Mathematics • Year 9 • Unit 3 • Lesson 14
Depression — Mixed Challenge
Pull together angle of depression, the depression-equals-elevation identity, inverse trig (Lesson 11), the complementary-angle rule (Lesson 12), and Pythagoras. Choose the right tool per problem, spot a plausible Year 9 mistake, and tackle an open-ended challenge.
1. Mixed problems — choose the right rule
Each question uses a different combination of depression, alternate-angles, Pythagoras and complementary angles. Decide which tool applies before you start. Show your working. 3 marks each
1.1 From a 90 m cliff, the angle of depression to a ship is 17°. (a) Find the ship's horizontal distance from the cliff base (nearest metre). (b) Find the LINE-OF-SIGHT distance from the cliff top to the ship (Pythagoras OR hyp = drop / sin θ).
1.2 A drone hovers at 80 m above the ground. From the drone's camera, a person on the ground is at angle of depression 50°. (a) Find the horizontal distance from the person to the point directly below the drone (2 d.p.). (b) State the angle of elevation of the drone as seen from the person, and explain how you know (no further calculation needed).
1.3 A lighthouse beam shines from a height of 35 m. Observer A is on the ground 100 m from the base. (a) Find the angle of depression of A from the lighthouse top (1 d.p.). (b) A second observer B stands twice as far from the base. Without recomputing, briefly justify whether B's angle of depression will be MORE or LESS than A's, then compute B's angle (1 d.p.) to confirm.
1.4 From the top of a 25 m tower, the depression to a car is 30°. (a) Find the horizontal distance to the car (2 d.p.). (b) The driver gets out and walks 10 m DIRECTLY TOWARD the tower. Find the new angle of depression from the tower top (1 d.p.).
1.5 A pilot at 2 km altitude sees a runway at depression 4°. (a) Find the horizontal distance to the runway (nearest metre). (b) Without re-calculating, state the angle the line of sight makes with the VERTICAL (not the horizontal). Explain.
1.6 From a cliff, two depression angles are observed: 40° to a kayaker and 25° to a boat further out (both on the same straight line out from the cliff base). The cliff is 60 m tall. (a) Find the kayaker's horizontal distance from the cliff base (2 d.p.). (b) Find the boat's horizontal distance (2 d.p.). (c) Find the distance between the kayaker and the boat (2 d.p.).
2. Find the mistake
A Year 9 student is asked: a 20 m tall lifeguard tower has a swimmer 35 m from its base. Find the angle of depression of the swimmer from the lifeguard. Their working is shown. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — depression of swimmer from 20 m tower, 35 m away:
Line 1: Set up: opp = 35 (distance to swimmer), adj = 20 (tower height).
Line 2: tan θ = opp / adj = 35 / 20 = 1.75
Line 3: θ = tan⁻¹(1.75) ≈ 60.3°
Line 4: The angle of depression is about 60.3°.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the correct angle of depression (1 d.p.).
Stuck? Re-draw the depression triangle with the angle at the lifeguard. From the lifeguard, the vertical drop to the water is the OPPOSITE (20 m), and the horizontal distance to the swimmer is the ADJACENT (35 m). The student has swapped them.3. Open-ended challenge — design a depression scenario
This question has more than one valid answer. 4 marks
3.1 Design TWO realistic everyday scenarios, each with a clear angle of depression. For EACH scenario:
(i) Describe the situation in 1–2 sentences (e.g. "a swimming pool lifeguard chair…", "a balcony on the third floor…").
(ii) State the height of the observer above the object, the angle of depression, and what is being asked.
(iii) Solve it — find either a horizontal distance or an angle.
(iv) Sense-check: is your answer realistic for the scenario you described?
Bonus: One scenario must ask for an angle (use inverse trig), and the other must ask for a distance (use forward trig). They must NOT just be the same problem with different numbers.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 90 m cliff, depression 17°
(a) adj = 90 / tan 17° ≈ 90 / 0.3057 ≈ 294 m.
(b) hyp = 90 / sin 17° ≈ 90 / 0.2924 ≈ 307.79 m line of sight. (Check with Pythagoras: √(90² + 294²) ≈ √(8100 + 86436) ≈ √94536 ≈ 307.47 m ✓ — small rounding difference.)
1.2 — Drone at 80 m, depression 50°
(a) adj = 80 / tan 50° ≈ 80 / 1.1918 ≈ 67.13 m.
(b) Elevation from the person = 50° — same value because the depression from the drone and the elevation from the person are ALTERNATE INTERIOR ANGLES on parallel horizontals.
1.3 — Lighthouse, observers A and B
(a) tan θ = 35 / 100 = 0.35. A's depression θ = tan⁻¹(0.35) ≈ 19.3°.
(b) B is FURTHER from the base, so B's depression angle is SMALLER (the shallower the angle, the further the object). B's depression = tan⁻¹(35 / 200) = tan⁻¹(0.175) ≈ 9.9° — confirms it's smaller than A's 19.3°. (Note: doubling the distance does NOT halve the angle — tan is non-linear.)
1.4 — 25 m tower, depression 30°, car moves 10 m closer
(a) Original distance: adj = 25 / tan 30° ≈ 25 / 0.5774 ≈ 43.30 m.
(b) New distance: 43.30 − 10 = 33.30 m. New depression: tan θ_new = 25 / 33.30 ≈ 0.7508. θ_new = tan⁻¹(0.7508) ≈ 36.9°.
1.5 — Pilot at 2 km, depression 4°
(a) Drop = 2000 m. adj = 2000 / tan 4° ≈ 2000 / 0.0699 ≈ 28615 m (≈ 28.6 km horizontally to the runway).
(b) Angle with vertical = 90° − 4° = 86° (the line of sight is mostly horizontal — only 4° below — so it's 86° from the vertical, using the complementary-angle rule).
1.6 — Two boats from a 60 m cliff
(a) Kayaker: adj = 60 / tan 40° ≈ 60 / 0.8391 ≈ 71.50 m.
(b) Boat: adj = 60 / tan 25° ≈ 60 / 0.4663 ≈ 128.67 m.
(c) Gap = 128.67 − 71.50 = 57.17 m between kayaker and boat.
2 — Find the mistake
(a) The mistake is on Line 1 (which makes Lines 2–4 wrong as a consequence).
(b) The student has swapped opp and adj. From the lifeguard's viewpoint (where the depression angle sits), the OPPOSITE side is the VERTICAL drop (20 m, tower height down to water) and the ADJACENT side is the HORIZONTAL distance (35 m, distance out to swimmer). The student has put 35 as opp and 20 as adj — backwards.
(c) Corrected working:
Set up: opp = 20 m (vertical drop), adj = 35 m (horizontal distance).
tan θ = opp / adj = 20 / 35 ≈ 0.5714.
θ = tan⁻¹(0.5714) ≈ 29.7°.
Sense check: 29.7° is much shallower than the original wrong answer of 60.3°, which makes sense for a swimmer well out from the tower.
3 — Open-ended challenge (sample solution)
Scenario 1 (find an angle). "A balcony on the 4th floor of an apartment is 12 m above the street. A delivery van pulls up 18 m from the building wall. Find the angle of depression from a person on the balcony to the van."
Solve: tan θ = 12 / 18 ≈ 0.6667. θ = tan⁻¹(0.6667) ≈ 33.7°.
Sense check: 33.7° is a moderately steep look-down, realistic for a 4-storey building and a kerbside van.
Scenario 2 (find a distance). "A drone hovers at 60 m above a flat sports oval. Its onboard camera detects a soccer player on the ground at angle of depression 22°. Find the player's horizontal distance from the spot directly below the drone."
Solve: adj = 60 / tan 22° ≈ 60 / 0.4040 ≈ 148.51 m.
Sense check: 148 m on a sports oval is realistic (a regulation soccer pitch is about 100 m long).
Marking: 2 marks per scenario (1 for clear realistic setup with numbers, 1 for correct solve + sense check). Up to 4 in total. One angle-finding + one distance-finding earns full marks; two of the same type earns 3/4.