt (dep, ft)
(b) A1 102.4 degrees awrt (ft their AC)
(b) Answer: 102.4 degrees (1 dp)

Question 16

(a) M1 14² = x² + (x+6)² - 2 x x x (x+6) x cos60 oe
(a) M1 correctly expands (x+6)² = x²+12x+36 and uses cos60=0.5 so 2x(x+6)(0.5)=x(x+6)=x²+6x
(a) M1 196 = x²+x²+12x+36-x²-6x, i.e. 196 = x²+6x+36 (dep)
(a) A1 x²+6x-160=0 cso (fully rearranged, answer given)
(a) Answer: x² + 6x - 160 = 0 (answer given)
(b) M1 solves x²+6x-160=0, e.g. (x+16)(x-10)=0 or quadratic formula
(b) A1 x=10 or x=-16 (both values seen)
(b) B1 rejects x=-16 since a length cannot be negative, so x=10 cm cao
(b) Answer: x = 10 cm

Question 17

(a) M1 AC² = 6² + 8² - 2 x 6 x 8 x cos100 oe
(a) M1 AC² = 117 awrt (dep)
(a) A1 10.8 cm awrt
(a) Answer: 10.8 cm (3 sf)
(b) M1 recognises right angles at B (VB perpendicular to plane), so VA² = 24²+6² (=612) and VC² = 24²+8² (=640), using Pythagoras
(b) M1 cos(AVC) = (612 + 640 - their AC²) / (2 x sqrt612 x sqrt640) oe ft
(b) M1 cos(AVC) = 0.907 awrt (dep, ft)
(b) A1 24.9 degrees awrt (ft their AC)
(b) Answer: 24.9 degrees (1 dp)

Question 18

M1 in right triangle ADC, h = b x sinA and k = b x cosA (using AC = b)
M1 states BD = c - k = c - b cosA (using AB = c) and applies Pythagoras in triangle BDC: a² = h² + BD²
M1 expands a² = (b sinA)² + (c - b cosA)² = b² sin^2A + c² - 2bc cosA + b² cos^2A
A1 uses sin^2A + cos^2A = 1 to simplify to a² = b² + c² - 2bc cosA cso (proof complete)
Answer: a² = b² + c² - 2bc cosA (proven)

The Cosine Rule - Worksheets, Questions and Revision

7.13 The Cosine Rule

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18