Key Idea: Angles of elevation and depression connect trigonometry to real-world measurements — heights of buildings, distances to ships, positions of aircraft. The core skill is drawing a clear diagram that identifies the horizontal baseline and the angle correctly, then selecting the right trig rule (usually SOH CAH TOA for right-angled problems, or the sine/cosine rule for more complex setups).
✅ Key definitions
Example: Example 1: A building is 40 m tall. From a point on the ground, the angle of elevation to the top is 28°. Find the horizontal distance. tan 28° = 40/d → d = 40/tan28° = 75.2 m Example 2: From the top of a 60 m cliff, the angle of depression to a boat is 18°. Find the distance from the boat to the base of the cliff. Angle at the boat = 18° (alternate angles). tan18° = 60/d → d = 60/tan18° = 185 m 3D example: A 10 m vertical mast stands on a flat field. A wire stretches from the top of the mast to a point 6 m from the base. Find the angle the wire makes with the ground. tan θ = 10/6 → θ = tan⁻¹(10/6) = 59.0°
Always draw the diagram first — every time. Mark the horizontal baseline and the angle from it. A mislabelled diagram is the most common source of error in these questions. For 3D questions: look for the right-angled triangle formed by the horizontal base, the vertical height, and the slant line. Use Pythagoras to find a missing side, then trig to find the angle.
Paper 1: You may need to express the answer exactly (e.g., tan⁻¹(5/3)) or leave as a fraction before evaluating. Paper 2 (GDC allowed): These problems can combine multiple steps. Organise clearly: label triangle 1, solve it, then label triangle 2. Show each step separately so you earn method marks even if the final answer is wrong.
IB-style question [6 marks]
A coastguard observation tower stands on a cliff. The top of the tower, T, is 75 m vertically above the sea. Let F be the point on the sea surface directly below T. (a) From T the angle of depression to a buoy A floating on the sea is 12°. Find the distance FA along the sea surface, correct to the nearest metre. (b) A second buoy B floats on the sea 90 m from A, in a direction at right angles to FA. Find the angle of depression from T to buoy B, correct to one decimal place.
Step by step:
(a) The angle of depression from T equals the angle of elevation from A (alternate angles), so the right triangle T–F–A has the 75 m height opposite the 12° angle and FA adjacent to it. Use tangent.
Rearrange to make FA the subject and evaluate.
(b) F, A and B lie on the sea surface with FA ⟂ AB, so FB is the hypotenuse of a right triangle on the water. Find FB by Pythagoras.
The angle of depression to B uses the vertical right triangle T–F–B: height 75 m opposite, FB adjacent.
Take the inverse tangent.
(a) FA ≈ 353 m. (b) Angle of depression to B ≈ 11.6°.