Mathematics Standard • Year 12 • Trigonometry • Lesson 7
Cosine Rule and Area Rule, Reasoning
Work backwards from an area, diagnose a worked solution that is wrong in two places, and run a full survey problem end to end. These are the Band 5 and 6 versions of the same three rules.
1. Working backwards from an area
4 marks
Q1.1 A triangle has two sides of 14 cm and 9 cm, and an area of 50 cm².
(a) Find the included angle, correct to the nearest minute.
(b) There is a second triangle with the same two sides and the same area. Find its included angle, and explain why the area rule has this second answer when the cosine rule never does.
2. Diagnose the solution
This student's work has two separate errors. Find both, name them, and give the correct answer. 4 marks
Problem. In triangle ABC, b = 9 cm, c = 12 cm and A = 40°. Find a, to 2 decimal places.
a² = 9² + 12² − 2(9)(12) cos40°
a² = 225 − 216 × 0.642788
a² = 225 − 138.84 = 86.16
a = 86.16 cm
3. A survey, end to end
A surveyor standing at A sights two corners of a property: B on a bearing of 072°T at 340 m, and C on a bearing of 145°T at 420 m. 7 marks
Q3.1 (a) Find the size of angle BAC. (1 mark)
(b) Find the distance BC, correct to 2 decimal places. (2 marks)
(c) Find the area of triangle ABC, correct to the nearest square metre. (2 marks)
(d) Explain why the angle at A could be read straight from the two bearings, but the angle at B could not. (2 marks)
4. Justify
3 marks
Q4.1 A student says: "The cosine rule is unnecessary. If I ever need it, I can just use the sine rule twice instead." Explain why this is wrong, using one of the two situations from this lesson to show it.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 (a), The included angle
Area = ½ ab sin C, so 50 = ½(14)(9) sin C = 63 sin C [1], giving sin C = 50 / 63 = 0.793651. Then C = inverse sin(0.793651) = 52.5280°, and 0.5280 × 60 = 31.7, so C = 52° 32′ [1].
Q1.1 (b), The second triangle, and why it exists
C = 180° − 52.5280° = 127.4720°, and 0.4720 × 60 = 28.3, so the second included angle is 127° 28′ [1].
It exists because the area rule contains a sine, and sin θ = sin(180° − θ), so two different angles give the same area from the same two sides. The cosine rule never has this problem because it contains a cosine, which is positive for acute angles and negative for obtuse ones, so its value names the angle uniquely [1]. This is the same ambiguity you met in the sine rule, arriving from a different direction.
Q2, The two errors
Error 1, wrong trigonometric value. The student used 0.642788, which is sin 40°, not cos 40° = 0.766044. The cosine rule takes the cosine of the included angle [1].
Error 2, answered with a squared instead of a. The final line reads a = 86.16 when 86.16 was the value of a squared. No square root was taken [1].
Correct answer. a² = 81 + 144 − 216(0.766044) = 225 − 165.466 = 59.534 [1], so a = √59.534 = 7.72 cm [1]. Sense check, a must lie between 12 − 9 = 3 and 12 + 9 = 21, and 86.16 was never a possible side length, which is a check the student could have run on their own answer.
Q3.1 (a), Angle BAC
Both bearings are measured from north at A, so the angle between the two lines of sight is their difference: 145° − 072° = 73° [1].
Q3.1 (b), Distance BC
Two sides and the angle between them, so the cosine rule: BC² = 340² + 420² − 2(340)(420) cos73° = 292000 − 285600(0.292372) = 292000 − 83501.36 = 208498.64 [1]. So BC = 456.62 m [1].
Q3.1 (c), Area
The same two sides and the same included angle, so the area rule applies with no extra work: Area = ½(340)(420) sin73° = 71400 × 0.956305 = 68280.16 [1], which is 68280 m², about 6.83 hectares [1].
Q3.1 (d), Why A is easy and B is not
Both bearings were taken from A, and a bearing is measured clockwise from north. Two lines of sight from the same point with north common to both means the angle between them is simply the difference of the two bearings [1].
At B, no bearing was recorded at all. The surveyor never stood there. To get the angle at B you would first have to find BC, then use the sine rule or the cosine rule on the completed triangle, which is calculation rather than reading [1].
Q4.1, Why the sine rule cannot replace the cosine rule
The sine rule equates fractions of the form side over sine of its opposite angle, so it can only start once you have one complete fraction, that is a side and the angle opposite it [1].
Take SSS, three sides and no angle at all. There is not a single angle to put in any fraction, so the sine rule cannot be written down, let alone used twice [1]. SAS fails for the same reason: the one angle you have is between the two known sides, not opposite either of them. In both cases the cosine rule is the only way to get the first angle, and only after that does the sine rule become available at all [1].