Mathematics • Year 9 • Unit 3 • Lesson 20
Trigonometry in the Real World
Apply the full Year 9 trig toolkit — Pythagoras, SOH-CAH-TOA, inverse trig, elevation/depression, bearings, multi-step — to authentic problems: a wind turbine, drone delivery, lighthouse spotting, surveyor tour, and skate-park ramp design. Then explain your tool-selection strategy.
1. Word problems
Each problem uses one or more of the six Year 9 trig tools (Pythagoras, sin/cos/tan, inverse trig, elevation/depression, bearings, multi-step). Show your working — a single final answer with no working only earns half marks.
1.1 — Wind turbine. A wind turbine stands on flat ground. From a viewing platform 80 m away, the angle of elevation to the very tip of the blade (highest point) is 38°, and the angle of elevation to the centre of the rotor (hub) is 28°.
(a) Find the height of the rotor hub above the ground (2 d.p.).
(b) Find the height of the tip of the blade above the ground (2 d.p.).
(c) Find the radius of the rotor (blade-tip minus hub height, 2 d.p.). 4 marks
1.2 — Drone delivery. A delivery drone leaves the warehouse and flies 1.5 km on true bearing 070°, then 0.8 km on true bearing 160° to the customer's house.
(a) Find the customer's position relative to the warehouse (km N and km E, 2 d.p.).
(b) Find the straight-line distance from warehouse to customer (2 d.p.).
(c) Find the true bearing on which the drone should fly DIRECTLY back to the warehouse (2 d.p.). 4 marks
1.3 — Lighthouse spotting. From the top of a 60 m lighthouse, the angle of depression to a boat is 25°.
(a) Find the horizontal distance from the lighthouse base to the boat (2 d.p.).
(b) The boat moves directly away from the lighthouse and the angle of depression becomes 15°. Find the new horizontal distance.
(c) How far has the boat moved? (2 d.p.) 3 marks
1.4 — Surveyor tour. A surveyor walks 200 m on true bearing 040° from point S to a benchmark M, then 150 m on true bearing 130° from M to a corner peg C.
(a) Find C's position relative to S (km N and km E, 2 d.p.).
(b) Find the straight-line distance SC (2 d.p.).
(c) Find the true bearing of C from S (2 d.p.). 4 marks
1.5 — Skate-park ramp design. A new launch ramp at the skate park is to have a horizontal run of 4.5 m and a slope of 22° up from the ground.
(a) Find the vertical rise (height) at the top of the ramp (2 d.p.).
(b) Find the length of the sloped surface (the part skaters actually roll on, 2 d.p.).
(c) Safety standards say a launch ramp should not exceed 25° for beginner use. Is this ramp within the limit? Justify in one sentence. 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate looks at a trig word problem and asks, "should I use sin, cos, tan or Pythagoras?" In your own words, describe a SIMPLE THREE-STEP procedure that always works for picking the right tool. (i) Describe the three steps. (ii) Use the procedure to explain which tool you would pick for: "I know an angle and the side opposite it; I want the hypotenuse." (iii) Use the procedure again for: "I know two sides; I want the third side." Refer to "given and wanted" somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Wind turbine (80 m away, blade-tip 38°, hub 28°)
(a) Hub height = 80 tan 28° ≈ 80 × 0.5317 ≈ 42.54 m.
(b) Blade-tip height = 80 tan 38° ≈ 80 × 0.7813 ≈ 62.50 m.
(c) Rotor radius = blade-tip − hub ≈ 62.50 − 42.54 ≈ 19.96 m.
Realistic: most onshore turbines have rotor radii in the 20-50 m range — this is a small-to-mid-size turbine.
1.2 — Drone delivery (1.5 km on 070, 0.8 km on 160)
Leg 1: N1 = 1.5 cos 70 ≈ 0.513; E1 = 1.5 sin 70 ≈ 1.410.
Leg 2: N2 = 0.8 cos 160 ≈ −0.752; E2 = 0.8 sin 160 ≈ 0.274.
(a) Net N ≈ 0.513 − 0.752 ≈ −0.24 km (0.24 km south); Net E ≈ 1.410 + 0.274 ≈ 1.68 km east.
(b) Distance = √(0.24² + 1.68²) ≈ 1.70 km.
(c) End-point in SE quadrant. Angle from S = tan−¹(1.68/0.24) ≈ 81.9°. True bearing warehouse → customer = 180 − 81.9 ≈ 98.1°. Return bearing = 98.1 + 180 = 278.10°.
1.3 — Lighthouse spotting (60 m, 25° then 15°)
(a) d1 = 60 / tan 25° ≈ 60 / 0.4663 ≈ 128.67 m.
(b) d2 = 60 / tan 15° ≈ 60 / 0.2679 ≈ 223.92 m.
(c) Boat moved = d2 − d1 ≈ 223.92 − 128.67 ≈ 95.25 m further from the lighthouse.
1.4 — Surveyor tour (200 m on 040, 150 m on 130)
Leg 1: N1 = 200 cos 40 ≈ 153.21; E1 = 200 sin 40 ≈ 128.56.
Leg 2: N2 = 150 cos 130 ≈ −96.42; E2 = 150 sin 130 ≈ 114.91.
(a) Net N ≈ 56.79 m north; Net E ≈ 243.47 m east.
(b) SC distance = √(56.79² + 243.47²) ≈ √(3225 + 59278) ≈ 250.00 m.
(c) NE quadrant, angle from N = tan−¹(243.47 / 56.79) ≈ 76.87°. True bearing of C from S ≈ 076.87°.
1.5 — Skate-park ramp (run 4.5 m, slope 22°)
(a) Vertical rise = 4.5 tan 22° ≈ 4.5 × 0.4040 ≈ 1.82 m.
(b) Sloped surface (hypotenuse) = 4.5 / cos 22° ≈ 4.5 / 0.9272 ≈ 4.85 m.
(c) Yes — the ramp slope is 22°, which is less than the 25° beginner limit, so the ramp meets the safety standard.
2.1 — Three-step tool-selection procedure (sample response)
The procedure: (1) Sketch the situation with a labelled right triangle. (2) Identify GIVEN and WANTED — write down every length/angle you know and what you're solving for. (3) Pick the tool by matching the given-wanted pair to one of: two sides + want third side → Pythagoras; angle + one side + want another side → sin/cos/tan; two sides + want angle → inverse trig.
Applied to "I know an angle and the side opposite it; I want the hypotenuse": given = angle + opposite, wanted = hypotenuse. The pair opp+hyp matches sin (SOH), so I'd write sin θ = opp / hyp and solve hyp = opp / sin θ.
Applied to "I know two sides; I want the third side": given = two sides, wanted = third side. That matches Pythagoras directly: c² = a² + b² (or a² = c² − b² if c is the hypotenuse).
Marking: 1 mark for stating the three-step procedure; 1 for correctly applying it to the sin-hypotenuse problem; 1 for correctly applying it to the Pythagoras problem; 1 for explicit "given and wanted" language and overall clarity.