Angles of Elevation and Depression
Draw the horizontal reference line first — always. The angle opens between that horizontal and your line of sight, never from the vertical. Get the diagram right and the trigonometry is straightforward.
Practise this lesson
Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.
You are standing on top of a cliff looking down at a boat. Your friend is in the boat looking up at you. You both measure the angle between the horizontal and your line of sight to each other. Will you get the same angle or different angles? Why?
Every elevation and depression problem starts with a horizontal reference line drawn from the observer's eye. The angle opens between that horizontal and the line of sight — never from the vertical, never from the slope.
Angle of elevation — measured upward from the horizontal to the line of sight when looking up at an object above.
Angle of depression — measured downward from the horizontal to the line of sight when looking down at an object below.
Key fact: The angle of elevation from A to B equals the angle of depression from B to A. Both equal $\theta$. This works because the two horizontal lines are parallel — the line of sight is a transversal, so the angles are alternate interior angles.
Key facts
- Angle of elevation: measured upward from horizontal
- Angle of depression: measured downward from horizontal
- Both angles are always referenced from the horizontal
- Elevation angle from A to B = depression angle from B to A
Concepts
- Why the horizontal reference line must appear in every diagram
- Why elevation and depression angles are equal (alternate angles)
- How two-observation problems share a common height
Skills
- Draw a correct labelled diagram from any word description
- Solve elevation and depression problems using SOHCAHTOA
- Apply the equal-angle relationship to simplify depression problems
- Set up and solve two-observation simultaneous problems
The angle of elevation is measured from the horizontal upward to the line of sight. It is never measured from the ground if the ground is sloped, and never from the vertical.
Elements of a correct elevation diagram:
- Observer at one end; horizontal reference line extending from observer
- Line of sight goes upward from observer to the object
- Angle of elevation sits between the horizontal and the line of sight
- Right angle at the foot of the vertical height of the object
What to write in your book
- Always draw the horizontal reference line from the observer before marking the angle.
- Elevation angle: between horizontal and line of sight going upward.
- Most common setup: O = height, A = horizontal distance → use $\tan\theta = O/A$.
- If the line-of-sight distance (hypotenuse) is involved, use $\sin$ or $\cos$ instead.
Did you get this? True or false: the angle of elevation is measured from the vertical to the line of sight.
Worked examples · 4 in a row, reveal as you go
From a point 45 m from the base of a vertical building, the angle of elevation to the top is 38°. Find the height of the building, correct to 2 decimal places.
From the top of a cliff 80 m high, a boat is observed at sea with an angle of depression of 24°. Find the horizontal distance from the base of the cliff to the boat, correct to 2 decimal places.
A 12 m antenna sits on top of a 40 m building. An observer stands 75 m from the base of the building. Find the angle of elevation to the top of the antenna, to the nearest minute.
What to write in your book
- For depression: always redraw using the equal elevation angle. Makes the triangle clearer.
- Two-observation strategy: Let $h = d\tan\theta_1 = (d+k)\tan\theta_2$. Set equal. Solve for $d$, then find $h$.
- The nearer observer always has the larger elevation angle. If algebra gives negative $d$, you have labelled the points backwards.
- Combined heights: add all vertical distances before substituting into the formula.
Quick check: From the top of a 50 m building, the angle of depression to a car is 28°. What is the horizontal distance to the car?
When two angles of elevation to the same object are given from different points, you have two expressions for the same height. Setting them equal produces an equation solvable for the unknown distance.
Strategy:
- Let height $= h$, distance from nearer point to base $= d$
- Write $h = d \cdot \tan\theta_1$ (nearer point — larger angle)
- Write $h = (d + k) \cdot \tan\theta_2$ (further point — smaller angle, $k$ = separation)
- Set equal and solve for $d$
- Substitute back to find $h$
From point A, the angle of elevation to the top of a tower is 52°. From point B, 30 m further from the tower on the same side, the angle of elevation is 38°. Find the height of the tower, correct to 2 decimal places.
From A: $h = d\tan 52°$ … (1)
From B: $h = (d+30)\tan 38°$ … (2)
$d\tan 52° = d\tan 38° + 30\tan 38°$
$d(\tan 52° - \tan 38°) = 30\tan 38°$
Common errors · the 3 traps that cost marks
What to write in your book
- Step 1 of every elevation/depression problem: draw the horizontal reference line.
- Elevation and depression angles are measured from the horizontal, not the vertical.
- Equal angles: $\theta_{\text{elev from A to B}} = \theta_{\text{dep from B to A}}$ — alternate interior angles.
- Two-observation: set $h = d\tan\theta_1 = (d+k)\tan\theta_2$; solve for $d$; then find $h$.
Fill the gap: The angle of depression from B down to A equals the angle of from A up to B, because the two horizontal lines are parallel — the angles formed are interior angles.
Quick-fire practice · 6 calculations
From a point 60 m from the base of a vertical tree, the angle of elevation to the top is 42°. Find the height of the tree to 2 d.p.
From the top of a 120 m cliff, the angle of depression to a boat is 31°. Find the horizontal distance from the cliff base to the boat, to 2 d.p.
An observer on the ground is 50 m from a building. The top is 35 m above the observer's eye level. Find the angle of elevation to the nearest minute.
A 15 m flagpole stands on top of a 25 m building. An observer is 80 m from the base. Find the angle of elevation to the top of the flagpole, to the nearest minute.
From point P, the angle of elevation to the top of a tower is 45°. From point Q, 20 m further from the tower than P, the angle of elevation is 30°. Find the height of the tower to 2 d.p.
From a lighthouse 65 m above sea level, a ship is observed with an angle of depression of 18°. Find the straight-line distance (line of sight) from the lighthouse to the ship, to 2 d.p.
Odd one out: Three of these statements are correct. Which one is wrong?
You and your friend on the cliff both measure the angle from the horizontal to your mutual line of sight. The answer: you get the same angle. The angle of depression from the cliff top to the boat equals the angle of elevation from the boat to the cliff top — they are alternate interior angles between the two parallel horizontal lines cut by the line of sight.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
SA 1. From a point on level ground, the angle of elevation to the top of a vertical cliff is 34°. The point is 220 m from the base of the cliff. (a) Draw a fully labelled diagram of this situation, including the horizontal reference line, angle, right angle, and all known measurements. (b) Find the height of the cliff correct to 2 decimal places. (3 marks)
SA 2. A ship is sailing toward a 45 m lighthouse at sea level. From the ship, the angle of elevation to the top of the lighthouse is 12°. (a) Find the horizontal distance from the ship to the base of the lighthouse, to the nearest metre. (b) After the ship travels closer, the angle of elevation becomes 28°. How far has the ship travelled? Give your answer to the nearest metre. (3 marks)
SA 3. From point A on flat ground, the angle of elevation to the top of a vertical tower is 55°. From point B, which is 40 m further from the tower than A and on the same side, the angle of elevation is 35°. (a) Let the horizontal distance from A to the base of the tower be $d$ metres and the height be $h$ metres. Write two equations for $h$ in terms of $d$. (b) Find the value of $d$ correct to 2 decimal places. (c) Hence find the height of the tower correct to 2 decimal places. (4 marks)
📖 Comprehensive answers (click to reveal)
Drill answers: 1: $h = 60\tan42° \approx \mathbf{54.05 \text{ m}}$ · 2: $d = 120/\tan31° \approx \mathbf{199.71 \text{ m}}$ · 3: $\tan^{-1}(35/50) \approx \mathbf{34°59'}$ · 4: total $= 40$ m; $\tan\theta = 40/80$; $\approx \mathbf{26°34'}$ · 5: $h = d = 20/(\sqrt{3}-1) \approx \mathbf{27.32 \text{ m}}$ · 6: $H = 65/\sin18° \approx \mathbf{210.34 \text{ m}}$
SA 1 (3 marks): (a) Horizontal ground; 34° elevation angle; horizontal reference from observer; right angle at cliff base; 220 m horizontal; height $x$ [1]. (b) $\tan34° = x/220$; $x = 220\tan34° = 220 \times 0.67450\ldots \approx \mathbf{148.39 \text{ m}}$ [2].
SA 2 (3 marks): (a) $d_1 = 45/\tan12° \approx 45/0.21256 \approx 211.73 \approx \mathbf{212 \text{ m}}$ [2]. (b) $d_2 = 45/\tan28° \approx 45/0.53171 \approx 84.64 \approx 85$ m; distance $= 212 - 85 = \mathbf{127 \text{ m}}$ [1].
SA 3 (4 marks): (a) $h = d\tan55°$ and $h = (d+40)\tan35°$ [1]. (b) $d\tan55° = (d+40)\tan35°$; $d(1.42814 - 0.70021) = 40 \times 0.70021$; $d \times 0.72793 = 28.008$; $d \approx \mathbf{38.48 \text{ m}}$ [2]. (c) $h = 38.48 \times \tan55° = 38.48 \times 1.42814 \approx \mathbf{54.95 \text{ m}}$ [1].
Five timed questions on elevation and depression. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering elevation and depression questions. Pool: lesson 16.
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