Mathematics • Year 9 • Unit 4 • Lesson 4
Elevation and Depression — Real World
Use angles of elevation and depression in real Aussie contexts: lifeguard tower, plane approaching Sydney Airport, drone above a beach, observation deck on a cliff, and a hot-air balloon over Canberra. Then explain your reasoning.
1. Word problems
Each problem uses an angle of elevation or depression. Draw a quick diagram with horizontal lines at the observer and the object, mark the angle, label the right triangle. Show your working. Calculator in degree mode. Round to 1 dp unless told otherwise.
1.1 — Lifeguard tower at Bondi. A lifeguard sits on the observation deck of a tower 6 m above the sand. They spot a swimmer in trouble at a depression angle of 8°.
(a) Draw the diagram (observer at top, swimmer on the sand) and label the 6 m height, depression 8°, and unknown distance d.
(b) Find the horizontal distance d from the base of the tower to the swimmer. 3 marks
1.2 — Plane approaching Sydney Airport. A plane is at an altitude of 1500 m, descending towards the runway. The pilot sees the start of the runway at a depression of 3°.
(a) Find the horizontal distance to the runway, to the nearest 100 m.
(b) Approximately how many KILOMETRES is this? 3 marks
1.3 — Drone above the beach. A drone hovers directly above the high-tide line. Its operator, 35 m inland on the dry sand, looks up at the drone at an angle of elevation of 42°.
(a) How high above the beach is the drone? (Ignore the operator's eye height.)
(b) What is the straight-line distance from the operator's eyes to the drone (the hypotenuse)? 3 marks
1.4 — Observation deck on a cliff. From the observation deck of a 120 m cliff in the Royal National Park, the angle of depression to a yacht is 15°. The yacht then sails directly towards the cliff. After a while, the angle of depression has increased to 35°.
(a) Find the yacht's first distance (at 15°) and second distance (at 35°), to the nearest metre.
(b) How far did the yacht sail? 3 marks
1.5 — Hot-air balloon over Canberra. A hot-air balloon is rising vertically above a launch point. A spectator standing 200 m from the launch point first sees the balloon at an elevation of 22°. Two minutes later the elevation has increased to 48° (the balloon is now higher).
(a) Find the balloon's first height (at 22°) and second height (at 48°), to the nearest metre.
(b) Find the balloon's rate of ascent in metres per minute. 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate says: "Angles of depression must be different from angles of elevation, because one looks down and the other looks up." In your own words, explain (i) WHY the two angles are actually EQUAL in size (even though they're measured from different starting points), (ii) what geometry rule lets you see this on a diagram, and (iii) how this fact lets you swap viewpoints in a problem (e.g. the boat sees the cliff at the same angle the cliff sees the boat). Refer to "alternate angles" and "parallel horizontal lines" somewhere.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Lifeguard tower at Bondi
Tower height = 6 m (opposite the 8° angle from the swimmer's viewpoint). Distance d = adjacent.
tan 8° = 6 / d → d = 6 / tan 8° = 6 / 0.1405 ≈ 42.7 m.
The shallow angle (8°) means the swimmer is much further away than the tower is tall.
1.2 — Plane approaching Sydney Airport
(a) tan 3° = 1500 / d → d = 1500 / tan 3° = 1500 / 0.0524 ≈ 28,600 m.
(b) About 28.6 km from the runway.
This is why planes start their final descent quite far from the airport — 3° is the standard ILS glide-slope.
1.3 — Drone above the beach
(a) tan 42° = h / 35 → h = 35 × tan 42° = 35 × 0.9004 ≈ 31.5 m.
(b) Hypotenuse = 35 / cos 42° = 35 / 0.7431 ≈ 47.1 m (or √(31.5² + 35²) ≈ √(992 + 1225) ≈ √2217 ≈ 47.1 m). ✓
1.4 — Observation deck on a cliff
(a) First distance (15°): d₁ = 120 / tan 15° = 120 / 0.2679 ≈ 448 m.
Second distance (35°): d₂ = 120 / tan 35° = 120 / 0.7002 ≈ 171 m.
(b) Distance sailed = d₁ − d₂ ≈ 448 − 171 = 277 m.
The angle of depression got STEEPER (15° → 35°) as the yacht got closer.
1.5 — Hot-air balloon over Canberra
(a) First height (22°): h₁ = 200 × tan 22° = 200 × 0.4040 ≈ 81 m.
Second height (48°): h₂ = 200 × tan 48° = 200 × 1.1106 ≈ 222 m.
(b) Rise = 222 − 81 = 141 m in 2 minutes → 70.5 m/min ascent rate.
A typical hot-air balloon ascent rate is around 60-100 m/min, so this is realistic.
2.1 — Explain your thinking (sample response)
The angles of elevation and depression are actually EQUAL because of geometry, not because of which way someone is looking. When the observer at the top of a tower looks down and a person on the ground looks up, both lines of sight are along the SAME straight line — that line acts as a transversal cutting two parallel horizontal lines (one through the observer's eyes at the top, one through the ground person's eyes at the bottom). The angle of depression (at the top, between the horizontal and the line going down) and the angle of elevation (at the bottom, between the horizontal and the line going up) are alternate angles formed by the transversal, and alternate angles between parallel lines are always equal. This means you can freely swap viewpoints in a problem — the boat sees the cliff at the same angle the cliff sees the boat — which often makes the right triangle easier to set up.
Marking: 1 mark for naming "alternate angles"; 1 for "parallel horizontal lines"; 1 for the transversal idea; 1 for stating the practical consequence (swap viewpoints).