IB Maths AA SL Topic 3 — Geometry & Trig Paper 1 & 2 ~9 min read

Angles of Elevation & Depression

Real-world trigonometry questions love elevation and depression problems — towers, planes, ships, lighthouses. The maths is just SOH CAH TOA, but most students lose marks on the setup, not the calculation. Get the diagram right and the answer almost falls out.

📘 What you need to know

Angle of elevation: measured upward from the horizontal to a point above you.
Angle of depression: measured downward from the horizontal to a point below you.
Both are always measured from the horizontal — never from the vertical.
The angle of elevation from A up to B equals the angle of depression from B down to A (alternate angles).
Once you’ve drawn the right-angled triangle, just apply SOH CAH TOA like normal.

Elevation vs depression

The two terms mean the obvious thing: looking up vs looking down. The “angle” is always measured from the imaginary horizontal line at your eye level — the angle between that horizontal and your line of sight.

📡 ⬆

ANGLE OF ELEVATION

Looking up — the angle between your horizontal and the line of sight going up.

⬇ 📡

ANGLE OF DEPRESSION

Looking down — the angle between your horizontal and the line of sight going down.

Both are measured from horizontal — not from the ground, not from a wall, not from the vertical. Always from the imaginary horizontal line at the observer’s eye level.

The “they’re equal” trick

Here’s a fact that simplifies a lot of two-observer problems: the angle of elevation looking up from A to B always equals the angle of depression looking down from B to A.

Elevation = Depression (alternate angles)

🤔 Why are they equal? — alternate interior angles

Both observers’ horizontal lines are parallel (they’re both horizontal). The line of sight between them is a transversal cutting both parallels. Geometry tells us that alternate interior angles formed this way are equal — that’s exactly what’s happening with the elevation and depression. So the two angles must be the same value.

If a question gives you the angle of depression but you’d rather work from the lower observer’s perspective, just relabel it as the angle of elevation — same number, easier to picture sometimes.

The 5-step method (works every time)

SKETCH

Draw a clear picture of the situation

HORIZONTAL

Draw a dashed horizontal line at the observer

ANGLE

Mark the angle from horizontal to line of sight

TRIANGLE

Identify the right-angled triangle and label sides

SOLVE

Apply SOH CAH TOA — pick S, C, or T

🧠 The most common setups

For an observer on the ground looking up at a tower: the height of the tower is opposite, the horizontal distance is adjacent, and the line of sight is the hypotenuse. Use tan when you have height and distance, since they’re the opposite and adjacent.

Worked examples

WE 1

Elevation — find the height of a building

From a point on horizontal ground, the angle of elevation to the top of a building is 32°. The point is 60 m from the base of the building. Find the height of the building to 3 s.f.

Step 1: Sketch and label Right-angled triangle: angle 32°, adjacent = 60 m (horizontal), opposite = height (unknown). Step 2: Pick the trig ratio Have A and want O → use tan (TOA). Step 3: Set up and solve tan 32° = height60 height = 60 × tan 32° = 60 × 0.6249… Height ≈ 37.5 m (3 s.f.) always sketch — even a rough triangle saves time!

WE 2

Elevation — find the distance to the base

A 25 m flagpole stands on horizontal ground. The angle of elevation from a person to the top of the flagpole is 18°. How far is the person from the base of the flagpole? Give your answer to 3 s.f.

Step 1: Sketch and label Angle 18°, opposite = 25 m (height of flagpole), adjacent = distance (unknown). Step 2: Pick the ratio Have O and want A → use tan. Step 3: Set up and rearrange tan 18° = 25distance distance = 25tan 18° = 250.3249… Distance ≈ 76.9 m (3 s.f.) when the unknown is on the bottom, divide instead of multiply

WE 3

Depression — find the distance to a ship

From the top of a 80 m cliff, the angle of depression to a ship at sea is 22°. Find the horizontal distance from the base of the cliff to the ship, to 3 s.f.

Step 1: Sketch — depression at top, equal angle of elevation at the ship The angle of depression from the cliff = angle of elevation from the ship = 22°. Step 2: Build the right-angled triangle (from ship up to top of cliff) Opposite = 80 m (cliff height), adjacent = horizontal distance (unknown). Step 3: Use tan tan 22° = 80distance distance = 80tan 22° = 800.4040… Distance ≈ 198 m (3 s.f.) depression from above = elevation from below — use whichever is easier

WE 4

Find the angle of elevation

A tree is 12 m tall. A bird is sitting on top, looking at a worm 18 m horizontally away on the ground. Find the angle of depression from the bird to the worm, to 3 s.f.

Step 1: Sketch and use the alternate-angles trick Depression from bird = elevation from worm. Opposite = 12 m, adjacent = 18 m. Step 2: Use tan (we have O and A) tan θ = 1218 = 23 Step 3: Apply inverse tan θ = tan⁻¹(23) = tan⁻¹(0.6667…) θ ≈ 33.7° (3 s.f.) finding angles → always inverse trig (tan⁻¹ here)

WE 5

Two-position problem (multi-step)

From a point P on horizontal ground, the angle of elevation to the top of a tower is 30°. After walking 50 m closer to the tower, the angle of elevation is now 45°. Find the height of the tower to 3 s.f.

Step 1: Set up — let h = height, d = distance from base at the second point From close point: tan 45° = h/d → h = d (since tan 45° = 1) Step 2: From the original point, distance = d + 50 tan 30° = hd + 50 Step 3: Substitute h = d tan 30° = dd + 50 0.5774… × (d + 50) = d 0.5774d + 28.87 = d 28.87 = d − 0.5774d = 0.4226d d = 28.87 / 0.4226 = 68.30… Height ≈ 68.3 m (3 s.f.) two angles, two equations — always set up before solving!

💡 Top tips

Always sketch the situation first. Even a rough drawing forces you to identify the horizontal, the line of sight, and the angle correctly.
Mark the horizontal as a dashed line. The angle is always between this and the line of sight — never anywhere else.
Use the equal-angles trick to switch between elevation and depression. Whichever is easier to picture, use that.
Tan is your friend. Most elevation/depression problems involve a height and a horizontal distance — that’s opposite and adjacent, so tan applies.
If the unknown is on the bottom of the trig ratio (e.g. tan θ = 25/d), rearrange so it’s on top: d = 25/tan θ.
For two-step problems (observer moves), set up two equations with two unknowns and solve simultaneously.
Calculator must be in degrees mode for SL elevation/depression problems unless the question explicitly uses radians.

⚠ Common mistakes

Measuring the angle from the wrong place. The angle is always between the horizontal and the line of sight — not between the vertical and the line of sight.
Treating the angle of depression as if it’s between the vertical wall and the line of sight. Easy mistake when the observer is on top of a cliff — always draw the horizontal!
Forgetting that elevation = depression for the same line of sight. Missing this trick makes some problems unnecessarily hard.
Calculator in the wrong mode. RAD vs DEG matters — sin 30° in the wrong mode gives the wrong answer.
Wrong trig ratio. Most of these problems use tan (height + horizontal distance), but if the line of sight is given, you might need sin or cos instead.
Confusing observer height with tower height. If the observer is standing tall, sometimes you need to add their eye-level height to the calculated value.
Misreading “at sea level” or “on horizontal ground” — these are clues telling you where the horizontal sits.

Elevation and depression problems are the bridge between abstract trig and real-world maths. Master the diagram setup and you’ll find these among the easiest marks in any IB paper.

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