Both angles start at the horizontal.
Not from the vertical
An angle of depression of 35° is measured down from the horizontal, not across from the cliff face. Measuring from the vertical gives 55° instead, and every answer after that is wrong.
- Home
- Grade 7-12 Math Review
- Angles of Elevation and Depression
Angles of Elevation and Depression
Both are measured from the horizontal — never from the vertical, and never from the slope.
Find the height
From 30 m away, the angle of elevation to the top of a tower is 50°. Roughly how tall is the tower?
The 30 m is adjacent to the angle and the height is opposite, so use tangent: tan 50° = h30, giving h = 30 tan 50° ≈ 35.8 m. A uses sine, treating 30 as the hypotenuse — but 30 m is measured along the ground, so it is the adjacent side. Since 50° is more than 45°, the height must exceed 30 m, which rules out A and B immediately.
Elevation and depression are equal.
a boat sees the cliff top at 35°
elevation
the cliff top sees the boat at 35°
depression
alternate interior angles
two parallel horizontals
The two horizontal lines are parallel and the line of sight is a transversal, so the angles are equal. Depression problems can be redrawn as elevation problems inside the triangle.
Drawing the triangle from the words.
| The phrase | In the triangle |
|---|---|
| "x metres from the base" | the adjacent side, along the ground |
| "height of the tower" | the opposite side, vertical |
| "length of the wire" or "line of sight" | the hypotenuse |
| "angle of elevation" | at the observer, from the ground |
A ground distance is almost always adjacent, not the hypotenuse. Reaching for sine when the problem says "from 30 m away" is the single most common error in these questions.
Worked, both directions.
30 m away, elevation 50°
find the height
tan 50° = h30
opposite over adjacent
h ≈ 35.8 m
taller than 30 ✓
a 12 m pole, from 20 m away
find the elevation
tan θ = 1220 = 0.6
the ratio
θ = tan⁻¹(0.6) ≈ 31°
less than 45°, and 12 < 20 ✓
Add the observer's height when the problem gives one.
eye level 1.6 m, 25 m away, elevation 42°
how tall is the building?
tan 42° = x25
x ≈ 22.5 m
but that is above eye level
not above the ground
height = 22.5 + 1.6 ≈ 24.1 m
add the observer
The triangle starts at the eye, not the feet
When a problem bothers to state someone's height, it is because the answer needs it. The trigonometry gives the part above eye level only.
When the unknown is on the bottom of the ratio.
If the side you want sits under the fraction bar, the answer is a division: known side ÷ the trig value.
A 25 m tower stands in sunlight. The sun's angle of elevation is 40°. How long is its shadow?
tan 40° = 25s
height opposite, shadow adjacent
s = 25 ÷ tan 40° ≈ 25 ÷ 0.8391
s was on the bottom
s ≈ 29.8 m
below 45°, so longer than 25 ✓
A ladder makes a 70° angle with the ground and reaches 5 m up a wall. How long is it?
sin 70° = 5L
height opposite, ladder the hypotenuse
L = 5 ÷ sin 70° ≈ 5 ÷ 0.9397
divide
L ≈ 5.32 m
a little longer than 5 ✓
Multiplying out of habit
25 × tan 40° ≈ 21.0 m makes the shadow shorter than the tower, but at an elevation below 45° the shadow must be longer. Likewise 5 × sin 70° ≈ 4.7 m would make the ladder shorter than the height it reaches.
Variations: a 45° angle, two observers, an angle from a gentle slope.
Special angles give exact answers; two triangles share one side.
From the top of a 24 m building the angle of depression to a parked car is 45°. How far is the car?
tan 45° = 24d
depression = elevation from the car
1 = 24d, so d = 24 m
tan 45° = 1
At 45° the triangle is isosceles: the distance equals the height exactly.
A 30 m tower is seen at 30° from point A and at 60° from point B, both on the same side. Find AB.
from A: d = 30 ÷ tan 30° = 30√3
≈ 51.96 m
from B: d = 30 ÷ tan 60° = 30√3 = 10√3
≈ 17.32 m
AB = 30√3 − 10√3 = 20√3 ≈ 34.6 m
subtract the distances
Two triangles share the tower's height. Find each ground distance, then the gap between them.
A ramp rises 1.2 m over a horizontal distance of 10 m. Find its angle of elevation.
tan θ = 1.210 = 0.12
rise over run
θ = tan⁻¹(0.12) ≈ 6.8°
a gentle slope
A small ratio gives a small angle; the answer should look shallow, as a ramp does.
On the exam
Expect a height from a distance and an angle, a distance from a cliff or plane using a depression angle, a ladder or kite string as the hypotenuse, and sometimes an observer's eye height to add. The fastest route is to draw the triangle, move a depression angle down to the far end as an equal elevation, and multiply or divide depending on where the unknown sits in the ratio. The 10-second check: below 45° the height is less than the ground distance, above 45° it is more.
Now you try
Practice problem
Find the distance
From the top of a 45 m lighthouse, the angle of depression to a boat is 30°. How far is the boat from the base?
The depression of 30° equals the elevation from the boat. With the height opposite and the distance adjacent, tan 30° = 45d, so d = 45tan 30° = 45√3 ≈ 77.9 m. B divides the wrong way round. Since 30° is a shallow angle, the boat must be far out — considerably further than the lighthouse is tall, which only A satisfies.
Hint: a shallow angle means a distant object. If your answer is smaller than the lighthouse's height, the ratio went in upside down.