Mathematics • Year 9 • Unit 4 • Lesson 4
Angles of Elevation and Depression
Build fluency reading "elevation" / "depression" out of word problems, drawing a labelled triangle, recognising that elevation = depression (alternate angles), and using tan to find the unknown height or distance. One step at a time.
1. I do — fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. Standing 25 m from the base of a tower on flat ground, you look up at the top of the tower at an angle of elevation of 32°. How tall is the tower? (Ignore your own height.)
Step 1 — Draw and label the right triangle.
Observer at one corner, base of tower 25 m horizontally away, top of tower directly above the base. The 32° angle of elevation is at the observer, measured UP from horizontal.
Reason: a clear diagram is half the work in elevation/depression problems.
Step 2 — Identify the sides relative to the 32° angle.
Opposite = tower height = h (unknown).
Adjacent = horizontal distance = 25 m.
Reason: vertical thing = opposite to the elevation angle; horizontal distance = adjacent.
Step 3 — Pick tan (opposite + adjacent → TOA).
tan 32° = h / 25
Reason: elevation/depression problems almost always use tan because you usually know the horizontal distance and want the vertical height (or vice versa).
Step 4 — Rearrange to make h the subject and evaluate.
h = 25 × tan 32° = 25 × 0.6249 ≈ 15.62 m
Reason: multiply both sides by 25 to free h.
Step 5 — Round and state the answer in context.
h ≈ 15.62 m → tower is about 15.6 m tall (1 dp).
Reason: always say what the number means in the real situation.
Answer: the tower is about 15.6 m tall.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. From a 60 m cliff, the angle of depression to a small boat at sea is 18°. How far is the boat from the base of the cliff?
Step 1 — Draw the triangle. Observer at top of cliff. Cliff is vertical, 60 m. Boat is horizontally out to sea, at unknown distance d. The 18° angle of depression is at the observer (looking DOWN from horizontal).
Step 2 — Use the equal-angle rule. Angle of depression from the cliff top = angle of ____________ from the boat = ____° (they are ____________ angles between parallel horizontal lines).
Step 3 — Set up tan from the boat's viewpoint. The cliff (60 m) is opposite the 18° angle; distance d is adjacent.
tan 18° = ____ / ____
Step 4 — Rearrange to make d the subject.
d = ____ / tan ____° = 60 / ____________ ≈ ____________ m
Step 5 — Round to the nearest metre and state the answer:
d ≈ ____________ m → the boat is about ____________ m from the cliff.
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (basic elevation OR depression). The middle two are standard (add observer height or pick the right side). The last two are extension (two-step or reasoning).
Foundation — basic elevation/depression
3.1 Standing 40 m from a tree on flat ground, the angle of elevation to the top is 28°. Find the tree height to 1 dp. (Ignore observer's height.) 1 mark
3.2 From the top of a 25 m lighthouse, the angle of depression to a swimmer is 30°. Find the swimmer's distance from the base of the lighthouse to 1 dp. 1 mark
3.3 A flagpole casts an 8 m shadow when the sun is at an elevation of 50°. Find the flagpole height to 1 dp. 1 mark
3.4 A 15 m building. From a point on the ground 10 m from its base, what is the angle of elevation to the top? Round to the nearest degree. 1 mark
Standard — combine ideas
3.5 A person 1.7 m tall stands 12 m from a building. The angle of elevation from their eyes to the top is 40°. Find the building's total height. 2 marks
3.6 A plane is flying at altitude 2000 m. The pilot looks down at a depression of 4° towards an airport runway. Find the horizontal distance from the plane to the runway, to the nearest 100 m. 2 marks
Extension — reasoning + multi-step
3.7 Explain in your own words WHY the angle of elevation from the ground to the top of a tower is equal to the angle of depression from the top of the tower to the same point on the ground. Use the words "horizontal", "parallel" and "alternate angles" somewhere in your answer. 2 marks
3.8 From a lookout on a 200 m cliff, the angle of depression to a boat is 12°. The boat sails directly towards the cliff and the new angle of depression becomes 25°. How far did the boat sail? Round to the nearest metre. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (60 m cliff, depression 18°)
Step 2: angle of elevation from the boat = 18° (they are alternate angles between parallel horizontal lines).
Step 3: tan 18° = 60 / d.
Step 4: d = 60 / tan 18° = 60 / 0.3249 ≈ 184.6 m.
Step 5: d ≈ 185 m → the boat is about 185 m from the cliff.
3.1 — Tree, 40 m away, elevation 28°
tan 28° = h / 40 → h = 40 × tan 28° ≈ 40 × 0.5317 ≈ 21.3 m.
3.2 — Lighthouse 25 m, depression 30°
tan 30° = 25 / d → d = 25 / tan 30° = 25 / 0.5774 ≈ 43.3 m.
3.3 — Flagpole shadow 8 m, sun elevation 50°
tan 50° = h / 8 → h = 8 × tan 50° ≈ 8 × 1.1918 ≈ 9.5 m.
3.4 — 15 m building, 10 m away
tan θ = 15 / 10 = 1.5 → θ = tan⁻¹(1.5) ≈ 56°.
3.5 — Person 1.7 m, building 12 m away, elevation 40°
Height above eye level: tan 40° = h / 12 → h = 12 × tan 40° ≈ 12 × 0.8391 ≈ 10.07 m.
Total building height = 10.07 + 1.7 ≈ 11.77 m.
Don't forget to add the observer's height!
3.6 — Plane at 2000 m, depression 4°
tan 4° = 2000 / d → d = 2000 / tan 4° ≈ 2000 / 0.0699 ≈ 28,600 m → 28,600 m (28.6 km).
A 4° depression is very shallow, so the horizontal distance is huge compared with altitude.
3.7 — Why elevation = depression
The ground and the horizontal line at the top of the tower are both horizontal, so they are parallel to each other. The line of sight between the observer at the top of the tower and the point on the ground acts as a transversal cutting both parallel lines. The angle of elevation (at the ground, looking up) and the angle of depression (at the top, looking down) are on opposite sides of this transversal and between the two parallel lines — so they are alternate angles, which are always equal.
3.8 — Boat sails towards cliff (200 m), depression 12° → 25°
First distance (depression 12°): tan 12° = 200 / d₁ → d₁ = 200 / tan 12° ≈ 200 / 0.2126 ≈ 940.7 m.
Second distance (depression 25°): tan 25° = 200 / d₂ → d₂ = 200 / tan 25° ≈ 200 / 0.4663 ≈ 428.9 m.
Distance sailed = d₁ − d₂ ≈ 940.7 − 428.9 ≈ 512 m.