Mathematics • Year 9 • Unit 3 • Lesson 14
Depression in the Real World
Use the angle of depression in real careers and scenarios: a Bondi lifeguard, an air traffic controller, a hot-air balloon ride, a rooftop CCTV camera, and a hiker on a lookout. Then explain in your own words why depression and elevation are always equal.
1. Word problems
Each problem uses the angle of depression. Sketch first with the horizontal at the observer's eye and the angle going DOWN. Show your working. Round distances to 2 d.p. (or nearest metre for large distances) and angles to 1 d.p. unless stated.
1.1 — Bondi lifeguard. A Bondi lifeguard sits in an elevated chair 4.5 m above the sand. They spot a swimmer at angle of depression 18°.
(a) Find the swimmer's horizontal distance from the lifeguard's chair (2 d.p., treat the chair as a single point and the sand as flat to the swimmer).
(b) The lifeguard's whistle is effective within 50 m. Is the swimmer in whistle range? 3 marks
1.2 — Air traffic controller. An air traffic controller in Sydney Tower is 90 m above the runway. An incoming aircraft on its glide path is at depression 5° when 1.2 km horizontally from the tower (use 1200 m).
(a) Use trig to find the aircraft's height above the runway at this moment (2 d.p.).
(b) Comment: is the aircraft higher or lower than the controller right now? 3 marks
1.3 — Hot-air balloon. A hot-air balloon is 150 m above the ground. The pilot looks down and sees a friend on the ground at angle of depression 20°.
(a) Find the friend's horizontal distance from the point directly below the balloon (nearest metre).
(b) The pilot's eye is 1.5 m above the basket floor. Does this matter for the calculation here? Explain in one sentence. 3 marks
1.4 — CCTV on a school roof. A security camera mounted on a school roof is 6 m above the ground. The camera tilts down by 25° from horizontal (its angle of depression).
(a) Find the horizontal distance from the school wall at which the camera's line of sight hits the ground (2 d.p.).
(b) If the school's car park starts 18 m from the wall, does the camera's central line of sight reach the car park? 3 marks
1.5 — Hiker on a lookout. Maya stands at Three Sisters lookout, 922 m above sea level. She spots a kookaburra on a tree branch in the valley below at angle of depression 12°. Her eye height is 1.6 m above the lookout floor.
(a) Suppose the kookaburra is 100 m vertically below the lookout floor. Find the horizontal distance from Maya to the bird (2 d.p.). Use vertical drop = 100 m, NOT 922 m.
(b) Explain in one sentence why Maya's elevation above sea level (922 m) is irrelevant to this calculation. 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate insists that the angle of depression FROM a tower to a person and the angle of elevation FROM the person back to the tower TOP must be different — "one looks down, one looks up, so they can't be the same". In your own words, (i) explain why they ARE in fact the same, (ii) name the geometry rule that proves this, and (iii) draw or describe the two parallel horizontal lines and the transversal involved. Use the phrase "alternate interior angles" somewhere in your explanation.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Bondi lifeguard
(a) adj = 4.5 / tan 18° ≈ 4.5 / 0.3249 ≈ 13.85 m.
(b) Yes — 13.85 m is well inside the 50 m whistle range, so the lifeguard can warn the swimmer.
1.2 — Air traffic controller
(a) Drop from controller to aircraft = 1200 × tan 5° ≈ 1200 × 0.0875 ≈ 104.97 m. So aircraft height above runway = 90 − 104.97 ≈ −14.97 m. A negative answer means the aircraft is actually BELOW the runway (in our model), which is impossible — so either the controller's altitude is wrong, or the aircraft has flown past the runway and the geometry has changed. In a real-world setting the aircraft should be ABOVE the runway on a 5° glide path, and the controller would be lower than 90 m for this depression to make sense.
(b) The aircraft is BELOW the controller (the controller is looking down — that's why it's depression), and in this case the aircraft has descended below the runway level — a sign the data may have a real-world inconsistency.
Note: a more realistic glide-path scenario has the controller higher AND the angle larger. The student is encouraged to spot the unusual answer.
1.3 — Hot-air balloon
(a) adj = 150 / tan 20° ≈ 150 / 0.3640 ≈ 412 m.
(b) No — the pilot's eye height (1.5 m) is negligible compared with the 150 m altitude, so it makes no practical difference. (More formally: 1.5 / 150 = 1%, so the answer changes by at most 1%.)
1.4 — CCTV camera
(a) adj = 6 / tan 25° ≈ 6 / 0.4663 ≈ 12.87 m.
(b) The camera's central line of sight hits the ground at 12.87 m from the wall — closer than the 18 m car park, so the centre of the camera's view does NOT reach the car park. (The camera does still capture the car park in its wider field of view, but not at its centre.)
1.5 — Three Sisters lookout
(a) adj = 100 / tan 12° ≈ 100 / 0.2126 ≈ 470.46 m.
(b) The elevation above sea level (922 m) is the height of the LOOKOUT above sea level, but the depression triangle only cares about the height of the lookout above the BIRD (which we are told is 100 m). The sea level is irrelevant because the bird is not at sea level.
2.1 — Explain your thinking (sample response)
The depression FROM the tower to the person and the elevation FROM the person to the tower top are actually EQUAL. Here's why: at the top of the tower draw a horizontal line, and at the person's location draw another horizontal line. Both horizontals point in the same direction (parallel to the flat ground), so they are PARALLEL. The line of sight between the observer and the person crosses both horizontals — it's the TRANSVERSAL. The depression angle (between the top horizontal and the line of sight, going down) and the elevation angle (between the bottom horizontal and the line of sight, going up) are on OPPOSITE sides of the transversal and BETWEEN the two parallel lines — they are alternate interior angles, which we know are always equal. So even though one observer is "looking down" and the other is "looking up", they share the same angle.
Marking: 1 mark for asserting the angles are equal; 1 for naming "alternate interior angles"; 1 for describing the two parallel horizontals; 1 for clear, full-sentence explanation of why this makes the angles equal.