Mathematics • Year 10 • Unit 3 • Lesson 18

Mixed Trig Problems — Real Australian Scenarios

Apply the full Lesson 18 toolkit (sine rule, cosine rule, area formula, Heron's) to multi-step real problems — three towns on a map, a survey from two observation points, a paddock with diagonal track, a bridge cable triangle, and a Bondi yacht course. The skill is picking the right tool at each step.

Apply · Real-World Maths

1. Word problems

For each problem: (i) draw and label the triangle, (ii) name the pattern (AAS / SAS / SSS), (iii) write each formula you apply, (iv) calculate, (v) state units. Round lengths to 1 dp and angles to 0.1°.

1.1 — Three towns on a map. Towns A, B, C on a regional NSW map have AB = 25 km, AC = 30 km, and ∠BAC = 60°.
(a) Find the distance BC. (Which pattern is this?)
(b) Find ∠ABC.
(c) Find the area of the triangular region ABC in km². 4 marks

Stuck on 1.1(a)? SAS → cosine rule. (b) Once you have BC, sine rule works.

1.2 — Surveyor from two observation points. Two surveyors stand 200 m apart on a flat field. The first measures the angle to a distant flagpole as 35°, and the second measures the angle to the same flagpole as 55°.
(a) What pattern of data is this?
(b) Find the third interior angle of the triangle.
(c) Find the distance from the first surveyor to the flagpole. 4 marks

Stuck? Two angles and the side between them = ASA. The third angle is 180° − 35° − 55°. Then use sine rule.

1.3 — Paddock with a diagonal track. A triangular paddock has all three sides measured: 80 m, 110 m, and 140 m. A farmer wants to know:
(a) The largest interior angle.
(b) The total area of the paddock (use Heron's). 4 marks

Stuck? SSS — cosine rule for the angle opposite the longest side, Heron's for the area.

1.4 — Bridge cable triangle. A diagonal cable runs from a pylon at point P down to a deck-anchor at Q. A second cable runs from P to a different deck-anchor at R. PQ = 18 m, PR = 22 m, and ∠QPR = 40°.
(a) Find the distance QR.
(b) Find the area of the triangle PQR in m². 3 marks

Stuck? SAS for QR; same SAS gives area = ½ × 18 × 22 × sin 40°.

1.5 — Bondi yacht course. A yacht sails from a start buoy on bearing 080° for 30 km, then turns and sails on bearing 150° for 40 km, reaching a finish buoy.
(a) The interior angle of the triangle at the turn is 180° − (150° − 80°) = ____° (fill in). Use it.
(b) Find the straight-line distance from start to finish.
(c) Find the area of the triangle formed by start, turn, finish (in km²). 4 marks

Stuck? Bearing change = 150° − 80° = 70° (external). Interior angle of triangle = 180° − 70° = 110°. Then SAS.

2. Explain your thinking

This question is about strategy. Use full sentences. 4 marks

2.1 A friend says: "If I always start with the sine rule, I will always get the answer eventually." Explain in 4-6 sentences (i) which specific pattern of data makes the sine rule impossible as a first step, (ii) what the correct first formula is in that case, (iii) what your follow-up formula would be once that first step is done, and (iv) the practical disadvantage of "always starting with the sine rule" even when it would work. Reference the words "SAS" and "SSS" in your answer.

Stuck? Revisit lesson § "Common Pitfalls" — the sine rule cannot start a pure SAS or pure SSS problem.

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Answers — Do not peek before attempting

1.1 — Towns A, B, C (AB = 25, AC = 30, ∠A = 60°)

(a) SAS → cosine rule. BC² = 25² + 30² − 2(25)(30) cos 60° = 625 + 900 − 1500 × 0.5 = 1525 − 750 = 775. BC = √775 ≈ 27.8 km.
(b) Sine rule: sin B / 30 = sin 60° / 27.8. sin B = 30 × 0.866 / 27.8 ≈ 0.935. B = sin⁻¹(0.935) ≈ 69.2°.
(c) Area = ½ × 25 × 30 × sin 60° = 375 × 0.866 ≈ 324.8 km².

1.2 — Two surveyors (200 m apart, angles 35°, 55°)

(a) ASA — two angles and the side between them.
(b) Third angle = 180° − 35° − 55° = 90°.
(c) Sine rule: distance₁ / sin 55° = 200 / sin 90°. distance₁ = 200 × 0.819 / 1 ≈ 163.8 m. (The flagpole is opposite the 55° angle measured from the second surveyor.)

1.3 — Paddock (80, 110, 140)

(a) Largest angle is opposite 140. cos C = (80² + 110² − 140²)/(2 × 80 × 110) = (6 400 + 12 100 − 19 600)/17 600 = −1 100/17 600 ≈ −0.0625. C ≈ 93.6°.
(b) s = (80 + 110 + 140)/2 = 165. A = √(165 × 85 × 55 × 25) = √19 284 375 ≈ 4 391.4 m².

1.4 — Bridge cables (PQ = 18, PR = 22, ∠P = 40°)

(a) QR² = 18² + 22² − 2(18)(22) cos 40° = 324 + 484 − 792 × 0.766 = 808 − 606.7 = 201.3. QR = √201.3 ≈ 14.2 m.
(b) Area = ½ × 18 × 22 × sin 40° = 198 × 0.6428 ≈ 127.3 m².

1.5 — Yacht course (30 km on 080°, 40 km on 150°)

(a) Interior angle = 180° − 70° = 110°.
(b) d² = 30² + 40² − 2(30)(40) cos 110° = 900 + 1600 − 2400 × (−0.342) = 2500 + 820.8 = 3320.8. d = √3320.8 ≈ 57.6 km.
(c) Area = ½ × 30 × 40 × sin 110° = 600 × 0.9397 ≈ 563.8 km².

2.1 — Explain your thinking (sample response)

The friend's "always start with the sine rule" rule fails on a pure SAS or pure SSS problem. The sine rule needs at least one matching angle-side pair (an angle A together with the side a opposite it), but in pure SAS you only know two sides and the included angle — there is no opposite-angle/opposite-side pair anywhere yet. In pure SSS you know no angles at all. The correct first formula in both cases is the cosine rule: it gives you a third side (SAS) or a first angle (SSS). Once that first step is done you do have a complete angle-side pair, so the sine rule becomes a perfectly good follow-up tool to find the remaining angles. Even when the sine rule would work, "always start with the sine rule" wastes time — for an area question with SAS, A = ½ ab sin C goes straight to the answer without needing any sine-rule step at all.

Marking: 1 for naming SAS/SSS as the failure case, 1 for explaining the missing angle-side pair, 1 for cosine rule as the correct first step and sine rule as the follow-up, 1 for the practical example (area formula short-cut).