Always measured from horizontal, never from vertical. If a problem says “the angle the line of sight makes with the vertical is 30°”, the angle of elevation is 60° (90° − 30°).
Worked examples
WE 1Elevation — find the height of a tree
A student stands 30 m from the base of a tall tree on level ground. The angle of elevation from the student’s eye level to the top of the tree is 30°. Find the height of the tree above the student’s eye level.
Sketch: right triangle, right angle at the tree base
horizontal = 30 (adjacent to 30°)
vertical = h (opposite — what we want)
Use TOA: tan 30° = h/30
h = 30 tan(30°)
= 30 × 1/√3 = 30/√3
= 10√3
≈ 17.3 m (3 sf)
height above eye = 10√3 ≈ 17.3 m
tan(30°) = 1/√3 exactly — a “special angle” that gives a clean exact form. If asked for the FULL tree height, add the student’s eye height (typically given separately).
WE 2Depression — find horizontal distance to a yacht
From the top of a 60 m lighthouse, the angle of depression to a yacht at sea is 32°. Find the horizontal distance from the yacht to the foot of the lighthouse.
Sketch: transfer angle to sea level by alternate angles
at yacht, elevation to lighthouse top = 32°
Right triangle: vertical 60, horizontal d
tan(32°) = opp/adj = 60/d
d = 60 / tan(32°)
≈ 60 / 0.6249
≈ 96.0 m (3 sf)
distance ≈ 96.0 m
depression FROM the top = elevation FROM the yacht. The angle drops into the right triangle at sea level — that’s how you use the depression angle without re-drawing.
WE 3Find the angle of elevation
A student stands 18 m from the base of a school building. The top of the building is 24 m above the student’s eye level. Find the angle of elevation from the student to the top of the building.
Right triangle: horizontal 18, vertical 24, find θ
tan θ = opp/adj = 24/18 = 4/3
Apply arctan
θ = tan⁻¹(4/3)
≈ 53.13°
θ ≈ 53.1° (3 sf)
(18, 24, 30) is the 3-4-5 triple scaled by 6 — so the line of sight has exact length 30 m. Spotting that gives an instant Pythagoras check.
WE 4Find the angle of depression
From the top of a 50 m cliff, a coastguard observes a swimmer 120 m horizontally from the base of the cliff. Find the angle of depression from the coastguard to the swimmer.
Transfer to sea level — angle of elevation at swimmer = same
vertical = 50, horizontal = 120
Use TOA
tan θ = 50/120 = 5/12
θ = tan⁻¹(5/12) ≈ 22.62°
depression ≈ 22.6°
(50, 120, 130) = (5, 12, 13) scaled by 10. The line-of-sight distance is exactly 130 m — useful if a later sub-part asks for it.
WE 5Two observation points — find the height
From point A on level ground, the angle of elevation to the top of a tower is 30°. The observer walks 20 m directly towards the tower to point B, where the angle of elevation is now 45°. Find the height h of the tower.
Let d = horizontal distance from B to the tower base
From B: tan(45°) = h/d → d = h (since tan 45° = 1)
From A: horizontal distance = d + 20
tan(30°) = h/(d + 20)
1/√3 = h/(h + 20) (since d = h)
Cross-multiply and solve
h + 20 = h√3
20 = h(√3 − 1)
h = 20 / (√3 − 1) = 10(√3 + 1)
≈ 27.3 m (3 sf)
h = 10(√3 + 1) ≈ 27.3 m
“two observation points” problems set up TWO equations involving the same height. Equate the expressions for h, then solve. The 30°-45° pair gives a clean exact answer.
WE 6Two elevation angles — find the flagpole length
A flagpole stands vertically on top of a building. From a point Q on the ground 50 m horizontally from the base of the building, the angle of elevation to the top of the building is 35° and the angle of elevation to the top of the flagpole is 42°. Find the length of the flagpole.
Building height (lower line of sight, 35°)
tan(35°) = h_b / 50
h_b = 50 tan(35°) ≈ 35.01 m
Total height to top of flagpole (upper line of sight, 42°)
tan(42°) = h_t / 50
h_t = 50 tan(42°) ≈ 45.02 m
Flagpole length = h_t − h_b
flagpole = 45.02 − 35.01 ≈ 10.0 m
flagpole ≈ 10.0 m
two elevations from ONE point give two heights that share the same horizontal. Subtract to get the piece between them — works for towers on hills, flagpoles on buildings, etc.
💡 Top tips
- Always draw a big diagram with dashed horizontals. Mark the angle in the RIGHT place — between the horizontal and the line of sight.
- Use the alternate-angle trick: a depression FROM A equals the elevation FROM B in the other triangle below. This puts the angle into a usable right triangle.
- tan is your friend: tan θ = vertical/horizontal handles 90% of these problems.
- Spot Pythagorean triples: (3, 4, 5), (5, 12, 13), (8, 15, 17). Often hidden inside elevation/depression questions for clean answers.
- For “walk closer to get a new angle” questions, set up TWO equations in the same height (one per observation point) and solve simultaneously.
⚠ Common mistakes
- Measuring the angle from the vertical: elevation and depression are ALWAYS from the horizontal. From-vertical = 90° − from-horizontal.
- Putting the angle of depression INSIDE the wrong triangle: it sits at the observer’s eye level, not at sea level. Transfer it down using alternate angles.
- Forgetting the observer’s eye height: if a question gives the height of the observer (e.g. “the boy’s eyes are 1.5 m above the ground”), add it to the calculated vertical to get the full height of the object.
- Wrong setup of tan: tan θ = opp/adj, not adj/opp. Opp is opposite the angle (vertical for elevation/depression), adj is adjacent (horizontal).
- GDC in radian mode: AI SL works in degrees. Set DEG before any tan / tan&supminus;¹ calculation.
Up next: Bearings & Constructions. Bearings are a directional language — always measured CLOCKWISE from NORTH, written as three digits. Combined with the sine and cosine rules, they let you find distances and directions for navigation problems — ships, planes, surveys, and similar.
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