Mathematics • Year 10 • Unit 3 • Lesson 16

Cosine Rule in the Real World

Apply the cosine rule to real Australian contexts — a cricket pitch diagonal, a Sydney sailing course, a triangular paddock, a roof rafter, and a bush-walking route. Decide SAS vs SSS, write the right form, and check whether the answer is reasonable.

Apply · Real-World Maths

1. Word problems

For each problem: (i) sketch a triangle with all known information labelled, (ii) decide whether it is SAS or SSS, (iii) write the correct cosine-rule form, (iv) calculate. Round lengths to 1 dp and angles to the nearest 0.1°.

1.1 — Sailing leg at Pittwater. A yacht sails from a marker for 4.5 km on one course, then turns through an external angle of 65° and sails 3.2 km. The internal angle of the triangle (between the two legs) is therefore 180° − 65° = 115°. How far is the yacht from the original marker in a straight line?    3 marks

Stuck? SAS → c² = a² + b² − 2ab cos C with C = 115°. Note cos 115° is negative.

1.2 — Triangular paddock at Dubbo. A farmer surveys a paddock whose three sides measure 240 m, 310 m, and 420 m. He needs the angle at the corner where the two shorter sides (240 m and 310 m) meet, so he can install a gate. Find that angle.    3 marks

Stuck? SSS → cos C = (a² + b² − c²) / (2ab) where C is the angle between the 240 m and 310 m sides, and c = 420 m is opposite it.

1.3 — Roof rafter on a Brisbane Queenslander. A symmetric A-frame roof has two rafters of length 5.4 m meeting at an apex angle of 70°. Find the width of the roof at the wall plate (i.e. the base of the triangle).    3 marks

Stuck? SAS with a = b = 5.4 m and the included apex angle = 70°.

1.4 — Bush-walking shortcut at Katoomba. Three lookouts P, Q, R form a triangle on a topographic map. PQ = 1.2 km, QR = 0.9 km, PR = 1.6 km. A ranger needs the angle at Q (the corner of the path she will turn through if she walks P → Q → R). Find ∠PQR.    3 marks

Stuck? ∠PQR sits between sides QP and QR. The side opposite it is PR = 1.6 km.

1.5 — Cricket ground diagonal. A boundary marker at the SCG is 65 m from the centre wicket and 78 m from a corner camera tower. The angle between these two sight-lines (at the marker) is 105°. How far apart are the centre wicket and the camera tower?    3 marks

Stuck? SAS with the 105° angle between the two known sides.

2. Explain your thinking

This question is about communication, not just numbers. Use full sentences. 4 marks

2.1 A classmate says: "If you have two sides and any angle, just use the cosine rule." Explain in 4-6 sentences (i) why this is wrong when the angle is not between the two known sides, (ii) which rule should be used instead in that case, and (iii) what specific pattern of information (SAS / SSS / SSA / AAS) actually triggers the cosine rule from Lesson 16. Reference the words "included angle" somewhere in your answer.

Stuck? Revisit lesson § "Choosing Between Sine and Cosine" — the cosine rule needs an included angle (SAS) or all three sides (SSS).

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Sailing leg (a = 4.5, b = 3.2, C = 115°)

SAS. c² = 4.5² + 3.2² − 2(4.5)(3.2) cos 115° = 20.25 + 10.24 − 28.8 × (−0.4226) = 30.49 + 12.17 = 42.66.
c = √42.66 ≈ 6.5 km. The obtuse angle makes c larger than √(4.5² + 3.2²) ≈ 5.5 km.

1.2 — Dubbo paddock (sides 240, 310, 420)

SSS. cos C = (240² + 310² − 420²) / (2 × 240 × 310) = (57 600 + 96 100 − 176 400) / 148 800 = −22 700 / 148 800 ≈ −0.1526.
C = cos⁻¹(−0.1526) ≈ 98.8°. Obtuse (as expected — 420 m is the longest side, so it faces the largest angle, and 240² + 310² < 420², so that angle exceeds 90°).

1.3 — Roof rafter (a = b = 5.4, apex C = 70°)

SAS. c² = 5.4² + 5.4² − 2(5.4)(5.4) cos 70° = 29.16 + 29.16 − 58.32 × 0.3420 = 58.32 − 19.95 = 38.37.
c = √38.37 ≈ 6.2 m. (The roof is about 6.2 m wide at the wall plate.)

1.4 — Katoomba lookouts (∠PQR)

SSS. At Q, the two sides are QP = 1.2 km and QR = 0.9 km, opposite side PR = 1.6 km.
cos Q = (1.2² + 0.9² − 1.6²) / (2 × 1.2 × 0.9) = (1.44 + 0.81 − 2.56) / 2.16 = −0.31 / 2.16 ≈ −0.1435.
∠PQR = cos⁻¹(−0.1435) ≈ 98.3°.

1.5 — SCG sight-lines (a = 65, b = 78, C = 105°)

SAS. c² = 65² + 78² − 2(65)(78) cos 105° = 4 225 + 6 084 − 10 140 × (−0.2588) = 10 309 + 2 624 = 12 933.
c = √12 933 ≈ 113.7 m. (Obtuse-angle case, so the opposite side is longer than √(65² + 78²) ≈ 101.5 m.)

2.1 — Explain your thinking (sample response)

The claim is not always correct. The cosine rule from Lesson 16 needs the angle to be the included angle — the one sitting between the two known sides. If the angle is opposite one of the known sides instead (the SSA case), the cosine rule cannot be applied directly because the formula c² = a² + b² − 2ab cos C expects C to be between a and b. In that situation the correct tool is the sine rule, which may give an ambiguous case (two possible triangles). So the patterns that actually trigger the cosine rule from Lesson 16 are: SAS (two sides and the included angle → find the third side) and SSS (all three sides → find any angle).

Marking: 1 for spotting the SSA exception, 1 for naming "included angle", 1 for naming the sine rule as the correct alternative, 1 for stating both SAS and SSS as the valid triggers.