Height and Distance Problems In Trigonometry

What are Heights and Distances?

Think about the last time you stood at the foot of a tall building and looked up. You had no way to physically measure its height, yet engineers, architects, and surveyors know its exact measurement down to the centimetre. This is precisely what heights and distances is about finding measurements you cannot directly take.

Heights and Distances is a real world application of trigonometry. It uses the relationship between angles and sides of a right-angled triangle to find unknown vertical heights or horizontal distances. Instead of climbing a tower or crossing a river to measure it, you stand at a safe point, measure an angle, and calculate what you need using simple formulas.

The concept rests on two foundational ideas the angle of elevation and the angle of depression.

One important fact worth knowing right away when two parallel horizontal lines are involved (the observer's eye level and the ground), the angle of elevation from below and the angle of depression from above are always equal. This equality comes from the property of alternate interior angles, and it is used constantly when solving problems.

How to Find Heights and Distances

The method is rooted in right-angled triangles. Whenever you draw a diagram of a height and distance problem, a right triangle almost always appears naturally. One side represents the height (the object you want to measure), another represents the horizontal distance (how far away you are), and the angle between the horizontal ground and the line of sight becomes the key input.

The three sides of this triangle have specific names based on their position relative to the angle you are working with:

In most height and distance problems, tangent (tan θ) is the ratio of choice. That is because the two values you usually deal with the height of an object and the horizontal distance from it are the opposite and adjacent sides of the triangle. The hypotenuse (the slant line of sight) rarely features directly in basic problems unless the length of a rope, ladder, or incline is involved.

Here is a straightforward approach to solving any height and distance problem:

Trigonometric Ratios of Standard Angles

Instead of calculating angles every time, mathematics provides a simple table of standard angles and their trigonometric ratios. These common angles 0°,30°,45°,60°,and90°are used in most height and distance problems in school and exams. Learning and remembering these values makes solving problems much easier and faster.

Heights and Distances Formulas

Every height and distance problem fits into one of a few standard situations. Once you recognise the situation, the formula follows naturally. Below are the core formulas covering most cases you will face, from the simplest single angle problems to the more layered two-observer scenarios.

Heights and Distances Example Questions

Example 1: Finding the Height of a Tower

Question: From a point on the ground 40 m away from the base of a tower, the angle of elevation of the top of the tower is 60°. Find the height of the tower.

60° h = ? 40 m Point Tower

Solution:

tan 60° = h / 40

        √3 = h / 40

      h = 40√3

Height of the tower = 40√3 ≈ 69.28 m

Example 2 : Finding Distance Using Angle of Depression

Question: From the top of a lighthouse 60 m high, the angle of depression of a ship at sea is 30°. Find the horizontal distance of the ship from the base of the lighthouse.

30° 60 m d = ? Lighthouse Ship

Solution: Using alternate angles, the angle of elevation from the ship = 30°.

tan 30° = 60 / d

  1/√3 = 60 / d

     d = 60√3

Horizontal distance of ship = 60√3 ≈ 103.92 m

Example 3 : Two Angles, Two Ships (Same Side)

Question: From the top of a cliff 100 m high, the angles of depression of two ships on the same side of the cliff are 30° and 45°. Find the distance between the two ships.

Cliff 45° 30° 100 m Distance = ? Ship 1 Ship 2

For Ship 1 (θ = 45°):  tan 45° = 100/d₁ ; d₁ = 100 m

For Ship 2 (θ = 30°):  tan 30° = 100/d₂ ; d₂ = 100√3 m

Distance between ships = d₂ − d₁ 

                         = 100√3 − 100

                         = 100(√3 − 1)

Distance between the two ships = 100(√3 − 1) ≈ 73.2 m.

Heights and Distances Practice Problems

Working through problems on your own is the Easy way to build confidence with this topic. These practice questions cover the full range from direct single step problems to multi angle scenarios.

1. A flagpole stands vertically on the ground. From a point 20 m away, the angle of elevation of the top of the pole is 45°. Find the height of the flagpole. Hint: tan 45° = 1, so height equals horizontal distance directly.

2. The angle of elevation of the top of a tower from a point on the ground is 30°. On moving 20 m closer to the tower, the angle becomes 60°. Find the height of the tower. Hint: Set up two equations using tan 30° and tan 60°, then solve simultaneously.

3. A kite is flying at a height of 75 m above the ground. The string attached to the kite makes an angle of 45° with the horizontal. Find the length of the string. Use sin 45° = height / string length.

4. Two poles of equal height stand on opposite sides of a road that is 80 m wide. From a point between them on the road, the angles of elevation of the tops of the poles are 60° and 30°. Find the height of each pole and the position of the point on the road. Hint: Let the point divide the road into x and (80 − x). Form two tan equations and equate heights.

5. From the top of a 50 m tall building, the angles of depression of the top and bottom of another building are 30° and 60° respectively. Find the height of the second building and its horizontal distance from the first. Hint: Use angle of