Measured up from the horizontal
Stand on level ground some way back from a tower and look at its top. Your line of sight rises from your eye to the top of the tower. The horizontal is the level line through your eye, running straight out across the ground.
The angle of elevation is the angle between the horizontal and the line of sight, when you look up. It is always measured from the horizontal, never from the tower and never from the vertical.
The further back you stand, the flatter the line of sight and the smaller the angle of elevation. Walk toward the tower and the angle grows.
The dashed line is the horizontal through the eye. The angle of elevation lies between it and the line of sight to the top of the tower.
The height of a tower
You stand 40 meters from the foot of a tower, and the angle of elevation of its top is 45°. The horizontal, the tower and the line of sight make a right triangle, with the right angle at the foot of the tower.
Name the sides from the 45° angle at your eye. The height of the tower, h, is across from it, so it is the opposite side. The 40 meters along the ground runs from the angle to the right angle, so it is the adjacent side. The line of sight is the hypotenuse.
The side you know is the adjacent side and the side you want is the opposite side. The ratio that links opposite and adjacent is the tangent, so .
Multiply both sides by 40: h = 40 × tan 45°. The exact value of tan 45° is 1, so h = 40 × 1 = 40 meters. At 45° the two shorter sides of the triangle are equal, so the tower is exactly as tall as you are far from it.
An angle of elevation of 45° from 40 m away: tan 45° = 1, so the height is 40 × 1 = 40 m.
θ = 30°: sin = 0.5, cos = 0.866, tan = 0.577; pull the corner outward and the triangle grows but not one ratio changes, because every side is scaled by the same factor
Swing the corner to 45° and read the three ratios
Swing the corner to 45°: the opposite and adjacent sides become equal, so tan 45° = 1. Pull the corner outward and the triangle grows, but the two sides stay equal at every size. Steeper than 45°, the opposite side is the longer one.
Measured down from the horizontal
Now stand at the top of a cliff and look down at a boat on the sea. The line of sight falls from your eye to the boat.
The angle of depression is the angle between the horizontal and the line of sight, when you look down. It is measured down from the horizontal through your eye, not from the face of the cliff.
This is the angle students most often put in the wrong place. The angle between the cliff face and the line of sight is a different angle: it is 90° minus the angle of depression, because the cliff face is vertical and makes a right angle with the horizontal.
The horizontal is drawn through the eye at the top of the cliff. The angle of depression lies between it and the line of sight down to the boat.
The same angle from the other end
The horizontal through your eye and the surface of the sea are both level, so they are parallel lines. The line of sight runs from one to the other, crossing both.
A line crossing two parallel lines makes equal alternate angles. The angle of depression, between the line of sight and the horizontal at the top, and the angle at the boat, between the line of sight and the sea, are alternate angles. So they are equal.
The angle at the boat is the angle of elevation of the cliff top, seen from the boat. So the angle of elevation of the cliff top from the boat always equals the angle of depression of the boat from the cliff top. That matters when you draw the triangle: the angle of depression sits outside the triangle, above the line of sight, and its equal partner at the boat is the angle inside it.
An angle of depression of 45° from a cliff 30 m high. The angle at the boat is also 45°, so the boat is 30 m out from the foot of the cliff.
How far out is the boat?
The cliff is 30 meters high and the angle of depression of the boat is 45°. Move the angle to the boat: the angle between the sea and the line of sight is also 45°.
From that angle, the height of the cliff is the opposite side and the distance d along the sea is the adjacent side. The ratio is the tangent again: .
Multiply both sides by d, then divide by tan 45°: meters.
A full problem, and the choice of ratio
The roof of a building is 25 meters above a level road. From the roof, the angle of depression of a car parked on the road is 32°. How far is the car from the foot of the building, and how long is the line of sight from the roof to the car?
Draw it first: the building, the road, and the line of sight from the roof to the car, with the angle of depression of 32° between the line of sight and the horizontal at the roof. By alternate angles, the angle at the car between the road and the line of sight is also 32°.
For the distance d along the road: from the 32° angle at the car, the height of 25 meters is the opposite side and d is the adjacent side. Opposite and adjacent call for the tangent, so and . With tan 32° = 0.6249, to 4 decimal places, meters, to 1 decimal place.
For the line of sight L: from the same angle, the height of 25 meters is the opposite side and L is the hypotenuse. Opposite and hypotenuse call for the sine, so and . With sin 32° = 0.5299, meters, to 1 decimal place.
Check with Pythagoras: , and to 1 decimal place, the same length. The two ratios were chosen by the same test: which two sides does this part of the question involve?
An angle of depression of 32° from a roof 25 m up. The distance along the road is , and the line of sight is , each to 1 decimal place.
Two angles from the same place
A flagpole stands on top of a building. From a point on the ground 50 meters from the building, the angle of elevation of the bottom of the flagpole is 30° and the angle of elevation of its top is 35°. How long is the flagpole?
Both lines of sight make right triangles with the same adjacent side, the 50 meters along the ground. The height to the bottom of the flagpole is 50 × tan 30° = 50 × 0.5774 = 28.87 meters. The height to the top is 50 × tan 35° = 50 × 0.7002 = 35.01 meters.
The flagpole is the difference between the two heights: 35.01 − 28.87 = 6.14 meters, which is 6.1 meters to 1 decimal place. The angle between the two lines of sight is 35° − 30° = 5°, but no triangle containing that 5° angle has a right angle in it, so the flagpole is found from the two right triangles, not from the 5°.
When the eye is above the ground
Most problems treat the eye as sitting on the ground. When the question gives the height of the observer’s eye, the horizontal runs through the eye, not along the ground, and the right triangle stops at that level.
A student whose eye is 1.6 meters above the ground stands 20 meters from a tree, and the angle of elevation of the top of the tree is 40°. The triangle gives the height of the tree above the student’s eye: 20 × tan 40° = 20 × 0.8391 = 16.78 meters. The tree is that height plus the height of the eye: 16.78 + 1.6 = 18.38 meters, which is 18.4 meters to 1 decimal place.
The usual mistakes
Measuring the angle from the vertical. An angle of depression of 35° is not the angle between the cliff face and the line of sight. That angle is 90° − 35° = 55°.
Putting the angle of depression inside the triangle at the top. It lies between the horizontal and the line of sight, outside the triangle. Carry it down to the other end of the line of sight, where the alternate angle is inside the triangle.
Choosing sine when the two sides involved are the height and the distance along the ground. Those are the two sides beside the right angle, the opposite and the adjacent, so the ratio is the tangent. Sine needs the line of sight.
Forgetting the height of the eye. When the eye is above the ground, the triangle only reaches the height above the eye, so the eye height is added at the end.
Worked example: A Lighthouse Keeper Watching a Boat Come In
Question The lamp L of a lighthouse is 42 m above the sea. The keeper sees a boat at P at an angle of depression of 35°. Later the boat has sailed straight toward the foot F of the lighthouse, to Q, and the angle of depression is 50°. Take tan 35° = 0.700 and tan 50° = 1.192, and give answers to 1 decimal place. (a) How far is the boat from the foot of the lighthouse at first? (b) How far does the boat sail between the two sightings?
1.Draw the horizontal through the lamp. The angle of depression of 35° lies between that horizontal and the line of sight LP. The horizontal and the sea are parallel, so the angle LPF at the boat is also 35°: they are alternate angles.
The angle of depression is below the horizontal through the lamp; the alternate angle at the boat is also 35°. 2.Triangle LFP is right-angled at F. The height LF = 42 m is opposite the 35° angle at P and the distance FP is adjacent to it, so tan 35° = 42FP.
Triangle LFP is right-angled at F, so tan 35° = 42FP. 3.(a) FP = 42tan 35° = 420.700 = 60.0 m.
(a) FP = 420.700 = 60.0 m. 4.At the second sighting the angle LQF is 50°, so FQ = 42tan 50° = 421.192 = 35.23 m.
At the second sighting FQ = 421.192 = 35.23 m. 5.(b) The boat sails PQ = 60.0 − 35.23 = 24.77 m, which is 24.8 m to 1 decimal place. Check: the steeper angle belongs to the nearer boat, and 35.23 m is less than 60.0 m.
(b) The boat sails 60.0 − 35.23 = 24.77, about 24.8 m.
Answer: (a) 60.0 m; (b) 24.8 m
Common mistakes
- Measuring the 35° from the lighthouse tower instead of from the horizontal. The angle between the tower and the line of sight is 90 − 35 = 55°; the angle of depression is the one below the horizontal.
- Using sin 35° = 42FP. Sine pairs the opposite side with the hypotenuse, which is the line of sight LP; the distance along the sea is the adjacent side, so tangent is the ratio that links it to the height.
Worked example: The Height of a Cliff Sighted at Two Angles of Elevation
Question A walker on level ground heads straight toward the foot of a vertical cliff. At point A the angle of elevation of the top of the cliff is 30°. She walks 40 m straight toward the cliff to point B, where the angle of elevation is 60°. Treat her eye as being at ground level, and give exact answers as well as answers to 1 decimal place. (a) How high is the cliff? (b) How far is B from the foot of the cliff?
1.Let the foot of the cliff be F and its top C. Let the height CF be h m and the distance BF be x m. Both triangles CBF and CAF are right-angled at F, and each angle of elevation is measured from the horizontal ground.
Both angles of elevation are measured up from the horizontal ground, and the cliff stands at a right angle to it. 2.In triangle CBF the height is opposite the 60° angle and BF is adjacent to it, so tan 60° = hx. The exact value is tan 60° = √3, so h = √3 x.
From B: tan 60° = hx, and tan 60° = √3, so h = √3 x. 3.In triangle CAF the distance AF is x + 40, so tan 30° = hx + 40. The exact value is tan 30° = 1√3, so h = x + 40√3.
From A: tan 30° = hx + 40, and tan 30° = 1√3, so h = x + 40√3. 4.Set the two expressions for h equal: √3 x = x + 40√3. Multiply both sides by √3 to get 3x = x + 40, so 2x = 40 and x = 20. (b) B is 20 m from the foot of the cliff.
(b) 3x = x + 40, so x = 20: B is 20 m from the foot of the cliff. 5.(a) The height is h = √3 × 20 = 20√3 m, which is 34.6 m to 1 decimal place. Check: A is 20 + 40 = 60 m from the foot, and 20√360 = √33 = 1√3, which is tan 30°.
(a) h = 20√3 ≈ 34.6 m.
Answer: (a) 20√3 m, which is 34.6 m to 1 decimal place; (b) 20 m
Common mistakes
- Writing tan 30° = h40. The 40 m is only the walk from A to B; the side adjacent to the angle at A is the whole distance from A to the foot of the cliff, x + 40.
- Rounding √3 to 1.7 at the start and carrying it through. The error grows at every step, so keep √3 exact until the last line and round once.