Why heights and distances matter in everyday life
Ever wondered how engineers figure out the height of a tower without climbing it? Or how you can guess the width of a river from the bank? That’s the magic of heights‑and‑distances problems – they let you measure the unseen with a little angle and some simple math.
💡 In Simple Words: When you know an angle of elevation or depression and one side of a right‑angled triangle, you can use sine, cosine or tangent to find the missing height or distance. It’s like using the tilt of a camera to guess how tall a building is.
What is a height‑and‑distance problem?
A height‑and‑distance problem is a word problem that involves a right‑angled triangle. The triangle is formed by the ground, a line of sight, and the object’s height or distance. The key angles are:
- Angle of elevation: the angle between the horizontal line of sight and the line looking up to the top of an object.
- Angle of depression: the angle between the horizontal line of sight and the line looking down to the bottom of an object.
Think of the angle of elevation like looking up at a kite, and the angle of depression like looking down at a car parked far away.
Key trigonometric ideas you need
Trigonometry deals with ratios of sides in a right‑angled triangle. The three basic ratios are:
- Sine (sin): opposite side ÷ hypotenuse.
- Cosine (cos): adjacent side ÷ hypotenuse.
- Tangent (tan): opposite side ÷ adjacent side.
Imagine water flowing through a pipe. The “opposite” side is like the water that goes up, the “adjacent” side is the water that runs straight, and the “hypotenuse” is the diagonal pipe that connects the two.
Step‑by‑step solving method
Most height‑and‑distance questions follow the same pattern. Follow these steps, and you’ll rarely get stuck.
- Draw a clear diagram. Label the known angles and sides.
- Identify which trigonometric ratio matches the known side and the side you need.
- Write the equation using sin, cos or tan.
- Solve for the unknown side.
- Check the answer by seeing if it makes sense (e.g., a tree can’t be 2 km tall).
Worked example 1: Finding the height of a pole
Problem: From a point 30 m away from the base of a pole, the angle of elevation to the top is 35°. What is the height of the pole?
Solution:
- Draw the right‑angled triangle. The horizontal ground is 30 m (adjacent side). The height we want is the opposite side.
- Since we have the angle and the adjacent side, use the tangent ratio:
tan 35° = opposite ÷ adjacent. - Rearrange: opposite = adjacent × tan 35° = 30 × tan 35°.
- Using a calculator, tan 35° ≈ 0.7002. Multiply: 30 × 0.7002 ≈ 21.0 m.
- So the pole is about 21 m tall.
Worked example 2: Distance across a river
Problem: A man stands on the near bank of a river. He spots a tree on the opposite bank at an angle of depression of 20°. He is 5 m above the water level on a small boat. How wide is the river?
Solution:
- Sketch the triangle. The vertical drop from the man to the water is 5 m (opposite side). The distance across the river is the adjacent side we need.
- Angle of depression equals the angle of elevation from the tree to the man, so we can use tan 20° = opposite ÷ adjacent.
- Rearrange: adjacent = opposite ÷ tan 20° = 5 ÷ tan 20°.
- tan 20° ≈ 0.36397. Divide: 5 ÷ 0.36397 ≈ 13.7 m.
- The river is roughly 13.7 m wide.
Quick reference table
| Situation | Known side | Needed side | Use |
|---|---|---|---|
| Angle of elevation, ground distance known | Adjacent (ground) | Opposite (height) | tan θ = opposite/adjacent |
| Angle of depression, height above ground known | Opposite (vertical) | Adjacent (distance) | tan θ = opposite/adjacent |
| Angle of elevation, height known | Opposite (height) | Adjacent (distance) | tan θ = opposite/adjacent → adjacent = opposite / tan θ |
| Angle of elevation, hypotenuse known | Hypotenuse | Opposite (height) | sin θ = opposite/hypotenuse |
📝 Likely Exam Questions
- Q1: From a point 50 m from a building, the angle of elevation to the top is 40°. Find the building’s height.
Answer: Height = 50 × tan 40° ≈ 42 m. - Q2: A lighthouse is 30 m tall. From a boat, the angle of elevation to the top is 15°. How far is the boat from the base?
Answer: Distance = 30 ÷ tan 15° ≈ 110 m. - Q3: A person 1.6 m tall stands 20 m from a hill. The angle of elevation to the hill’s peak is 25°. Find the hill’s height above the ground.
Answer: Height = 20 × tan 25° + 1.6 ≈ 10.5 m + 1.6 = 12.1 m. - Q4: From a balcony 10 m above ground, the angle of depression to a car on the road is 30°. How far is the car from the base of the building?
Answer: Distance = 10 ÷ tan 30° ≈ 17.3 m.