IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~6 min read

Angles of Elevation & Depression

Look up at something — that’s an angle of elevation. Look down — that’s an angle of depression. Both are measured from the horizontal. Once you draw the right triangle, it’s just SOH CAH TOA — usually tan.

📘 What you need to know

Elevation vs depression

Elevation
looking UP ↗
angle from horizontal up to the object
Depression
looking DOWN ↘
angle from horizontal down to the object
The two are equal: if you’re on a cliff looking down at a boat with depression 30°, the boat looking up at you sees an elevation of 30°. Same triangle, same angle — measured between parallel horizontals.

🧭 Recipe — every elevation/depression problem

  1. Draw a sketch. Mark the horizontal, the line of sight, and the angle between them.
  2. Identify the right triangle. The horizontal distance and the vertical height form the two legs.
  3. Label H, O, A relative to the angle.
  4. Pick the trig ratio — usually tan, since you usually have/want height (O) and ground distance (A).
  5. Solve — multiply or divide to find a length, or use tan−1 to find an angle.

Worked examples

WE 1

Find a height from an angle of elevation

A person stands 30 m from the foot of a building. The angle of elevation from the person to the top of the building is 42°. Find the height of the building, correct to 3 s.f.

Step 1: Sketch — right triangle with 42° at ground level A = 30 m (horizontal); O = h (height); angle = 42° Step 2: Use tan (have A, want O) tan 42° = h/30 h = 30 × tan 42° h = 30 × 0.9004… = 27.013… height ≈ 27.0 m (3 s.f.)
WE 2

Find a distance from an angle of depression

From the top of a tower 50 m high, the angle of depression to a boat at sea is 28°. Find the horizontal distance from the boat to the foot of the tower, correct to 3 s.f.

Step 1: Sketch — depression 28° at top equals elevation 28° at boat O = 50 m (height); A = d (distance); angle = 28° Step 2: Use tan tan 28° = 50/d d = 50/tan 28° d = 50/0.5317… = 94.05… d ≈ 94.1 m (3 s.f.) depression at the top = elevation at the bottom (alternate angles)
WE 3

Find an angle of elevation

A bird is 25 m above the ground. A person stands 40 m from the point on the ground directly below the bird. Find the angle of elevation from the person to the bird, correct to 3 s.f.

Step 1: Right triangle — height 25, base 40 O = 25; A = 40 Step 2: Use inverse tan tan θ = 25/40 = 0.625 θ = tan⁻¹(0.625) = 32.005…° θ ≈ 32.0° (3 s.f.)
WE 4

Two angles from one point — find the flagpole

A flagpole stands on top of a building. From a point 80 m from the foot of the building, the angle of elevation to the top of the building is 25° and the angle of elevation to the top of the flagpole is 32°. Find the height of the flagpole, correct to 3 s.f.

Step 1: Find the building height tan 25° = h_b/80 h_b = 80 × tan 25° = 80 × 0.4663… = 37.30… Step 2: Find total height (top of flagpole) tan 32° = h_t/80 h_t = 80 × tan 32° = 80 × 0.6249… = 49.99… Step 3: Subtract to get the flagpole alone flagpole = 49.99… − 37.30… = 12.68… flagpole ≈ 12.7 m (3 s.f.) two angles from same observation point — find each height, then subtract
WE 5

Two boats from one lighthouse

An observer at the top of a lighthouse 45 m high sees two boats in line directly out at sea. The angles of depression to the boats are 27° (closer boat) and 18° (farther boat). Find the distance between the two boats, correct to 3 s.f.

Step 1: Distance to closer boat (depression 27°) tan 27° = 45/d₁ d₁ = 45/tan 27° = 45/0.5095… = 88.30… Step 2: Distance to farther boat (depression 18°) tan 18° = 45/d₂ d₂ = 45/tan 18° = 45/0.3249… = 138.49… Step 3: Distance between boats d = d₂ − d₁ = 138.49 − 88.30 = 50.19… distance ≈ 50.2 m (3 s.f.) smaller depression → farther boat (line of sight is more horizontal)
WE 6

Two observation points — find the cliff height

From point A on level ground, the angle of elevation to the top of a cliff is 35°. From point B, which is 50 m closer to the cliff than A on the same line, the angle of elevation is 50°. Find the height of the cliff, correct to 3 s.f.

Step 1: Let x = distance from B to base of cliff; h = height from A: tan 35° = h/(x + 50) → h = (x+50) tan 35° from B: tan 50° = h/x → h = x tan 50° Step 2: Set equal and solve for x x tan 50° = (x+50) tan 35° x(tan 50° − tan 35°) = 50 tan 35° x = 50 tan 35° / (tan 50° − tan 35°) x = 50 × 0.7002 / (1.1918 − 0.7002) x = 35.01 / 0.4916 = 71.21… Step 3: Find h h = 71.21 × tan 50° = 71.21 × 1.1918 = 84.86… cliff height ≈ 84.9 m (3 s.f.) two-equation simultaneous setup — both equations express h, set them equal and solve

💡 Top tips

⚠ Common mistakes

Next note: Bearings & Constructions — same trig, but now with compass directions. North-based, clockwise, three-digit angles. The setup changes, the maths stays the same.

Need help with Elevation & Depression?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →