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

14 original exam-style questions - 4 pages of questions with a full mark scheme - free printable PDF.

Download PDF Jump to mark scheme (page 5)The Sine Rule, Cosine Rule and Area of a Triangle revision guide

This topic is chapter 3 of IGCSE Maths Practice Book 2.

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3.2 The Sine Rule, Cosine Rule and Area of a Triangle

EDEXCEL 4MA1 · Calculator allowed · about 60 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Show full working for multi-step questions. Write answers in clear sentences for questions where a reason is asked. Suggested time 75 minutes.

In triangle ABC (not right-angled) angle A = 50 degrees, side a = BC = 12 cm and side b = AC = 15 cm. Use the sine rule to find the possible value(s) of angle B in degrees, giving answers to the nearest degree. State if two values are possible.

(Total for Question 1 is 2 marks)

Triangle PQR has sides p = QR = 9 cm, q = RP = 14 cm and angle P = 40 degrees. Calculate the length r = PQ, correct to 2 decimal places, using the cosine rule.

(Total for Question 2 is 2 marks)

In triangle ABC, a = BC = 10 cm and angle A = 30 degrees. No other sides or angles are given. State briefly whether the sine rule can be used to find side b = AC uniquely, and explain if an ambiguous case may arise.

(Total for Question 3 is 1 mark)

Triangle DEF has sides DE = 8 cm, EF = 7 cm and DF = 10 cm. Use the cosine rule to find angle E (the angle at vertex E between DE and EF), giving the answer to the nearest degree.

(Total for Question 4 is 2 marks)

A triangle has two sides of lengths 13 cm and 14 cm with included angle 58 degrees between them. Work out the area of this triangle in square centimetres, correct to 1 decimal place, using the area formula Area = 1/2 ab sin C.

(Total for Question 5 is 3 marks)

Triangle ABC has sides AB = 10 cm, AC = 14 cm and angle A = 47 degrees. First use the cosine rule to find BC to 2 decimal places, then find the area of triangle ABC to 1 decimal place using 1/2 ab sin C (you will need an appropriate angle).

(Total for Question 6 is 3 marks)

Show that for triangle MNO with sides MN = 7 cm, NO = 13 cm and MO = 15 cm, the angle at N is obtuse. Use the cosine rule and give the angle to the nearest degree.

(Total for Question 7 is 2 marks)

Triangle GHI has GH = 11 cm, HI = 9 cm and angle G = 46 degrees (angle at G between GH and GI). Calculate the third side GI to 2 decimal places using the cosine rule.

(Total for Question 8 is 3 marks)

A boat sails from point A on a bearing of 075 degrees for 8.5 km to point B. From B it sails on a bearing of 200 degrees for 6 km to point C. Work out the straight-line distance AC, correct to 2 decimal places, by extracting the triangle ABC and using the cosine rule.

(Total for Question 9 is 2 marks)

In triangle XYZ, side x = YZ = 7 cm, side y = ZX = 12 cm and angle Z = 38 degrees. Use the sine rule to find angle X in degrees, to the nearest degree. Show any ambiguous case consideration and give the correct final angle(s).

(Total for Question 10 is 3 marks)

A triangular prism has an equilateral triangular cross-section with side 6 cm. A triangle is formed by one edge of length 6 cm on a triangular face and a line of length 8 cm from one end of that edge to a point elsewhere on the prism, with an angle of 60 degrees between the 6 cm edge and the 8 cm line. Work out the area of that triangle using 1/2 ab sin C and give your answer correct to 1 decimal place.

(Total for Question 11 is 3 marks)

A triangular sail has sides 5.2 m and 7.4 m with included angle 112 degrees. A designer needs the area to the nearest 0.1 m², and also the length of the third side to 2 decimal places. First work out the area using 1/2 ab sin C, then find the third side using the cosine rule.

(Total for Question 12 is 4 marks)

Triangle RST has RS = 9 cm, ST = 16 cm and angle R = 41 degrees. Use the sine rule to find angle T to the nearest degree, then find the area of triangle RST to 1 decimal place.

(Total for Question 13 is 4 marks)

A surveyor measures a triangle formed by three stakes A, B and C. He finds AB = 21 m, AC = 13 m and angle A = 102 degrees. (a) Calculate BC to 2 decimal places using the cosine rule. (b) Calculate the area of triangle ABC to 1 decimal place. Show all working.

(Total for Question 14 is 6 marks)

Mark scheme · 3.2 The Sine Rule, Cosine Rule and Area of a Triangle

Question 1

M1 substitutes into sine rule: sin B / 15 = sin 50 / 12 and evaluates sin B = 15*sin50/12
A1 gives both possible angles B = 73 degrees and B = 107 degrees, correct to nearest degree, with statement that two values are possible
Answer: B = 73 degrees or 107 degrees (to nearest degree), two possible values

Question 2

M1 uses r² = p² + q² - 2*p*q*cos P with correct substitution: r² = 9² + 14² - 2*9*14*cos40
A1 r = 9.16 cm to 2 decimal places, cao
Answer: r = 9.16 cm

Question 3

B1 correct statement that b cannot be determined uniquely from only a and A and that an ambiguous case may arise
Answer: Not unique: the sine rule does not give b from only a and A; an ambiguous case can arise and additional information is needed

Question 4

M1 uses cos E = (DE² + EF² - DF²) / (2*DE*EF) with correct substitution
A1 E = 83 degrees (nearest degree), cao
Answer: E = 83 degrees

Question 5

M1 substitutes into area formula: Area = 1/2 * 13 * 14 * sin 58
M1 evaluates sine and multiplies correctly to a suitable intermediate accuracy
A1 gives final area = 77.2 cm² correct to 1 decimal place, cao
Answer: 77.2 cm²

Question 6

M1 uses cosine rule to find BC² = AB² + AC² - 2*AB*AC*cos A with correct substitution
M1 evaluates BC to 2 dp correctly
A1 uses area formula with two sides and included angle (e.g. 1/2 * AB * AC * sin A) and gives area to 1 dp correctly
Answer: BC = 10.25 cm (to 2 dp), Area = 51.2 cm² (to 1 dp)

Question 7

M1 uses cosine rule: cos N = (MN² + NO² - MO²) / (2*MN*NO) with correct substitution
A1 evaluates cos N and finds N = 92 degrees (nearest degree), concluding angle N is obtuse
Answer: Angle N = 92 degrees (to nearest degree), so it is obtuse

Question 8

M1 applies cosine rule to form equation: HI² = GH² + GI² - 2*GH*GI*cos G and rearranges to a quadratic in GI
M1 solves the quadratic correctly for GI (algebraic or using quadratic formula)
A1 gives GI = 11.93 cm and GI = 3.35 cm to 2 decimal places, or notes the appropriate physically valid value(s) with correct rounding
Answer: GI = 11.93 cm or GI = 3.35 cm (to 2 dp)

Question 9

M1 correctly finds the included angle at B between BA and BC: the back-bearing from B to A is 075 + 180 = 255 degrees, and the bearing from B to C is 200 degrees, so angle ABC = 255 - 200 = 55 degrees
A1 uses cosine rule AC² = AB² + BC² - 2*AB*BC*cos55 and gives AC = 7.05 km to 2 dp, cao
Answer: AC = 7.05 km

Question 10

M1 recognises this is an SAS configuration (angle Z is included between sides x = YZ = 7 and y = ZX = 12), so the cosine rule must be used first to find the third side z = XY: z² = x² + y² - 2xy cos Z = 7² + 12² - 2*7*12*cos38
M1 evaluates z = XY = 7.79 cm (to 2 dp), then applies the sine rule sin X / x = sin Z / z to get sin X = 7*sin38/7.79
A1 gives X = 34 degrees (to nearest degree), cao
Answer: X = 34 degrees (to nearest degree)

Question 11

M1 identifies the triangle sides to use: face edge = 6 cm, line to the third point = 8 cm, and the given angle of 60 degrees between them
M1 substitutes into area formula Area = 1/2 * 6 * 8 * sin60 and evaluates correctly
A1 gives area = 20.8 cm² correct to 1 decimal place, cao
Answer: 20.8 cm²

Question 12

M1 uses area formula: 1/2 * 5.2 * 7.4 * sin112 with correct substitution
M1 evaluates area to required accuracy
M1 uses cosine rule to form third side² = 5.2² + 7.4² - 2*5.2*7.4*cos112
A1 gives third side = 10.52 m (to 2 dp) and area = 17.8 m² (to 1 dp), cao
Answer: Area = 17.8 m², third side = 10.52 m

Question 13

M1 uses sine rule sin T / RS = sin R / ST or equivalent correctly and forms sin T = RS*sin R / ST = 9*sin41/16
M1 evaluates sin T = 0.369, giving angle T = 22 degrees (nearest degree); notes this is the only valid solution since RS < ST means angle T < angle R, so the supplementary angle would make the angle sum exceed 180 degrees
M1 finds angle S = 180 - 41 - 22 = 117 degrees, then uses area formula 1/2*RS*ST*sin S with correct substitution
A1 gives angle T = 22 degrees and area = 64.0 cm² (answers to required accuracy), cao
Answer: T = 22 degrees, Area = 64.0 cm²

Question 14

M1 uses cosine rule BC² = AB² + AC² - 2*AB*AC*cos A with correct substitution
M1 evaluates BC to 2 dp correctly
M1 uses area formula Area = 1/2 * AB * AC * sin A with correct substitution
M1 evaluates sine and multiplies to a suitable intermediate accuracy
A1 gives BC = 26.90 m to 2 dp, cao
A1 gives area = 133.5 m² to 1 dp, cao
Answer: BC = 26.90 m, Area = 133.5 m²

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

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3.2 The Sine Rule, Cosine Rule and Area of a Triangle

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