Mathematics • Year 9 • Unit 4 • Lesson 4
Elevation & Depression — Mixed Challenge
Pull together the trig from Lessons 1-4: SOH CAH TOA, the Pythagorean identity, solving triangles, and elevation/depression problems with observer height, double-angle setups and reversed viewpoints. You'll spot a mistake and tackle an open-ended challenge.
1. Mixed problems — choose the right approach
Each question uses a different combination of the elevation/depression ideas. Draw the diagram first. Show your working. 3 marks each
1.1 From a 50 m cliff, the angle of depression to a swimmer is 14°. How far is the swimmer from the base of the cliff? (Nearest metre.)
1.2 A 1.6 m tall hiker stands 8 m from a tree and looks up at the top at an elevation of 60°. Find the tree's total height (1 dp).
1.3 From a point on the ground, the angle of elevation to the top of a flagpole is 30°. After walking 10 m DIRECTLY TOWARDS the flagpole, the angle of elevation becomes 60°. Find the flagpole's height. (Hint: write tan 30° and tan 60° in terms of the unknown height and unknown original distance d.)
1.4 A drone is 80 m above a beach. It sees two surfers in the water at depressions of 30° and 45° (both in the same direction). How far apart are the two surfers?
1.5 A 25 m vertical antenna is fixed on the roof of a building that is 40 m tall. From a point on the ground (in line with the antenna), the angle of elevation to the BASE of the antenna (top of building) is 35°. Find the angle of elevation to the TOP of the antenna, to 1 dp.
1.6 A 100 m cliff. A boat is observed at depression 20°. Then a SECOND boat, 80 m beyond the first (further out), is observed too. Find the depression to the second boat to 1 dp.
2. Find the mistake
Another student has tried to find the building height from the elevation problem: "A 1.7 m person stands 20 m from a building. The angle of elevation to the top is 38°. Find the building's total height." Their working is below. Exactly one line contains a mistake. Spot it, explain why, then re-do the working correctly. 3 marks
Student's working:
Line 1: Set up: tan 38° = h / 20 (h = height above eye level).
Line 2: h = 20 × tan 38° = 20 × 0.7813 ≈ 15.6 m.
Line 3: So building height above the person's eyes ≈ 15.6 m.
Line 4: Building total height = 15.6 m.
Line 5: Final answer: 15.6 m.
(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 corrected final answer.
Stuck? The 15.6 m is only the height ABOVE THE PERSON'S EYES (above 1.7 m). What's the height from the GROUND?3. Open-ended challenge — design a problem
This question has more than one valid answer. 4 marks
3.1 Design two different word problems based on real Australian scenarios — one using angle of elevation and one using angle of depression — where the final answer to each is exactly 50 m (height or distance).
For each problem you design:
(i) Write the word problem (2-3 sentences setting up a realistic scenario).
(ii) Show the diagram with all measurements and the angle labelled.
(iii) Show the trig setup that makes the answer come out to 50 m.
(iv) Solve the problem to verify the answer is 50 m (to within 0.5 m).
Bonus: One problem must have the answer be a HEIGHT, the other must have the answer be a HORIZONTAL DISTANCE.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 50 m cliff, depression 14°
tan 14° = 50 / d → d = 50 / tan 14° ≈ 50 / 0.2493 ≈ 201 m.
1.2 — Hiker 1.6 m, tree 8 m away, elevation 60°
Height above eye level: 8 × tan 60° = 8 × 1.7321 ≈ 13.86 m.
Total tree height = 13.86 + 1.6 ≈ 15.5 m.
1.3 — Flagpole, elevation 30° → 60° after walking 10 m closer
Let h = flagpole height, d = distance from second observation point to flagpole.
tan 60° = h / d → h = d × √3.
tan 30° = h / (d + 10) → h = (d + 10) × (1/√3) = (d + 10) / √3.
Set equal: d × √3 = (d + 10) / √3 → 3d = d + 10 → 2d = 10 → d = 5.
So h = 5 × √3 ≈ 8.66 m.
1.4 — Drone 80 m up, depressions 30° and 45°
Distance to surfer at 30°: 80 / tan 30° = 80 / (1/√3) = 80√3 ≈ 138.6 m.
Distance to surfer at 45°: 80 / tan 45° = 80 / 1 = 80 m.
Distance apart = 138.6 − 80 ≈ 58.6 m.
1.5 — 40 m building + 25 m antenna, elevation to base = 35°
Horizontal distance from observer: tan 35° = 40 / d → d = 40 / tan 35° ≈ 40 / 0.7002 ≈ 57.13 m.
Total height to top of antenna = 40 + 25 = 65 m.
Elevation to top: tan θ = 65 / 57.13 ≈ 1.1378 → θ = tan⁻¹(1.1378) ≈ 48.7°.
1.6 — 100 m cliff, first boat at depression 20°, second boat 80 m further
First boat distance: d₁ = 100 / tan 20° = 100 / 0.3640 ≈ 274.7 m.
Second boat distance: d₂ = 274.7 + 80 = 354.7 m.
Depression to second boat: tan φ = 100 / 354.7 ≈ 0.2819 → φ = tan⁻¹(0.2819) ≈ 15.7°.
Sanity check: further boat → shallower angle. 15.7° < 20°. ✓
2 — Find the mistake
(a) The mistake is on Line 4.
(b) The student forgot to ADD the observer's eye height of 1.7 m. The 15.6 m is only the height of the building ABOVE the person's eyes, not above the ground.
(c) Corrected working:
tan 38° = h / 20 → h = 20 × tan 38° ≈ 15.6 m (above eye level).
Total building height = 15.6 + 1.7 = 17.3 m (above the ground).
This is the exact pitfall flagged in card 3 of the lesson: "forget to add the observer's height".
3 — Open-ended challenge (sample solutions)
Many valid answers. Here are two:
Problem A — elevation, answer is a HEIGHT: "A surveyor stands 50 m from the base of a flagpole at Parramatta High School. She measures the angle of elevation to the top as 45°. How tall is the flagpole? (Ignore her eye height.)" Setup: tan 45° = h / 50, h = 50 × 1 = 50 m. ✓ Answer = 50 m.
Problem B — depression, answer is a HORIZONTAL DISTANCE: "From the top of a 50 m cliff at Bondi, a lifeguard sees a swimmer at a depression of exactly 45°. How far is the swimmer from the base of the cliff?" Setup: tan 45° = 50 / d, d = 50 / 1 = 50 m. ✓ Answer = 50 m.
Other valid approaches: use 30°/60° with adjusted other dimension. e.g. "From 50√3 ≈ 86.6 m away, elevation 30°: h = 86.6 × tan 30° = 50 m" or "from 50/√3 ≈ 28.87 m away, elevation 60°: h = 28.87 × tan 60° = 50 m".
Marking: 2 marks per problem (1 for a realistic scenario with diagram, 1 for correct setup and answer of 50 m ± 0.5). Up to 4 in total. Bonus that one is height and one is distance.