Mathematics Standard • Year 12 • Trigonometry • Lesson 7
The Cosine Rule and the Area Rule, Skill Drill
Build fluency in the three moves of this lesson: the cosine rule for an unknown side, the cosine rule rearranged for an unknown angle, and the area rule from two sides and the angle between them. Then practise deciding which of the three a question is asking for.
1. Quick recall
Answer in the space provided. 1 mark each
Q1.1 Write the cosine rule for finding a side.
Q1.2 What does "included angle" mean, and which two sides does it sit between?
Q1.3 If the cosine rule gives a negative value for cos C, then angle C is ____________.
Q1.4 Write the area rule for a triangle from two sides and the angle between them.
2. Worked example, unknown side
Follow each line. Every step has its reason.
Problem. In triangle ABC, b = 8 cm, c = 11 cm and A = 47°. Find a.
Step 1, name the included angle.
Sides b and c meet at A, so A is included and a is opposite it.
Reason: every letter in the rule shifts together. The side you are finding is always opposite the angle you are given.
Step 2, substitute before simplifying.
a² = 8² + 11² − 2(8)(11) cos47° = 185 − 176(0.681998)
Reason: the two squares give 185 and the correction term is one product. Tidying early is where sign errors get in.
Step 3, finish the square, then square root.
a² = 185 − 120.0317 = 64.9683, so a = √64.9683 = 8.06 cm
Reason: the rule delivers a squared, not a. Stopping at 64.97 is the single most common lost mark in this topic. Sense check, a must lie between 11 − 8 = 3 and 11 + 8 = 19.
3. Unknown side (SAS)
Two sides and the angle between them. Give answers to 2 decimal places. 1 mark each
Q3.1 b = 7, c = 9, A = 50°. Find a.
Q3.2 a = 12, b = 15, C = 35°. Find c.
Q3.3 a = 6.5, c = 9.2, B = 105°. Find b. Say what the obtuse angle does to the correction term.
4. Unknown angle (SSS)
Three sides and no angle. Rearrange to cos C = (a² + b² − c²) / 2ab. 2 marks each
Q4.1 a = 5, b = 6, c = 8. Find angle C, to the nearest minute.
Q4.2 a = 11, b = 13, c = 14. Find angle A, to the nearest degree.
5. Area rule
Two sides and the angle between them, with no height anywhere. 2 marks each
Q5.1 Two sides of 9 cm and 13 cm meet at 47°. Find the area, to 2 decimal places.
Q5.2 A triangle has sides 8 m, 10 m and 12 m. Find the angle between the 8 m and 10 m sides, then find the area to 2 decimal places.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, The cosine rule
c² = a² + b² − 2ab cos C, where C is the angle included between sides a and b, and c is the side opposite C.
Q1.2, Included angle
The angle formed where the two named sides meet. For sides a and b the included angle is C, and it is the only angle that both of those sides touch.
Q1.3, Negative cosine
Obtuse. Cosine is positive for acute angles and negative for obtuse ones, so the sign settles it before you touch the inverse cosine, and the calculator returns the obtuse angle directly.
Q1.4, The area rule
Area = ½ ab sin C, using the two sides that form the angle C. No perpendicular height is needed.
Q3.1, b = 7, c = 9, A = 50°
a² = 49 + 81 − 2(7)(9) cos50° = 130 − 126(0.642788) = 130 − 80.9913 = 49.0087, so a = 7.00. Check: 130 − 49.0087 = 80.99, and 80.99 ÷ 126 = 0.6428 = cos50°.
Q3.2, a = 12, b = 15, C = 35°
c² = 144 + 225 − 2(12)(15) cos35° = 369 − 360(0.819152) = 369 − 294.8947 = 74.1053, so c = 8.61. Sense check, 35° is well under a right angle so c comes in below √369 = 19.21, and well below because the angle is small.
Q3.3, a = 6.5, c = 9.2, B = 105°
b² = 42.25 + 84.64 − 2(6.5)(9.2) cos105° = 126.89 − 119.6(−0.258819) = 126.89 + 30.9548 = 157.8448, so b = 12.56 [1]. Because cos105° is negative, the correction term is added, and b comes out longer than the Pythagoras value √126.89 = 11.26 [1]. An obtuse included angle always opens the triangle up.
Q4.1, a = 5, b = 6, c = 8
cos C = (25 + 36 − 64) / (2 × 5 × 6) = −3 / 60 = −0.05 [1]. The cosine is negative, so C is obtuse. C = inverse cos(−0.05) = 92.866°, and 0.866 × 60 = 52.0, so C = 92° 52′ [1]. Nothing is subtracted from 180° here; the cosine rule returns the obtuse angle on its own.
Q4.2, a = 11, b = 13, c = 14
For angle A the opposite side is a, so cos A = (b² + c² − a²) / 2bc = (169 + 196 − 121) / (2 × 13 × 14) = 244 / 364 = 0.670330 [1]. A = 47.91° = 48° [1].
Q5.1, Sides 9 cm and 13 cm at 47°
Area = ½(9)(13) sin47° = 58.5 × 0.731354 = 42.78 cm² [2]. Sense check, it must be under ½(9)(13) = 58.5, which is the area you would get at a full 90°.
Q5.2, Sides 8 m, 10 m and 12 m
No angle is given, so find one first. The angle between the 8 m and 10 m sides is opposite the 12 m side: cos θ = (64 + 100 − 144) / (2 × 8 × 10) = 20 / 160 = 0.125, so θ = 82.82° [1]. That angle is included between the two sides used, so Area = ½(8)(10) sin82.82° = 40 × 0.992134 = 39.69 m² [1]. Finding the angle and then forgetting to apply the area rule is the usual half-finished answer here.