Mathematics • Year 9 • Unit 3 • Lesson 10

Sin, Cos or Tan? Five Real-World Decisions

Each problem deliberately mixes the side-pair so you have to PICK the right ratio for each context: lighthouse line-of-sight, paragliding wing, train track gradient, surveying a paddock and reverse engineering a 3-4-5 ladder.

Apply · Real-World Maths

1. Word problems

Each part requires you to FIRST identify the two sides involved (opp/adj/hyp) and choose the right ratio. Show working. Calculator DEG mode. Round to 2 d.p. unless told.

1.1 — Lighthouse line-of-sight. Standing on a beach, you look up at the top of a 30 m lighthouse at an angle of 12° above horizontal. The line of sight from your eyes to the top of the lighthouse is the hypotenuse.

(a) Side-pair? Name the ratio. Find your horizontal distance from the lighthouse base (adj).
(b) Side-pair? Name the ratio. Find the length of the line of sight (hyp).    3 marks

Stuck on (a)? You know opp = 30, want adj — that's opp+adj → tan. For (b), opp + hyp → sin.

1.2 — Paragliding wing. A paraglider's wing line is taut and 12 m long; it makes a 70° angle with the horizontal ground below.

(a) Side-pair? Find the pilot's height above the ground (opp).
(b) Side-pair? Find the horizontal distance from the pilot to the point directly below the wing's top anchor (adj).    3 marks

Stuck on (a)? Known hyp, want opp → sin. For (b), known hyp, want adj → cos.

1.3 — Train track gradient. The Sydney–Newcastle rail line includes a section that rises at 2.1° over a track length of 850 m (track measured along the slope, i.e. the hypotenuse).

(a) Side-pair? Find the vertical rise (opp).
(b) Side-pair? Find the horizontal distance covered (adj).    3 marks

Stuck? hyp + θ given. opp = hyp × sin θ; adj = hyp × cos θ.

1.4 — Surveying a triangular paddock. A farmer wants to fence a triangular paddock. He measures the longest side (a creek bank, hyp) as 180 m, and finds that the creek bank meets the western fence at 25°.

(a) Side-pair? Find the length of the western fence (adj).
(b) Side-pair? Find the length of the southern fence (opp).    3 marks

Stuck? Both parts use hyp = 180; one wants adj (cos), the other opp (sin).

1.5 — 3-4-5 ladder problem. A 5 m ladder is set against a wall so that its foot is exactly 3 m from the wall.

(a) Without using a calculator, find the angle θ the ladder makes with the ground in degrees. (Hint: this is a 3-4-5 triangle. cos θ = adj/hyp = 3/5 = 0.6. Use the lesson's table or your calculator's cos⁻¹ to find θ to the nearest degree.)
(b) Side-pair? Find the height the ladder reaches up the wall (the opp).    3 marks

Stuck on (b)? In a 3-4-5 triangle the third side is 4. So height = 4 m exactly — no calculator needed. Or use opp = hyp × sin θ.

2. Explain your thinking

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

2.1 Your classmate sees a trig problem and starts by writing "I'll use tan because I memorised that first." In your own words, explain (i) why this is the wrong strategy, (ii) what should ALWAYS happen first in a trig side problem, (iii) which lesson concept (a single phrase from L10) makes the choice automatic. Refer to "the two sides involved" and "decision flowchart" somewhere in your explanation.

Stuck? Revisit lesson § "The Three-Question Test" — angle, known side, wanted side, in that order.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Lighthouse

(a) Side-pair opp + adj → tan. adj = 30 / tan 12° ≈ 30 / 0.2126 ≈ 141.12 m.
(b) Side-pair opp + hyp → sin. hyp = 30 / sin 12° ≈ 30 / 0.2079 ≈ 144.30 m.
Cross-check: hyp² ≈ adj² + opp² → 144.30² ≈ 141.12² + 30² = 19,914.65 + 900 ≈ 20,815 ≈ 20,822 ✓.

1.2 — Paraglider

(a) Side-pair opp + hyp → sin. opp = 12 sin 70° ≈ 12 × 0.9397 ≈ 11.28 m above ground.
(b) Side-pair adj + hyp → cos. adj = 12 cos 70° ≈ 12 × 0.3420 ≈ 4.10 m horizontally.

1.3 — Train track 850 m at 2.1°

(a) Side-pair opp + hyp → sin. Rise = 850 sin 2.1° ≈ 850 × 0.0366 ≈ 31.16 m.
(b) Side-pair adj + hyp → cos. Horizontal = 850 cos 2.1° ≈ 850 × 0.9993 ≈ 849.43 m.
For a shallow 2.1° gradient, horizontal is almost the full track length.

1.4 — Paddock

(a) Side-pair adj + hyp → cos. Western fence = 180 cos 25° ≈ 180 × 0.9063 ≈ 163.14 m.
(b) Side-pair opp + hyp → sin. Southern fence = 180 sin 25° ≈ 180 × 0.4226 ≈ 76.07 m.
Total fencing for the triangle = 180 + 163.14 + 76.07 ≈ 419.21 m — useful for ordering materials.

1.5 — 3-4-5 ladder

(a) cos θ = 3/5 = 0.6, so θ = cos⁻¹(0.6) ≈ 53°.
(b) Side-pair opp + hyp → sin. height = 5 sin 53° ≈ 5 × 0.7986 ≈ 3.99 ≈ 4 m — matches the 4-side of the 3-4-5 triangle exactly (the tiny error is rounding).
This problem is the classic "is my ladder set up safely?" calculation builders do every day.

2.1 — Explain your thinking (sample response)

My classmate's strategy is wrong because picking tan automatically (or any single ratio by habit) only works for one third of all problems — the third where the side-pair happens to be opp + adj. For the other two thirds, tan gives the wrong setup. What should ALWAYS happen first is to identify the two sides involved in the problem: one is the side you know, the other is the side you want. Once you have that pair, the choice is automatic via the L10 decision flowchart: opp + hyp → sin, adj + hyp → cos, opp + adj → tan. That single phrase — "the two sides involved" — turns every Year 9 trig side problem into the same three-step routine (list pair, pick ratio, rearrange and compute) and removes guesswork entirely.

Marking: 1 mark for explicitly saying "habit gives the wrong setup most of the time"; 1 mark for naming "the two sides involved" as the first step; 1 mark for citing the decision flowchart with all three rules; 1 mark for a clear full-sentence explanation.

HSCScience • Mathematics • Year 9 • Unit 3 • Lesson 10 • Worksheet 2 (Apply)