Angles of elevation and depression — Mathematics

Before this

Finding missing angles using inverse trig
Finding missing sides using trigonometry
Angles in parallel lines — F, Z and C angles · coming soon

This leads to

Two students, one lighthouse, two answers

Here is a question from a test.

A lighthouse is 45 m tall. From the top, the angle of depression of a boat is 28°. How far is the boat from the foot of the lighthouse?

Two students solve it. Both draw a right-angled triangle. Both use the tangent. Both use a calculator in degree mode. Rania gets about 84.6 m. Tomás gets about 23.9 m.

They cannot both be right. And the mistake is not in the trigonometry. Both rearranged their equations correctly. The mistake happened earlier, at the moment they turned the words into a picture. One of them put the 28° in the wrong place.

This lecture adds no new ratio and no new button. What is new is a pair of words, elevation and depression, and one angle that does not sit inside the triangle at all. That is where the difficulty lives. So the lecture spends most of its time on the picture, not the calculation.

Every one of these problems hides the same triangle

Imagine you are standing on flat ground, looking at the top of a building. Three lines describe the scene.

The horizontal. A level line through your eye, parallel to flat ground. It is where you look when you look straight ahead.
The vertical. The upright line of the building, at 90° to the horizontal.
The line of sight. The straight line from your eye to the thing you are looking at.

These three lines make a right-angled triangle. The right angle is where the horizontal meets the vertical. The line of sight is the longest side, so it is the hypotenuse.

The angle of elevation θ is measured at the eye, up from the horizontal to the line of sight. The triangle's vertical side is h, the height above eye level. The full height of the building is h + e, where e is the eye height.

The angle of elevation is the angle between the horizontal and the line of sight, when you look up. It sits at your eye, inside the triangle. From that angle, the vertical side is the opposite, and the horizontal is the adjacent. So the tangent links the height and the distance, exactly as in the ramp and ladder problems of the last two lectures.

Look at the figure again. The horizontal starts at your eye, not at your feet. So the vertical side of the triangle is only the part of the building above your eye. This detail does not matter when the angle is measured from the ground, for example by an instrument lying on the ground. It matters a lot when a question gives your eye height, and Worked Example 4 shows how.

The angle that sits outside the triangle

Now climb to the top of a cliff and look down at a boat. You look straight ahead first, along the horizontal. Then you lower your eyes until you see the boat. The angle you turned through, from the horizontal down to the line of sight, is the angle of depression.

Here is the part that catches almost everyone. The horizontal at the top of the cliff is not a side of the triangle. The triangle's sides are the cliff, the sea and the line of sight. The horizontal at the top sticks out over the sea. So the angle of depression lies between a side of the triangle and a line outside it. It is outside the triangle.

The angle of depression θ is at the top, between the dashed horizontal and the line of sight: outside the triangle. The angle of elevation from the boat is at the bottom, inside the triangle. The arrow marks show that the two horizontals are parallel, so the two angles are equal. Inside the triangle at the top, the angle is 90° − θ.

So where does the angle of depression go in your working? The answer comes from a fact you met with parallel lines. The horizontal at the top of the cliff is parallel to the sea. The line of sight crosses both of them. Two lines crossed by a third make alternate angles (the "Z angles"), and alternate angles are equal. The angle of depression at the top and the angle of elevation at the boat make exactly this Z shape. So they are equal.

Elevation and depression

The angle of elevation is measured up from the horizontal to the line of sight. The angle of depression is measured down from the horizontal to the line of sight. Both are always measured from the horizontal, never from the vertical.

The angle of depression from A down to B equals the angle of elevation from B up to A, because the two horizontals are parallel (alternate angles).

That is it. That is the whole idea. Move the angle of depression down to the bottom of the triangle, and the problem becomes an angle of elevation problem. Then everything from the last three lectures works without change.

Now you can settle the lighthouse. Rania moved the 28° down to the boat. From there, the lighthouse (45 m) is the opposite, and the distance $d$ is the adjacent:

tan 2 8^{\circ} = \frac{45}{d} \Rightarrow d = \frac{45}{tan 2 8 ^{\circ}} \approx 84.6 m

Tomás put the 28° inside the triangle at the top, between the lighthouse and the line of sight. From there, $d$ is the opposite and 45 m is the adjacent, so he wrote $tan 2 8^{\circ} = d \div 45$ and got $d \approx 23.9$ m. His algebra was perfect. His angle was in the wrong place.

See the difference for yourself below. Move the boat and watch the three angles.

Explore — move the boat

The cliff is 45 m high. Move the boat closer or further. Watch the angle of depression at the top, the angle of elevation at the boat, and the grey angle inside the triangle at the top.

Try:

Horizontal distance to the boat d: 85 m

This table shows a cliff 45 m high and a boat at several horizontal distances d. The angle of depression θ comes from tan θ = 45 ÷ d. It equals the angle of elevation from the boat. The angle inside the triangle at the top is 90° − θ.

Distance d (m)	Depression = elevation θ	Angle inside at the top
20	66.0°	24.0°
45	45°	45°
85	27.9°	62.1°
150	16.7°	73.3°

As the boat moves away, the angle of depression gets smaller and the angle inside the triangle at the top gets bigger. The two always add up to 90°. They are equal only when the boat is 45 m away, the same as the height of the cliff.

You should have seen three things. First, the angle of depression and the angle of elevation are always the same number. Second, the grey angle inside the triangle at the top is a different angle: it is 90° minus the angle of depression. Third, the two can only be confused without harm at one distance, 45 m, where both are 45°. Everywhere else, using the wrong one changes the answer.

Stop and think

Tomás says: "I like working from the top of the lighthouse, where the observer is. Is it wrong to use the angle at the top at all?"

Show the answer

No, it is not wrong. The angle inside the triangle at the top is a perfectly good angle to use. But it is not 28°. It is $9 0^{\circ} - 2 8^{\circ} = 6 2^{\circ}$ . From that angle, $d$ is the opposite and 45 m is the adjacent, so $tan 6 2^{\circ} = d \div 45$ , and $d = 45 \times tan 6 2^{\circ} \approx 84.6$ m. That is Rania's answer, by a second route. Tomás's only mistake was the label. He called the inside angle 28°, when 28° is the angle outside the triangle, measured from the horizontal. This gives you a useful check for every depression problem: work it once from the bottom with θ, and once from the top with 90° − θ.

The method: sketch first, calculate second

The words change from problem to problem. The triangle does not.

1. Sketch the scene. Draw the ground (or the sea) as a horizontal line. Draw the tall object as a vertical line. Mark where the observer stands.
2. Draw the line of sight. Join the observer's eye to the object being looked at. This line is the hypotenuse.
3. Mark the angle from a horizontal. Elevation: at the lower point, up from the ground. Depression: draw a dashed horizontal at the top first, then mark the angle down from it.
4. Move a depression angle to the bottom. Write "alternate angles" beside it. Now every problem has its angle at the lower end of the line of sight.
5. Name the sides and choose the ratio. Name the hypotenuse first. Then opposite and adjacent, from your angle. The pair you know, and the side you want, choose sine, cosine or tangent.
6. Solve and check. Write the equation, solve it, and check the size of the answer.

One size check is very quick, and it would have saved Tomás. If the angle is below 45°, the horizontal distance is longer than the height. If it is above 45°, the height is longer. This is because $tan 4 5^{\circ} = 1$ , so the opposite equals the adjacent at 45°. The lighthouse angle, 28°, is below 45°, so the boat must be further away than 45 m. Tomás's 23.9 m fails the check at once.

Worked examples

Worked Example 1 — The height of a building

From a point on level ground 30 m from the foot of a building, the angle of elevation of the top is 42°. The angle is measured at ground level. Find the height of the building.

Think first. The angle is at the ground, 30 m away. From it, the height $h$ is the opposite and 30 m is the adjacent. Opposite and adjacent means tangent. The unknown is on top, so multiply.

tan 4 2^{\circ} = \frac{h}{30} \Rightarrow h = 30 \times tan 4 2^{\circ} \approx 27.0 m (3 s.f.)

Check. The angle, 42°, is a little below 45°, so the height should be a little less than the distance, 30 m. It is. ✓

Worked Example 2 — The lighthouse, set out in full

A lighthouse is 45 m tall. The angle of depression of a boat from the top is 28°. Find the horizontal distance from the foot of the lighthouse to the boat.

Think first. Sketch, with a dashed horizontal at the top. The 28° is between that horizontal and the line of sight. By alternate angles, the angle of elevation of the top from the boat is also 28°. From the boat, 45 m is the opposite and the distance $d$ is the adjacent. The unknown is on the bottom, so divide.

tan 2 8^{\circ} = \frac{45}{d} \Rightarrow d = \frac{45}{tan 2 8 ^{\circ}} \approx 84.6 m (3 s.f.)

Check. 28° is below 45°, so the distance should be more than 45 m. It is. ✓ The second route from the top gives $45 \times tan 6 2^{\circ} \approx 84.6$ m too. ✓

Worked Example 3 — Finding the angle

A straight ramp rises 3 m over a horizontal distance of 12 m. Find its angle of elevation.

Think first. Both sides are known and the angle is wanted, so this needs an inverse function, as in the last lecture. The rise is the opposite and the 12 m is the adjacent.

tan θ = \frac{3}{12} = 0.25 \Rightarrow θ = tan^{- 1} (0.25) \approx 14. 0^{\circ} (1 d.p.)

Check. The rise is much shorter than the run, so the angle should be well below 45°. It is. ✓

Worked Example 4 — Do not forget the eye

Kwame stands 50 m from a tree on level ground. The angle of elevation of the top of the tree, from his eye, is 35°. His eye is 1.6 m above the ground. Find the height of the tree.

Think first. The horizontal starts at Kwame's eye, as in the first figure. So the triangle's vertical side is only the part of the tree above his eye. Call it $h$ . The 1.6 m below his eye is outside the triangle, and it must be added at the end.

tan 3 5^{\circ} = \frac{h}{50} \Rightarrow h = 50 \times tan 3 5^{\circ} \approx 35.01 m

Height of the tree $= 35.01 + 1.6 \approx 36.6$ m (3 s.f.).

Check. 35° is below 45°, so $h$ should be less than 50 m. It is. ✓ The tree is a little taller than $h$ , by exactly the eye height. ✓

Here is one more lighthouse-type problem, solved by a student. One line holds a mistake. Find it.

Spot the error — a cliff and a boat

A cliff is 70 m high. From the top, the angle of depression of a boat is 35°. Find the horizontal distance d from the foot of the cliff to the boat. Tap the line with the mistake.

This box shows a student's working for a cliff 70 m high and an angle of depression of 35°, with the mistake explained.

The mistake is in the second line. The student put the 35° inside the triangle, at the top, between the cliff and the line of sight. But the angle of depression is measured from the horizontal, so it lies outside the triangle. The angle inside at the top is 90° − 35° = 55°. Moving the 35° down to the boat by alternate angles gives tan 35° = 70 ÷ d, so d = 70 ÷ tan 35° ≈ 100.0 m. The student's 49.0 m also fails the size check: 35° is below 45°, so d must be more than 70 m.

Notice that every line after the mistake was correct. That is why this error is so hard to see in your own work: the algebra looks perfect. Only the sketch, or the 45° check, catches it.

Three traps

Trap 1 — Putting the angle of depression inside the triangle

The angle of depression is between the horizontal and the line of sight, so it lies outside the triangle. If you write it in the corner between the vertical and the line of sight, you are using $9 0^{\circ} - θ$ under the wrong name. Draw the dashed horizontal at the top every time, and move the angle to the bottom by alternate angles.

Trap 2 — Forgetting the eye height, or adding it twice

When an angle of elevation is measured from someone's eye, the triangle only reaches up from eye level. Add the eye height at the end, as in Worked Example 4. But when the angle is measured from the ground, there is no eye height to add. Read the question to see where the angle is measured from.

Trap 3 — Mixing up the two distances

"How far is the boat from the foot of the lighthouse?" asks for the horizontal distance, the adjacent. "How far is the boat from the top?" asks for the line of sight, the hypotenuse. These are different lengths. For the lighthouse, the line of sight is $45 \div sin 2 8^{\circ} \approx 95.9$ m, not 84.6 m. Read which two points the question names.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

A vertical pole 1.8 m tall casts a shadow 2.5 m long on level ground. Find the angle of elevation of the sun, to 1 d.p.

Hint 1
The sun's rays travel along the line of sight from the tip of the shadow to the top of the pole. Sketch the triangle.

Hint 2
From the tip of the shadow, the pole is the opposite and the shadow is the adjacent: $tan θ = \frac{1.8}{2.5}$ .

Full answer
$θ = tan^{- 1} (\frac{1.8}{2.5}) \approx 35.7 5^{\circ} \approx 35. 8^{\circ}$ (1 d.p.). Check: the pole is shorter than the shadow, so the angle should be below 45°. It is. ✓ Put it back: $2.5 \times tan 35.7 5^{\circ} \approx 1.80$ m. ✓
Aisha looks down from a window 12 m above level ground at a car parked 40 m away horizontally. (a) Find the angle of depression of the car, to 1 d.p. (b) How far is the car from the window, in a straight line?

Hint 1
Move the angle of depression down to the car. From the car, which side is 12 m and which is 40 m?

Hint 2
(a) $tan θ = \frac{12}{40}$ . (b) The straight-line distance is the line of sight, the hypotenuse. Pythagoras works, and so does sine.

Full answer
(a) By alternate angles, the angle of depression equals the angle of elevation from the car. $θ = tan^{- 1} (\frac{12}{40}) \approx 16.7 0^{\circ} \approx 16. 7^{\circ}$ . (b) $1 2^{2} + 4 0^{2} = 1744 \approx 41.8$ m. Check: $12 \div sin 16.7 0^{\circ} \approx 41.8$ m agrees. ✓ The line of sight is longer than both other sides, as a hypotenuse must be. ✓ The angle inside the triangle at the window is $9 0^{\circ} - 16.7 0^{\circ} \approx 73. 3^{\circ}$ , which is not the angle of depression.
Lucía flies a kite on a straight string 50 m long. The string makes an angle of elevation of 55° at her hand, which is 1.2 m above the ground. How high is the kite above the ground?

Hint 1
The string is the hypotenuse. Which side do you want, from the 55° angle at her hand?

Hint 2
The height above her hand is the opposite: $sin 5 5^{\circ} = \frac{h}{50}$ . Then think about her hand.

Full answer
$h = 50 \times sin 5 5^{\circ} \approx 40.96$ m above her hand. Her hand is 1.2 m above the ground, so the kite is $40.96 + 1.2 \approx 42.2$ m above the ground (3 s.f.). Check: 55° is above 45°, so the height above her hand should be more than the horizontal distance, $50 \times cos 5 5^{\circ} \approx 28.7$ m. It is. ✓ And $h$ is less than the 50 m string. ✓
A stretch. Ravi stands at a point A on level ground and measures the angle of elevation of the top of a tower as 28°. He walks 20 m straight towards the tower, to a point B, and measures the angle again: 40°. Both angles are measured at ground level. Find the height of the tower.

Hint 1
There are two right-angled triangles, sharing the tower. Neither has a known horizontal distance. Call the height $h$ and write each distance in terms of $h$ .

Hint 2
From B, the distance is $\frac{h}{tan 4 0 ^{\circ}}$ . From A, it is $\frac{h}{tan 2 8 ^{\circ}}$ . The two distances differ by 20 m.

Full answer
$\frac{h}{tan 2 8 ^{\circ}} - \frac{h}{tan 4 0 ^{\circ}} = 20$ . Take $h$ out as a factor: $h (\frac{1}{tan 2 8 ^{\circ}} - \frac{1}{tan 4 0 ^{\circ}}) = 20$ . The bracket is about $1.8807 - 1.1918 = 0.6890$ , so $h \approx 20 \div 0.6890 \approx 29.0$ m (3 s.f.). Check: from B the distance is $29.03 \div tan 4 0^{\circ} \approx 34.6$ m, and from A it is $29.03 \div tan 2 8^{\circ} \approx 54.6$ m. The difference is 20.0 m. ✓ The angle is bigger at B, the nearer point, as it should be. ✓ This is how surveyors find a height when they cannot reach the foot of the object: two angles and one measured walk.

The bigger idea

In the lecture on finding missing sides, you met the Great Trigonometrical Survey of India and the peak it called Peak XV. Its height came from angles of elevation. Between 1849 and 1850, the surveyor James Nicolson measured the angle up to the summit from six stations in the plains. Each station was between 108 and 118 miles (about 174 to 190 km) away. (Sources: H. L. Crosthwait, "Mount Everest", Nature, 1933; "Two Memoranda: A Brief History of Mount Everest's Altitude", American Alpine Journal, 1995.)

From that far away, the summit sits only about 3° above the horizontal. And the simple triangle of this lecture is no longer enough. The triangle assumes flat ground. But the Earth curves. Over 174 km, a level line rises above the Earth's surface by about 2.4 km, from the Earth's radius of about 6371 km. That is more than a quarter of the mountain's height. Light also bends a little as it passes through the air, which changes the angle the observer sees. So the survey's computers corrected every angle for the Earth's curvature and for this bending, before announcing 29 002 feet in 1856.

This is how mathematics works in the real world. A simple model does almost all the work. Then you ask what the model leaves out, and how big that is. For a tree 50 m away, the Earth's curvature is far too small to matter. For a mountain 174 km away, it is the difference between a right answer and a wrong one.

Key concepts in this lesson

The angle of elevation is measured up from the horizontal to the line of sight, and the angle of depression is measured down from the horizontal to the line of sight.
The angle of depression lies outside the triangle, between the horizontal at the top and the line of sight.
The angle of depression from A to B equals the angle of elevation from B to A, because the two horizontals are parallel and the angles are alternate angles.
Every elevation or depression problem is a right-angled triangle made of a horizontal, a vertical and a line of sight, so sketching the scene comes before any calculation.
When the angle is below 45°, the horizontal distance is longer than the height, which gives a quick check on every answer.
When an angle is measured from an observer's eye, the eye height must be added to the height found from the triangle.