Angles of Elevation and Depression: WAEC Worked Examples

Measuring Heights You Cannot Climb

How tall is a telecommunications mast, a church tower or a palm tree? You could climb it with a tape measure, or you could stand some distance away, measure one angle, and let trigonometry do the rest. Surveyors, engineers and navigators have worked this way for centuries. Questions on angles of elevation and depression appear in nearly every WAEC and NECO mathematics paper, usually worth a lot of marks. This tutorial explains what the angles are, how to draw a clear diagram, and how to solve one-triangle and two-triangle problems involving towers, cliffs, kites, aeroplanes and shadows, with worked examples. It assumes you know SOH CAH TOA from the tutorial on trigonometric ratios.

Definitions

The angle of elevation of an object is the angle between the horizontal and the line of sight when you look up at the object.
The angle of depression of an object is the angle between the horizontal and the line of sight when you look down at the object.

Both angles are always measured from the horizontal, never from the vertical. Instruments called clinometers and theodolites measure them in practice.

The key fact: they are equal

If a person at A looks down at B with an angle of depression θ, then a person at B looking up at A sees an angle of elevation θ. The horizontal lines through A and B are parallel, and the line of sight AB cuts across them, so the two angles are alternate angles and therefore equal. This fact lets you move the angle of depression to the bottom of the diagram, inside a right-angled triangle.

Strategy for Every Problem

Draw a large, clear diagram. Mark the horizontal, the vertical height and the line of sight.
Find the right-angled triangle. Heights are vertical and the ground is horizontal, so there is a right angle at the foot of the tower or cliff.
Place the angle correctly. For an angle of depression, use the equal angle of elevation at the bottom.
Label the sides relative to the angle: opposite, adjacent and hypotenuse.
Choose the ratio (sin, cos or tan) linking the side you know and the side you want, and solve.

Most problems on heights and distances use tan, because they involve a height (opposite) and a horizontal distance (adjacent).

One-Triangle Problems

Worked example 1: height of a tower

From a point on level ground 50 m from the foot of a tower, the angle of elevation of the top of the tower is 32°. Find the height of the tower.

In the right-angled triangle, the height h is opposite the 32° angle, and the 50 m distance is adjacent.

tan 32° = h/50, so h = 50 × tan 32° = 50 × 0.6249 ≈ 31.2 m.

Worked example 2: boat from a cliff

From the top of a cliff 80 m high, the angle of depression of a boat is 25°. How far is the boat from the foot of the cliff?

The angle of elevation of the cliff top from the boat is also 25°. The height 80 m is opposite and the distance d is adjacent.

tan 25° = 80/d, so d = 80 ÷ tan 25° = 80 ÷ 0.4663 ≈ 171.6 m.

Notice that d appears in the denominator, so you divide. Many students multiply here by mistake.

Worked example 3: a kite

A kite is flown on a straight string 60 m long, which makes an angle of elevation of 40° with the ground. Find the height of the kite.

The string is the hypotenuse, and the height is opposite the angle, so use sine.

h = 60 × sin 40° = 60 × 0.6428 ≈ 38.6 m.

Worked example 4: including the observer's height

A man whose eyes are 1.6 m above the ground stands 30 m from a tree. The angle of elevation of the top of the tree is 35°. Find the height of the tree.

The angle is measured from eye level, so the triangle gives only the part of the tree above eye level: 30 × tan 35° ≈ 21.0 m.

Height of tree = 21.0 + 1.6 ≈ 22.6 m. Forgetting to add the observer's eye height is a common error.

Worked example 5: finding the angle

A vertical pole 12 m high casts a shadow 9 m long. Find the angle of elevation of the sun.

tan θ = 12/9 = 1.333, so θ = tan⁻¹ 1.333 ≈ 53.1°.

Two-Triangle Problems

Harder questions involve two angles and two right-angled triangles that share a side, usually the height. Write each unknown distance in terms of the height, then form an equation.

Worked example 6: moving closer

From a point A on level ground, the angle of elevation of the top of a building is 30°. After walking 20 m directly towards the building to point B, the angle of elevation is 45°. Find the height of the building.

Let the height be h and the distance from B to the foot of the building be x.

From B: tan 45° = h/x, so x = h (since tan 45° = 1).

From A: tan 30° = h/(x + 20). Since tan 30° = 1/√3, this gives x + 20 = h√3.

Substitute x = h: h + 20 = h√3, so h(√3 − 1) = 20 and h = 20 ÷ (√3 − 1) = 20 ÷ 0.732 ≈ 27.3 m.

Worked example 7: two cars seen from a tower

From the top of a tower 60 m high, the angles of depression of two cars on the same side of the tower, in a straight line with its foot, are 30° and 45°. Find the distance between the cars.

The nearer car has the larger angle of depression (45°). Its distance from the foot = 60 ÷ tan 45° = 60 m.

The farther car: distance = 60 ÷ tan 30° = 60√3 ≈ 103.9 m.

Distance between the cars = 103.9 − 60 ≈ 43.9 m.

Worked example 8: a flagpole on a building

A flagpole stands on top of a building. From a point 40 m from the foot of the building, the angles of elevation of the bottom and top of the flagpole are 38° and 45°. Find the length of the flagpole.

Height to the top = 40 × tan 45° = 40 m. Height to the bottom = 40 × tan 38° ≈ 31.25 m.

Length of the flagpole = 40 − 31.25 ≈ 8.75 m.

Combining with Bearings

Some WAEC questions combine an angle of elevation with bearings: for example, a mast is observed from two points that are on different bearings. Find each horizontal distance with tan, then use Pythagoras' theorem, or the sine and cosine rules, in the horizontal plane. The tutorial on sine and cosine rules covers bearings in more detail.

Exam Tips and Common Mistakes

Measure from the horizontal. If a question gives the angle a ladder makes with a wall (the vertical), subtract from 90° first.
Move the angle of depression to the bottom of the triangle using alternate angles.
Divide when the unknown is in the denominator, as in d = h ÷ tan θ.
Add the observer's height when the angle is measured from eye level.
Keep full calculator values until the final step, then round to the accuracy asked for, usually 1 decimal place or 3 significant figures.
Use exact values where possible: tan 45° = 1, tan 30° = 1/√3 and tan 60° = √3.

Angles of Elevation and Depression: Practice Questions

From a point 30 m from the foot of a tower, the angle of elevation of its top is 60°. Find the height of the tower.
From the top of a lighthouse 45 m high, the angle of depression of a ship is 18°. How far is the ship from the lighthouse?
A tree 15 m tall is observed from a point 20 m from its foot. Find the angle of elevation of the top.
An aeroplane flying at a height of 1500 m is seen at an angle of elevation of 25°. Find its horizontal distance from the observer.
Two observers stand on opposite sides of a tower 50 m high, in a straight line with its foot. Their angles of elevation of the top are 30° and 45°. How far apart are they?

Answers: 1. 30 × tan 60° = 30√3 ≈ 52.0 m. 2. 45 ÷ tan 18° ≈ 138.5 m. 3. tan⁻¹(15/20) ≈ 36.9°. 4. 1500 ÷ tan 25° ≈ 3217 m. 5. 50 ÷ tan 30° + 50 ÷ tan 45° = 86.6 + 50 ≈ 136.6 m.

Conclusion

Angles of elevation and depression are both measured from the horizontal, and they are equal for two points looking at each other, because they are alternate angles. Every problem reduces to one or two right-angled triangles: draw a clear diagram, place the angle carefully, and use tan for height and horizontal distance, or sine and cosine when a slanting length is involved. In two-triangle problems, express each distance in terms of the common height and form an equation.