Mathematics • Year 9 • Unit 3 • Lesson 14
Angles of Depression
Build fluency with the angle of depression — the angle measured from the horizontal at your eye DOWN to an object below. Use the alternate-angles rule (depression = elevation from below) and walk from one worked example through one guided fill-in to eight independent practice problems.
1. I do — fully worked example
Read every step. The standard pattern for depression problems: sketch with a horizontal line at the observer's eye, mark the angle going DOWN, label sides, choose ratio.
Problem. A lifeguard on a 12 m tower sees a swimmer at angle of depression 28°. How far is the swimmer from the base of the tower (2 d.p.)?
Step 1 — Sketch the right triangle.
Draw the horizontal at the lifeguard's eye. Mark the 28° angle going DOWN from that horizontal. Draw the vertical drop of 12 m to the water; the horizontal distance to the swimmer is the unknown.
Reason: depression is measured at the observer's eye from the horizontal, going DOWN.
Step 2 — Use the alternate-angles rule to re-label (if it helps).
By alternate interior angles, the swimmer would see the lifeguard at an elevation of 28°. So we can place the 28° angle at the SWIMMER instead. From the swimmer's view: opp = 12 m (lifeguard height above water), adj = ? (horizontal distance).
Reason: depression = elevation by alternate angles, because the horizontal at the eye and horizontal at the swimmer are parallel.
Step 3 — Pick the ratio: opp + adj → tan.
tan 28° = 12 / adj
Reason: TOA.
Step 4 — Rearrange and solve.
adj = 12 / tan 28° ≈ 12 / 0.5317 ≈ 22.57 m
Reason: divide both sides by tan 28° because the unknown is in the denominator.
Step 5 — State answer with units.
Swimmer is ≈ 22.57 m from the base of the tower.
Answer: ≈ 22.57 m.
2. We do — fill in the missing steps
Same structure as Section 1, but the working is faded. Fill in each blank. 4 marks
Problem. An aircraft at 1500 m altitude sights an airport at depression 8°. Find the horizontal distance to the airport (to the nearest metre).
Step 1 — Sketch: draw the horizontal at the pilot's eye and a depression angle of ________ ° going down. The aircraft is ________ m above the airport (vertical drop) and the horizontal distance is the unknown.
Step 2 — Label: opp = ________ m, θ = ________ °, adj = ?
Step 3 — Ratio: opp + adj → ________ .
Step 4 — Set up and solve:
tan 8° = ________ / adj
adj = ________ / tan 8° ≈ ________ / ________ ≈ ________ m
Step 5 — State the answer: Aircraft is ≈ ________ m from the airport horizontally.
3. You do — independent practice
Show working under each problem. Problems 3.1–3.4 are foundation (find distance given drop and angle). Problems 3.5–3.6 are standard (find the angle of depression). Problems 3.7–3.8 are extension (vertical drop or alternate-angles application).
Foundation — find horizontal distance
3.1 From a 10 m tower, the angle of depression to a person is 40°. Find their horizontal distance from the base (2 d.p.). 1 mark
3.2 From a 50 m cliff, the depression to a kayaker is 35°. Find the kayaker's horizontal distance from the cliff base (2 d.p.). 1 mark
3.3 A balcony 8 m above the street has a depression angle to a dog of 45°. Find the dog's distance from the building base. 1 mark
3.4 A drone 100 m above ground sees a target at depression 30°. Find the horizontal distance (2 d.p.). 1 mark
Standard — find the angle of depression
3.5 A tower is 15 m tall and a swimmer is 25 m from the base. Find the angle of depression of the swimmer from the top (1 d.p.). 2 marks
3.6 A ship is 200 m from the base of a 50 m cliff. Find the angle of depression of the ship from the cliff top (1 d.p.). 2 marks
Extension — find the drop, or use alternate angles
3.7 From the top of a tower, the depression to a car at horizontal distance 60 m is 22°. Find the height of the tower (2 d.p.). 2 marks
3.8 A pilot sees an airport at depression 12°. The airport is 800 m below the pilot's altitude. (a) Find the horizontal distance to the airport (nearest metre). (b) Without re-calculating, state the angle of elevation of the pilot's aircraft as seen from the airport. Explain why. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (aircraft, 1500 m, depression 8°)
Step 1: depression angle 8°, aircraft 1500 m above the airport.
Step 2: opp = 1500 m, θ = 8°.
Step 3: opp + adj → tan.
Step 4: tan 8° = 1500 / adj; adj = 1500 / tan 8° ≈ 1500 / 0.1405 ≈ 10676 m (nearest metre).
Step 5: aircraft is ≈ 10676 m from the airport horizontally.
3.1 — 10 m tower, 40°
adj = 10 / tan 40° ≈ 10 / 0.8391 ≈ 11.92 m.
3.2 — 50 m cliff, 35°
adj = 50 / tan 35° ≈ 50 / 0.7002 ≈ 71.41 m.
3.3 — 8 m balcony, 45°
adj = 8 / tan 45° = 8 / 1 = 8 m. (At 45° drop equals distance — clean check.)
3.4 — 100 m drone, 30°
adj = 100 / tan 30° ≈ 100 / 0.5774 ≈ 173.21 m.
3.5 — 15 m tower, swimmer 25 m
tan θ = 15 / 25 = 0.6. θ = tan⁻¹(0.6) ≈ 31.0°.
3.6 — 200 m, 50 m cliff
tan θ = 50 / 200 = 0.25. θ = tan⁻¹(0.25) ≈ 14.0°.
3.7 — Tower top, car at 60 m, depression 22°
opp = 60 × tan 22° ≈ 60 × 0.4040 ≈ 24.24 m — the tower is about 24.24 m tall.
3.8 — Aircraft sights airport at 12°, 800 m drop
(a) adj = 800 / tan 12° ≈ 800 / 0.2126 ≈ 3763 m horizontal distance.
(b) Angle of elevation from the airport = 12° — same as the depression, because the horizontals at the aircraft and at the airport are parallel and the line of sight is a transversal, making the two angles ALTERNATE INTERIOR ANGLES (which are equal).