Heights and Distances — Trigonometry Formulas & Examples

TL;DR

Heights and distances is the branch of trigonometry that finds the height of an object or the distance to it using a measured angle and one known length — without ever climbing or pacing it out. This article covers the angle of elevation and angle of depression, the line-of-sight idea, the three-step method, the formulas, and six worked examples on towers, poles, and buildings.

How Surveyors Measure a Mountain They Cannot Climb

The summit of Mount Everest was fixed at 8,848 metres in 1856 — by men standing on the plains of India more than 150 kilometres away, who never set foot on it. They used theodolites to measure angles, a known baseline distance, and trigonometry. That is the whole promise of heights and distances: if you can see the top of something and measure the angle your eye makes with the horizontal, you can compute its height from solid ground. No ladder, no drone, no climb.

What Are Heights and Distances in Trigonometry?

Heights and distances is the application of trigonometric ratios to find the height of a distant object or the horizontal distance to it, using a measured angle and one known length. It is the practical face of trigonometry — often taught under the heading Some Applications of Trigonometry — and it rests entirely on the right triangle formed by the object, the observer, and the ground.

Three terms carry the whole topic, and each must be clear before any problem makes sense.

• Line of sight — the straight line from the observer's eye to the object being viewed.
• Angle of elevation — the angle the line of sight makes with the horizontal when the object is above the observer (you look up).
• Angle of depression — the angle the line of sight makes with the horizontal when the object is below the observer (you look down).

In every case the object, the observer's position, and the foot of the object form a right triangle, and the six trigonometric ratios connect its angle to its sides. That is the engine; the rest is choosing the right ratio.

Angle of elevation versus angle of depression

The single most useful fact about these two angles: the angle of depression from a high point to a low point equals the angle of elevation from the low point back up to the high point. They are alternate angles between two parallel horizontals cut by the same line of sight. So a depression problem can always be redrawn as an elevation problem — handy when the right angle is easier to spot from the ground. The full definitions live in angle of elevation and angle of depression.

How Do You Find Heights and Distances? The Method

Every heights-and-distances problem yields to the same three steps.

1. Draw the right triangle. Mark the object as the vertical side, the ground as the horizontal side, and the line of sight as the hypotenuse. Label the given angle and the given length.
2. Pick the ratio that links what you know to what you want. If you know the angle and the base and want the height, use tan(opposite over adjacent). If you know the angle and the hypotenuse, use sin or cos.
3. Substitute the standard value and solve. The angle is almost always 30°, 45°, or 60°, so its ratio is an exact surd you can read off the table.

The standard-angle values you will reach for constantly:

The full set is in the trigonometric table. The most common formula, by far, is the elevation case:

( an(\text{angle of elevation}) = \frac{\text{height of object}}{\text{horizontal distance}})

where the height is the side opposite the angle and the distance is the side adjacent to it.

Examples of Heights and Distances

Example 1

The angle of elevation of the top of a tower from a point 30 m away on level ground is 30°. Find the height of the tower.

Draw the right triangle: tower is the opposite side (height h), ground distance 30 m is the adjacent side, angle is 30°.

(\tan 30° = \frac{h}{30})

(\frac{1}{\sqrt{3}} = \frac{h}{30} \implies h = \frac{30}{\sqrt{3}} = 10\sqrt{3} \approx 17.32) m.

Final answer: h = 10√3 ≈ 17.32 m.

Example 2

A 12 m ladder leans against a wall and makes a 60° angle with the ground. How high up the wall does it reach?

The height is opposite the 60° angle, and the side we know is the hypotenuse — thus,

(\sin 60° = \frac{\text{height}}{12} \implies \text{height} = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3} \approx 10.39) m.

Final answer: height ≈ 10.39 m.

Example 3

A bird sits on top of a tree. From a point 40 m from the base of the tree, the angle of elevation of the bird is 45°. How high is the bird?

At 45°, opposite and adjacent are equal.

(\tan 45° = \frac{h}{40} \implies h = 40) m.

Final answer: the bird is 40 m high.

Example 4

From the top of a 50 m cliff, the angle of depression of a boat at sea is 30°. How far is the boat from the foot of the cliff?

(\tan 30° = \frac{50}{d} \implies d = 50\sqrt{3} \approx 86.6 m.

Final answer: the boat is approximately 86.6 m from the foot of the cliff.

Example 5

Two poles of equal height stand on opposite sides of an 80 m wide road. From a point between them on the road, the angles of elevation of the tops are 60° and 30°. Find the height of the poles and the position of the point.

Let the point be x m from the foot of the first pole, so it is (80−x) m from the second. Setting up the equations:

1. (\tan 60° = \frac{h}{x} \implies h = x\sqrt{3})
2. (\tan 30° = \frac{h}{80 - x} \implies h = \frac{80 - x}{\sqrt{3}} )

Setting these equal,

(x\sqrt{3} = \frac{80 - x}{\sqrt{3}} \implies 3x = 80 - x \implies 4x = 80 \implies x = 20.)

Then h = 20√3 ≈ 34.64 m.

Final answer: each pole is approximately 34.64 m tall, and the point is 20 m from the first pole.

Example 6

A person standing on the ground finds the angle of elevation of the top of a building to be 45°. On walking 20 m toward the building, the angle becomes 60°. Find the height of the building.

Setting the distances:

1. (\tan 60° = \frac{h}{x} \implies h = x\sqrt{3})
2. (\tan 45° = \frac{h}{x + 20} \implies h = x + 20)

Setting these equal,

(x\sqrt{3} = x + 20 \implies x(\sqrt{3} - 1) = 20 \implies x = \frac{20}{\sqrt{3} - 1} = 10(\sqrt{3} + 1) \implies h = 30 + 10\sqrt{3} \approx 47.32 m.

Final answer: height ≈ 47.32 m.

Why Heights and Distances Matter Beyond the Exam

Heights and distances exists because there are countless objects we need to measure but cannot reach — and an angle plus a known length is almost always cheaper to obtain than a direct measurement.

• Surveying and mapping.
• Construction and safety.
• Navigation and aviation.

Key Takeaways

• Heights and distances uses one measured angle and one known length to find an unreachable height or distance through a right triangle.
• The angle of elevation is measured up from the horizontal; the angle of depression is measured down.
• The method is always: draw the triangle, choose the ratio that links the known and unknown sides, substitute the standard value, solve.

Practice Before Moving On

1. The angle of elevation of the top of a 15 m pole from a point on the ground is 30°. How far is the point from the foot of the pole?
2. From the top of a 20 m building, the angle of depression of a car is 45°. How far is the car from the building?
3. A kite is flying at a height of 60 m. The string makes a 60° angle with the ground. Find the length of the string (assume it is straight).

Answer to Question 1: d ≈ 25.98 m. Answer to Question 2: 20 m. Answer to Question 3: string ≈ 69.28 m.