Mathematics • Year 10 • Unit 3 • Lesson 4
Elevation & Depression — Mixed Challenge
Pull together every idea from Lesson 4: angle definitions, diagram conventions, alternate-angles transfer, observer-height adjustments, and two-step "walk closer" problems. Spot the elevation-vs-depression labelling error, then design two depression scenarios that share the same horizontal distance.
1. Mixed problems
Each question pulls a different idea from Lesson 4. Sketch first, then choose the ratio, then calculate. Round lengths to 1 d.p. and angles to the nearest degree unless told otherwise. 2-4 marks each
1.1 From a point 60 m from a vertical pole, the angle of elevation to the top of the pole is 36°. Find the height of the pole. 2 marks
1.2 A helicopter is 350 m above the ground. The pilot sees a car at an angle of depression of 28°. Find the horizontal distance from the helicopter (directly above a point on the ground) to the car. 2 marks
1.3 From the top of a 90 m cliff, a fishing boat at sea is seen at an angle of depression of 14°. Find the horizontal distance from the foot of the cliff to the boat. 2 marks
1.4 A 20 m-tall lamp post is 30 m horizontally from a parked car. Find the angle of depression from the top of the lamp post to the car. 2 marks
1.5 A photographer 1.7 m tall stands 18 m from the base of a statue. The angle of elevation from her eyes to the top of the statue is 28°. Find the total height of the statue. 3 marks
1.6 From the top of a lighthouse, two boats on the same straight line out at sea are seen at angles of depression 35° and 18° respectively. The lighthouse is 50 m tall and the boats are on the same side. Find the distance between the two boats. 4 marks
2. Find the mistake
A Year 10 student attempts the following: "From a 30 m lookout, the angle of depression to a kayaker is 22°. Find the horizontal distance from the foot of the lookout to the kayaker." Their working is shown. Exactly one line contains a mistake. Spot it, explain why, and redo correctly. 3 marks
Student's working:
Line 1: Lookout height = 30 m (vertical = O at the top of the lookout).
Line 2: Angle of depression = 22°, which is inside the triangle at the top of the lookout.
Line 3: tan 22° = 30 / d → d = 30 / tan 22° = 30 / 0.4040 = 74.25 m.
Line 4: So the kayaker is about 74.3 m from the foot of the lookout.
(a) Which line contains the mistake?
(b) Explain in one or two sentences what is wrong with that line and why it does not change the final answer in this case (but would mislead in many others).
(c) Write the corrected line.
Stuck? The 22° angle of depression sits between the horizontal at the top of the lookout and the line of sight. It is outside the triangle. The angle inside the triangle at the top of the lookout (between the vertical lookout and the line of sight) is 90° − 22° = 68°. But the angle of elevation at the kayaker is also 22° (alternate angles), which is what makes the answer here come out the same.3. Open-ended challenge — design two depression scenarios
This question has many valid answers. Be creative but show every number. 4 marks
3.1 Design two different elevation/depression scenarios in which the horizontal ground distance from observer to object is exactly 100 m, but the heights and angles differ.
For each scenario:
(i) Describe the context in one sentence (e.g. "A drone observer looks at a parked car…").
(ii) State the angle (elevation or depression).
(iii) Calculate the matching vertical height to 1 d.p.
(iv) Verify by working back: tan (angle) should equal (vertical / 100).
Bonus: Choose angles that are not standard (no 30°, 45°, 60°). Make one elevation and one depression.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Pole height
tan 36° = h / 60 → h = 60 × tan 36° = 60 × 0.7265 = 43.59 ≈ 43.6 m.
1.2 — Helicopter to car
tan 28° = 350 / d → d = 350 / tan 28° = 350 / 0.5317 = 658.31 ≈ 658.3 m.
1.3 — Cliff to fishing boat
tan 14° = 90 / d → d = 90 / tan 14° = 90 / 0.2493 = 361.04 ≈ 361.0 m.
1.4 — Lamp post to car
tan θ = 20 / 30 = 0.6667. θ = tan⁻¹(0.6667) = 33.69° ≈ 34°.
1.5 — Statue total height
Height above eyes: tan 28° = h / 18 → h = 18 × tan 28° = 18 × 0.5317 = 9.57 m.
Total statue height = 9.57 + 1.7 = 11.3 m (to 1 d.p.).
1.6 — Two boats from a lighthouse
Nearer boat (35° depression): tan 35° = 50 / d₁ → d₁ = 50 / tan 35° = 50 / 0.7002 = 71.41 m.
Further boat (18° depression): tan 18° = 50 / d₂ → d₂ = 50 / tan 18° = 50 / 0.3249 = 153.89 m.
Distance between boats = 153.89 − 71.41 = 82.5 m (to 1 d.p.).
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The angle of depression is measured from the horizontal at the observer, downward. It sits outside the triangle — between the horizontal at the top of the lookout and the line of sight. The angle inside the triangle at the top of the lookout (between the vertical lookout and the line of sight) is the complement, i.e. 90° − 22° = 68°. The final answer is still 74.3 m because, by alternate angles between parallel horizontals, the angle of elevation at the kayaker equals the angle of depression at the observer (both 22°) — and 22° is inside the triangle at the kayaker's vertex, which is what tan 22° = 30 / d uses.
(c) Corrected Line 2: "The angle of depression sits outside the triangle at the top of the lookout. By alternate angles, the angle of elevation at the kayaker is also 22°, and this is the angle inside the triangle at the kayaker."
This is exactly the "where is the angle, really?" issue flagged in Lesson 4.
3 — Open-ended (sample solutions)
Scenario A — Elevation, 23°: A photographer 100 m from the base of a small tower measures an angle of elevation of 23° to the top. Vertical height = 100 × tan 23° = 100 × 0.4245 = 42.5 m. Verify: tan 23° = 42.5 / 100 = 0.425. ✓
Scenario B — Depression, 51°: A drone hovering above a parking lot observes a car directly 100 m horizontally away at an angle of depression of 51°. Vertical height of drone = 100 × tan 51° = 100 × 1.2349 = 123.5 m. Verify: tan 51° = 123.5 / 100 = 1.235. ✓
Marking: 2 marks per scenario (1 for clear context + angle, 1 for correct height + verification). Any pair satisfying the constraints is acceptable.