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GCSE · Trigonometry

G5 Sine Rule, Cosine Rule and Area of a Triangle

AQA 8365 · Calculator allowed · about 105 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

This question tests recall of the key formulae for solving non-right-angled triangles. In each part, let triangle ABC have sides a, b and c opposite angles A, B and C respectively.

(a)State the sine rule.(1)

(b)State the cosine rule for finding the length of side a.(1)

(c)State the formula for the area of triangle ABC in terms of sides a and b and the angle C between them.(1)

(Total for Question 1 is 3 marks)

A triangle has two known angles and one known side, and that side is not between the two given angles (an AAS triangle).

(a)Which rule should be used first to find one of the unknown sides?(1)

A) Sine rule
B) Cosine rule
C) Area = 1/2 * a * b * sin(C)
D) Pythagoras' theorem

(Total for Question 2 is 1 mark)

In triangle ABC, angle A = 52 degrees, angle B = 71 degrees, and side a (opposite angle A, i.e. BC) = 9.6 cm.

(a)Calculate the length of side c (opposite angle C, i.e. AB), giving your answer correct to 3 significant figures.(3)

(Total for Question 3 is 3 marks)

In triangle ABC, b = 8.2 cm, c = 6.5 cm and the angle between them, angle A, = 78 degrees.

(a)Calculate the length of side a, giving your answer correct to 3 significant figures.(3)

(Total for Question 4 is 3 marks)

A triangle has two sides of length 5.4 cm and 7.8 cm, with an included angle of 63 degrees between them.

(a)Calculate the area of the triangle, giving your answer correct to 3 significant figures.(2)

(Total for Question 5 is 2 marks)

In triangle ABC, angle A = 35 degrees, side a = 11 cm and side b = 7.5 cm.

(a)Calculate angle B, giving your answer correct to 1 decimal place.(3)

(b)Hence calculate angle C.(1)

(Total for Question 6 is 4 marks)

A triangle has sides a = 13 cm, b = 15 cm and c = 20 cm.

(a)Calculate the size of the largest angle of the triangle, giving your answer correct to 1 decimal place.(3)

(Total for Question 7 is 3 marks)

A triangle has two sides of length 8 cm and 9 cm with an included angle of 30 degrees.

(a)Show that the area of the triangle is exactly 18 cm^2.(3)

(Total for Question 8 is 3 marks)

A triangle has all three side lengths known but none of its angles are known.

(a)Explain why the cosine rule, rather than the sine rule, must be used to find one of the angles.(2)

(Total for Question 9 is 2 marks)

In triangle ABC, angle A = 48 degrees, side a (opposite angle A, i.e. BC) = 13 cm and side b (opposite angle B, i.e. AC) = 10 cm.

(a)Calculate angle B, giving your answer correct to 1 decimal place.(3)

(b)Hence calculate angle C.(1)

(c)Hence calculate the area of triangle ABC, giving your answer correct to 3 significant figures.(3)

(Total for Question 10 is 7 marks)

A symmetrical roof truss forms an isosceles triangle. The two equal sloping sides each have length 4.2 m, and the angle between them at the apex is 100 degrees.

(a)Calculate the length of the base of the truss, giving your answer correct to 3 significant figures.(3)

(b)Calculate the area enclosed by the truss, giving your answer correct to 3 significant figures.(2)

(Total for Question 11 is 5 marks)

A triangle has sides a = 8 cm, b = 10 cm and c = 13 cm, where angle C is opposite side c.

(a)Calculate the size of angle C, giving your answer correct to 1 decimal place.(3)

(b)Hence calculate the area of the triangle, giving your answer correct to 3 significant figures.(3)

(Total for Question 12 is 6 marks)

Quadrilateral ABCD has diagonal AC = 14 cm. In triangle ABC, AB = 9 cm and angle BAC = 35 degrees. In triangle ACD, AD = 11 cm and angle DAC = 50 degrees, with B and D on opposite sides of the diagonal AC.

(a)Calculate the area of triangle ABC, giving your answer correct to 3 significant figures.(2)

(b)Calculate the area of triangle ACD, giving your answer correct to 3 significant figures.(2)

(c)Hence calculate the total area of quadrilateral ABCD, giving your answer correct to 3 significant figures.(1)

(Total for Question 13 is 5 marks)

In triangle ABC, angle A = 38 degrees, side a = 7 cm and side b = 10 cm.

(a)There are two possible values for angle B. Calculate both, giving your answers correct to 1 decimal place.(4)

(b)Hence find the two corresponding values of angle C.(2)

(Total for Question 14 is 6 marks)

A circle has centre O and radius 9 cm. Points X and Y lie on the circumference such that angle XOY = 140 degrees.

(a)Calculate the length of chord XY, giving your answer correct to 3 significant figures.(3)

(b)Calculate the area of triangle OXY, giving your answer correct to 3 significant figures.(2)

(Total for Question 15 is 5 marks)

A ship sails from port A on a bearing of 065 degrees for 12 km to point B. At B it changes course and sails on a bearing of 130 degrees for 9 km to reach point C.

(a)Show that angle ABC = 115 degrees.(2)

(b)Calculate the distance AC, giving your answer correct to 3 significant figures.(3)

(c)Calculate the bearing of C from A, giving your answer to the nearest degree.(3)

(Total for Question 16 is 8 marks)

A triangle has two sides of length 10 cm and 12 cm with an included angle of 45 degrees.

(a)Show that the exact area of the triangle is 30*√2 cm^2.(3)

(Total for Question 17 is 3 marks)

A vertical flagpole TF stands on horizontal ground at foot F. Points A and B lie on the ground such that angle AFB = 76 degrees, FA = 15 m and FB = 20 m. The angle of elevation of the top T of the flagpole from A is 25 degrees.

(a)Calculate the height TF of the flagpole, giving your answer correct to 3 significant figures.(2)

(b)Calculate the distance AB, giving your answer correct to 3 significant figures.(3)

(c)Calculate the angle of elevation of T from B, giving your answer correct to 1 decimal place.(2)

(Total for Question 18 is 7 marks)

A field is in the shape of quadrilateral ABCD. AB = 14 m, BC = 11 m, angle ABC = 100 degrees, CD = 16 m and DA = 21 m. The diagonal AC divides the field into triangles ABC and ACD.

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

(b)Calculate the size of angle ADC, giving your answer correct to 1 decimal place.(3)

(c)Hence calculate the total area of the quadrilateral field ABCD, giving your answer correct to 3 significant figures.(4)

(Total for Question 19 is 10 marks)

Mark scheme · G5 Sine Rule, Cosine Rule and Area of a Triangle

Question 1

(a) B1 a/sin(A) = b/sin(B) = c/sin(C) oe (reciprocal form also accepted)
(a) Answer: a/sin(A) = b/sin(B) = c/sin(C)
(b) B1 a² = b² + c² - 2bc*cos(A) oe
(b) Answer: a² = b² + c² - 2bc*cos(A)
(c) B1 Area = 1/2 * a * b * sin(C) oe
(c) Answer: Area = (1/2)*a*b*sin(C)

Question 2

(a) B1 A cao
(a) Answer: A) Sine rule

Question 3

(a) M1 angle C = 180 - 52 - 71 = 57 (angle sum of a triangle)
(a) M1 correct sine rule setup: c/sin(57) = 9.6/sin(52) oe
(a) A1 c = 10.2 cm awrt
(a) Answer: c = 10.2 cm

Question 4

(a) M1 correct substitution a² = 8.2² + 6.5² - 2(8.2)(6.5)cos(78)
(a) M1 a² = 109.49 - 22.16 = 87.3 (awrt) oe
(a) A1 a = 9.34 cm awrt
(a) Answer: a = 9.34 cm

Question 5

(a) M1 Area = 1/2 * 5.4 * 7.8 * sin(63)
(a) A1 18.8 cm² awrt
(a) Answer: 18.8 cm²

Question 6

(a) M1 correct sine rule setup sin(B)/7.5 = sin(35)/11 oe
(a) M1 sin(B) = 7.5*sin(35)/11 = 0.391 (awrt)
(a) A1 B = 23.0 degrees awrt
(a) Answer: B = 23.0 degrees
(b) B1 C = 180 - 35 - 23.0 = 122.0 degrees ft from part (a)
(b) Answer: C = 122.0 degrees

Question 7

(a) M1 identify that the largest angle, C, is opposite the longest side, c = 20
(a) M1 correct rearranged cosine rule: cos(C) = (13² + 15² - 20²)/(2*13*15) oe
(a) A1 C = 90.9 degrees awrt
(a) Answer: C = 90.9 degrees

Question 8

(a) M1 Area = 1/2 * 8 * 9 * sin(30)
(a) M1 sin(30) = 1/2 used exactly oe
(a) A1 Area = 36 * (1/2) = 18 cm², fully shown cso
(a) Answer: 18 cm² (shown)

Question 9

(a) B1 states that the sine rule needs at least one known angle-side pair (an angle and its opposite side) to set up a valid ratio oe
(a) B1 states that the cosine rule, rearranged to find an angle, only requires the three side lengths, which are all known here oe
(a) Answer: The sine rule needs a known angle to form a ratio with its opposite side; here no angles are known, so it cannot be used. The cosine rule, rearranged as cos(A) = (b²+c²-a²)/(2bc), needs only the three sides, all of which are given, so it can be used to find an angle directly.

Question 10

(a) M1 correct sine rule setup sin(B)/10 = sin(48)/13 oe
(a) M1 sin(B) = 10*sin(48)/13 = 0.572 (awrt)
(a) A1 B = 34.9 degrees awrt
(a) Answer: B = 34.9 degrees
(b) B1 C = 180 - 48 - 34.9 = 97.1 degrees ft from part (a)
(b) Answer: C = 97.1 degrees
(c) M1 Area = 1/2 * 13 * 10 * sin(97.1), ft from part (b)
(c) M1 = 65*sin(97.1) = 65*0.9923 (awrt)
(c) A1 Area = 64.5 cm² awrt
(c) Answer: 64.5 cm²

Question 11

(a) M1 correct substitution base² = 4.2² + 4.2² - 2(4.2)(4.2)cos(100)
(a) M1 = 35.28 - 35.28*(-0.17365) = 41.4 (awrt) oe
(a) A1 base = 6.43 m awrt
(a) Answer: 6.43 m
(b) M1 Area = 1/2 * 4.2 * 4.2 * sin(100)
(b) A1 8.69 m² awrt
(b) Answer: 8.69 m²

Question 12

(a) M1 correct rearranged cosine rule cos(C) = (8² + 10² - 13²)/(2*8*10) oe
(a) M1 cos(C) = -5/160 = -0.03125 (awrt)
(a) A1 C = 91.8 degrees awrt
(a) Answer: C = 91.8 degrees
(b) M1 Area = 1/2 * 8 * 10 * sin(91.8), ft from part (a)
(b) M1 = 40*sin(91.8) = 40*0.9995 (awrt)
(b) A1 Area = 40.0 cm² awrt
(b) Answer: 40.0 cm²

Question 13

(a) M1 Area = 1/2 * 9 * 14 * sin(35)
(a) A1 36.1 cm² awrt
(a) Answer: 36.1 cm²
(b) M1 Area = 1/2 * 11 * 14 * sin(50)
(b) A1 59.0 cm² awrt
(b) Answer: 59.0 cm²
(c) B1 95.1 cm² ft from parts (a) and (b)
(c) Answer: 95.1 cm²

Question 14

(a) M1 correct sine rule setup sin(B)/10 = sin(38)/7 oe
(a) M1 sin(B) = 10*sin(38)/7 = 0.8795 (awrt), giving B = 61.6 (awrt)
(a) M1 recognises the ambiguous case applies since b > a and b*sin(A) < a < b, giving a second value B = 180 - 61.6
(a) A1 B = 61.6 degrees or B = 118.4 degrees (both required) awrt
(a) Answer: B = 61.6 degrees or 118.4 degrees
(b) M1 C = 180 - 38 - B for each value of B, ft from part (a)
(b) A1 C = 80.4 degrees or C = 23.6 degrees (both required) awrt
(b) Answer: C = 80.4 degrees or 23.6 degrees

Question 15

(a) M1 correct substitution XY² = 9² + 9² - 2(9)(9)cos(140)
(a) M1 = 162 - 162*(-0.76604) = 286 (awrt) oe
(a) A1 XY = 16.9 cm awrt
(a) Answer: 16.9 cm
(b) M1 Area = 1/2 * 9 * 9 * sin(140)
(b) A1 26.0 cm² awrt
(b) Answer: 26.0 cm²

Question 16

(a) B1 correct reasoning: the bearing of A from B is 065+180 = 245 degrees oe
(a) B1 angle ABC = 245-130 = 115 degrees, fully shown cso
(a) Answer: 115 degrees (shown)
(b) M1 correct substitution AC² = 12² + 9² - 2(12)(9)cos(115), ft from part (a)
(b) M1 = 225 - 216*(-0.42262) = 316 (awrt) oe
(b) A1 AC = 17.8 km awrt
(b) Answer: 17.8 km
(c) M1 correct sine rule setup sin(BAC)/9 = sin(115)/17.8, ft from part (b)
(c) M1 sin(BAC) = 9*sin(115)/17.8 = 0.459 (awrt), giving angle BAC = 27.3 (awrt)
(c) A1 bearing = 065 + 27.3 = 092 degrees (nearest degree) awrt
(c) Answer: 092 degrees

Question 17

(a) M1 Area = 1/2 * 10 * 12 * sin(45)
(a) M1 sin(45) = √2/2 used exactly oe
(a) A1 Area = 60*√2/2 = 30*√2 cm², fully shown cso
(a) Answer: 30*√2 cm² (shown)

Question 18

(a) M1 TF = FA*tan(25) = 15*tan(25) oe
(a) A1 TF = 6.99 m awrt
(a) Answer: 6.99 m
(b) M1 correct substitution AB² = 15² + 20² - 2(15)(20)cos(76)
(b) M1 = 625 - 600*0.24192 = 480 (awrt) oe
(b) A1 AB = 21.9 m awrt
(b) Answer: 21.9 m
(c) M1 tan(angle) = TF/FB = 6.9946/20, ft from part (a)
(c) A1 19.3 degrees awrt
(c) Answer: 19.3 degrees

Question 19

(a) M1 correct substitution AC² = 14² + 11² - 2(14)(11)cos(100)
(a) M1 = 317 - 308*(-0.17365) = 370 (awrt) oe
(a) A1 AC = 19.2 m awrt
(a) Answer: 19.2 m
(b) M1 correct rearranged cosine rule in triangle ACD: cos(ADC) = (16² + 21² - AC²)/(2*16*21), ft from part (a)
(b) M1 cos(ADC) = (697-370.48)/672 = 0.486 (awrt)
(b) A1 ADC = 60.9 degrees awrt
(b) Answer: 60.9 degrees
(c) M1 Area of ABC = 1/2 * 14 * 11 * sin(100) = 75.8 (awrt), ft
(c) M1 Area of ACD = 1/2 * 16 * 21 * sin(60.9), ft from part (b)
(c) M1 Area of ACD = 168*sin(60.9) = 147 (awrt) oe
(c) A1 total area = 75.8 + 146.9 = 223 m² awrt
(c) Answer: 223 m²

Sine Rule, Cosine Rule and Area of a Triangle - Worksheets, Questions and Revision