IB Maths AI HL Trigonometry Paper 1 & 2 Elevation & depression ~8 min read

Angles of Elevation & Depression

An angle of elevation looks up from the horizontal; an angle of depression looks down. Both are measured from the horizontal line at the observer’s eye. Once you sketch the right triangle, the rest is SOHCAHTOA — with tan doing most of the work because problems usually give vertical and horizontal distances.

📘 What you need to know

Angle of elevation: angle between the horizontal and the line of sight when looking up at an object.
Angle of depression: angle between the horizontal and the line of sight when looking down at an object.
Alternate-angles rule: the depression angle from a high point to a low point equals the elevation angle from the low point back up to the high point (parallel horizontal lines).
Always sketch the right triangle: vertical (height), horizontal (ground distance), hypotenuse (line of sight). The angle sits at the observer’s eye.
tan is the default ratio: tan(θ) = vertical / horizontal. Use sin or cos only if the hypotenuse is involved.
For two observation points, set up two equations — one per triangle — and eliminate to solve.

Sketching the right triangle

Every elevation/depression problem becomes a right triangle once you draw it correctly. The horizontal line at the observer’s eye is one side of the angle; the line of sight is the other. The vertical wall between observer and object (or between the foot of the object and ground level beneath the observer) is the third side, and it’s perpendicular to the horizontal — that’s where the right angle sits. The angle of elevation at a low observer always equals the angle of depression from the corresponding high observer (alternate angles between parallel horizontals).

Elevation and depression are alternate angles across the two parallel horizontal lines (eye level and top-of-object). The right triangle has vertical h, horizontal d, and the line of sight as the hypotenuse.

Default ratio tan(θ) = oppositeadjacent = verticalhorizontal · sin(θ) = verticalline of sight angle of elevation at low point = angle of depression at high point (alternate angles)

Two-observer and two-step problems

When the question gives two observation points along the same line (often two people looking up at the same object, or one person observing two markers on a single tall structure), each observer generates a separate right triangle. Set up one trig equation per triangle and either eliminate a shared unknown, or solve sequentially. The two-observer setup is the most common multi-step pattern in Paper 2 questions on this topic.

Two-observer setup: if observers P and Q are d apart on level ground, both looking up at a point above the line PQ, then tan(elevation at P) = h/(d + x) and tan(elevation at Q) = h/x, where x is the horizontal from the closer observer to the point directly below the object. Eliminate x to solve for h.

🧭 Recipe — elevation & depression

Sketch with the horizontal at the observer’s eye; show the right triangle clearly.
Label vertical, horizontal, and angle; mark the right angle where vertical meets horizontal.
Use alternate angles to move the angle to whichever vertex is convenient.
Pick the ratio: tan if you have vertical + horizontal; sin/cos if the line of sight (hypotenuse) is involved.
Multi-step: for two observers/markers, write one equation per triangle and combine.

Worked examples

WE 1

Find the height — basic elevation

From a point on the ground 45 m from the base of a vertical tower, the angle of elevation to the top of the tower is 32°. Find the height of the tower, to 3 s.f.

label the right triangle opp = h, adj = 45, angle = 32° use tan tan(32°) = h / 45 h = 45 · tan(32°) = 45 × 0.6249 = 28.119… h ≈ 28.1 m (3 s.f.)

WE 2

Find horizontal distance — depression

A drone hovers 80 m directly above level ground. It observes a person on the ground at an angle of depression of 22°. Find the horizontal distance from the point directly below the drone to the person, to 3 s.f.

alternate angles: elevation from person = 22° right triangle: opp = 80, adj = x, angle = 22° tan(22°) = 80 / x x = 80 / tan(22°) = 80 / 0.4040 = 198.0… x ≈ 198 m (3 s.f.) depression at drone = elevation at person — same angle in the triangle.

WE 3

Find the angle of elevation

A vertical tree of height 14 m casts a shadow of length 9 m on level ground at noon. Find the angle of elevation of the sun, to 1 d.p.

opp = tree height = 14, adj = shadow = 9 tan(θ) = 14 / 9 = 1.5556… inverse tan θ = tan⁻¹(1.5556…) = 57.265° θ ≈ 57.3° (1 d.p.) the shorter the shadow, the steeper the sun — quick sanity check.

WE 4

Two-step — mast with a beacon

A vertical communications mast of total height 60 m stands on level ground. From a point B on the ground, the angle of elevation to the top T of the mast is 25°. A warning beacon at point M sits part-way up the mast, and the angle of elevation from B to M is 12°. Find: (a) the distance BF from B to the foot F of the mast; (b) the height of the beacon M above the ground. Give both answers to 3 s.f.

(a) right triangle BFT tan(25°) = 60 / BF BF = 60 / tan(25°) = 60 / 0.4663 = 128.67… BF ≈ 129 m (3 s.f.) (b) right triangle BFM, same BF tan(12°) = h / BF h = 128.67 × tan(12°) = 128.67 × 0.2126 = 27.35… h ≈ 27.3 m (3 s.f.) use the FULL value of BF (not rounded 129) when computing h.

WE 5

Hot-air balloon — height from depression

A passenger in a hot-air balloon hovering directly above point B observes a landmark L on level ground at an angle of depression of 38°. The horizontal distance from B to L along the ground is 220 m. Find the height of the balloon above the ground, to 3 s.f.

alternate angles: elevation at L = 38° right triangle: opp = h, adj = 220 tan(38°) = h / 220 h = 220 · tan(38°) = 220 × 0.7813 = 171.88… h ≈ 172 m (3 s.f.)

WE 6

Two observers on a runway — find aircraft height

An aircraft is approaching an airport. From a point P on the runway, the angle of elevation to the aircraft is 8°. From a point Q, located 200 m further along the runway towards the aircraft, the angle of elevation is 14°. Assuming the aircraft lies directly above the line PQ, find the height of the aircraft above the runway, to 3 s.f.

let h = aircraft height, x = horizontal from Q to point below two right triangles tan(14°) = h / x → x = h / tan(14°) tan(8°) = h / (x + 200) → x + 200 = h / tan(8°) subtract to eliminate x 200 = h / tan(8°) − h / tan(14°) 200 = h · (cot 8° − cot 14°) 200 = h · (7.1154 − 4.0108) = h · 3.1046 solve h = 200 / 3.1046 = 64.42… h ≈ 64.4 m (3 s.f.) eliminate the shared unknown (x) by subtracting the two equations.

💡 Top tips

Sketch first, formula second — a clearly labelled diagram solves half the problem.
tan is your default: almost every basic elevation/depression problem uses tan(θ) = vertical / horizontal.
Use alternate angles freely: move the angle to wherever the triangle is easiest to label.
Track which side is which: the angle sits at the observer; the “opposite” is the vertical wall between observer and object.
For two-observer setups, write one equation per triangle and eliminate the shared horizontal distance.

⚠ Common mistakes

Confusing elevation and depression — both are measured from the horizontal, not from the vertical.
Measuring from the wrong reference line — depression starts at the horizontal at the high point, NOT at the vertical drop.
Using the wrong ratio — you almost always want tan; sin or cos only if the slant (line of sight) is given or required.
Rounding mid-calculation in two-step problems — carry full calculator accuracy until the final answer.
Forgetting eye-height if the observer’s height is given — usually it’s ignored unless explicitly stated, but read the question carefully.

Next up — Bearings & Constructions. Same triangle techniques, but now applied to north-clockwise compass directions and multi-leg journeys.

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