Mathematics • Year 9 • Unit 4 • Lesson 17

Trigonometry in the Real World

Apply SOH CAH TOA, inverse trig and the elevation/depression checklist to authentic Y9 contexts: surveying, construction, drone flight, sailing and bushwalking. Then explain how to verify a trig calculation in your own words.

Apply · Real-World Maths

1. Word problems

For each: sketch a right triangle, label sides relative to the given angle, identify known/wanted, pick the ratio, solve. Don't forget observer height where needed.

1.1 — Surveying a lighthouse. A boat is 100 m from the base of a lighthouse. From the deck (5 m above sea level), the angle of elevation to the top of the lighthouse is 15°.

(a) Find the height of the lighthouse above the deck.
(b) Find the total height of the lighthouse above sea level.
(c) State which trig ratio you used and why. 4 marks

Stuck on (a)? You know the adjacent (100 m) and the angle, you want the opposite — that's tan.

1.2 — Building a wheelchair ramp. Australian standards say a wheelchair ramp must not exceed an inclination of 4.8° (1:12 ratio). A builder makes a ramp that rises 0.45 m over a horizontal distance of 5.0 m.

(a) Find the angle of inclination to 1 d.p.
(b) Does the ramp meet the Australian standard? Justify with numbers.
(c) If not, what is the shortest horizontal length that would meet the standard for a 0.45 m rise? 4 marks

Stuck on (a)? tan θ = rise ÷ run, then use tan⁻¹.

1.3 — Drone search-and-rescue. A rescue drone is hovering at 80 m altitude. It spots a hiker on the ground at an angle of depression of 25°.

(a) Find the horizontal distance from a point directly below the drone to the hiker.
(b) Find the slant (line-of-sight) distance from the drone to the hiker.
(c) The drone has a top speed of 12 m/s. How long would it take to fly directly to the hiker along the slant? 4 marks

Stuck on (a)? Angle of depression = angle of elevation at the hiker (alternate angles). Then use tan with opposite = 80 m.

1.4 — Bushwalking pole height. A bushwalker (eye-level 1.6 m) stands 25 m from the base of a power pole. She measures the angle of elevation to the top of the pole as 42°.

(a) Find the height of the pole above eye-level.
(b) Find the total height of the pole above the ground.
(c) The pole is supposed to be 25 m tall. Within what percentage is your answer? 4 marks

Stuck on (b)? Add the 1.6 m eye-level — see lesson § "Elevation and Depression Checklist" Step 6.

1.5 — Sailing bearings. A sailing boat leaves harbour and sails 12 km on a bearing — the boat's path makes a 35° angle with the north-south line.

(a) Find how far east of the harbour the boat is (the side opposite the 35°).
(b) Find how far north of the harbour the boat is (the side adjacent to the 35°).
(c) Check your answers with Pythagoras: east² + north² should be close to 12². 4 marks

Stuck? Both legs in a right-triangle bearing problem are 12 × sin 35° and 12 × cos 35° — the hypotenuse is the actual distance sailed.

2. Explain your thinking

Use full sentences. 4 marks

2.1 A classmate calculates that a 25 m ladder leaning against a wall reaches 30 m up the wall. Without doing any calculations of your own, explain in your own words:

(i) Why this answer must be wrong (refer to the role of the hypotenuse).
(ii) Which "reasonableness check" from the lesson's Checking Answers card it fails.
(iii) What you would tell the classmate to do next (one specific suggestion using a particular trig ratio).

Stuck? Revisit lesson § "Checking Answers" — "Can you verify with a different method?"

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Lighthouse

(a) Height above deck = 100 × tan 15° ≈ 26.8 m.
(b) Total above sea = 26.8 + 5 = 31.8 m.
(c) Used tan because we knew the adjacent (100 m) and angle (15°), and wanted the opposite (height).

1.2 — Wheelchair ramp

(a) tan θ = 0.45 ÷ 5.0 = 0.09 → θ = tan⁻¹(0.09) ≈ 5.1°.
(b) Fails — 5.1° is above the 4.8° maximum.
(c) Shortest horizontal length: tan 4.8° = 0.45 ÷ L → L = 0.45 ÷ tan 4.8° ≈ 0.45 ÷ 0.084 ≈ 5.36 m. So the ramp needs to be at least 5.4 m long horizontally.

1.3 — Rescue drone

(a) Horizontal distance d: tan 25° = 80 ÷ d → d = 80 ÷ tan 25° ≈ 171.6 m.
(b) Slant distance = 80 ÷ sin 25° ≈ 189.3 m. (Or check with Pythagoras: √(80² + 171.6²) ≈ 189.3.)
(c) Time = 189.3 ÷ 12 ≈ 15.8 seconds.

1.4 — Bushwalking pole

(a) Height above eye-level = 25 × tan 42° ≈ 22.5 m.
(b) Total pole height = 22.5 + 1.6 = 24.1 m.
(c) Difference from 25 m: |25 − 24.1| ÷ 25 = 0.9 ÷ 25 = 0.036 = 3.6% — within about 4%, very reasonable for a real-world field measurement.

1.5 — Sailing bearings

(a) East displacement = 12 × sin 35° ≈ 6.88 km.
(b) North displacement = 12 × cos 35° ≈ 9.83 km.
(c) Check: 6.88² + 9.83² ≈ 47.3 + 96.6 = 143.9, and 12² = 144. Match (small rounding). ✓

2.1 — Explain your thinking (sample response)

(i) The classmate's answer of 30 m is impossible because the height the ladder reaches up the wall is the opposite side of the right triangle, and the hypotenuse (the 25 m ladder itself) is always the longest side. You cannot reach higher than the length of the ladder. (ii) This fails the lesson's "Is the answer physically reasonable?" check, and specifically the rule that "if you calculate a side longer than the hypotenuse, recheck." (iii) I'd tell them to redo the calculation using sine (sin θ = opposite ÷ hypotenuse): height = 25 × sin(angle), where the angle should be the one the ladder makes with the ground. That will always give a number smaller than 25.

Marking: 1 for the hypotenuse rule; 1 for the correct check; 1 for naming sine; 1 for clear, full-sentence writing.