Some Applications of Trigonometry
This chapter puts trigonometry to work in the real world — finding the heights of towers and the distances of objects without measuring them directly. The key idea is to draw a right triangle from the situation and use the angle of elevation or depression with a trig ratio. The problems are visual and, once the diagram is right, quick to solve.
Learning objectives
- Define line of sight, angle of elevation and angle of depression.
- Draw a correct right-triangle diagram from a word problem.
- Choose the right trig ratio (sin, cos or tan) for the given sides.
- Solve heights-and-distances problems using standard angles.
- Handle problems with two observation points or angles.
Key concepts
Line of sight and angles
The line of sight is the line from the observer's eye to the object. The angle of elevation is the angle this line makes with the horizontal when the object is above; the angle of depression is the angle below the horizontal when the object is lower.
Setting up the triangle
Represent the height as the vertical side and the ground distance as the horizontal side of a right triangle, with the angle of elevation/depression at the observation point. Then height/distance = tan(angle).
Choosing the ratio
Use tan when the horizontal distance and height are involved, sin when the hypotenuse (e.g. a ladder or rope) and the opposite side are involved, and cos when the hypotenuse and the adjacent side are involved.
Depression equals elevation
The angle of depression of a point from an observer equals the angle of elevation of the observer from that point (they are alternate angles), which is useful in two-point problems.
Important formulas
Height from elevation
height = distance × tan(angle of elevation)
Distance from height
distance = height ÷ tan(angle of depression)
Using hypotenuse
opposite = hypotenuse × sin(angle)
Key definitions
- Angle of elevation
- The angle between the horizontal and the line of sight to an object above the observer.
- Angle of depression
- The angle between the horizontal and the line of sight to an object below the observer.
- Line of sight
- The straight line from the observer's eye to the object being viewed.
Solved examples
Q1. The angle of elevation of the top of a pole from a point 30 m away on the ground is 30°. Find the height of the pole.
Solution: height = distance × tan 30° = 30 × (1/√3) = 30/√3 = 10√3 m ≈ 17.3 m.
Q2. From the top of a 50 m building, the angle of depression of a car is 45°. How far is the car from the foot of the building?
Solution: tan 45° = height/distance ⇒ 1 = 50/distance ⇒ distance = 50 m.
Q3. A ladder leaning against a wall makes a 60° angle with the ground, and its foot is 2 m from the wall. Find the length of the ladder.
Solution: cos 60° = adjacent/hypotenuse = 2/ladder ⇒ 1/2 = 2/ladder ⇒ ladder = 4 m.
Common mistakes to avoid
- Confusing elevation (looking up) with depression (looking down).
- Measuring the angle of depression from the vertical instead of the horizontal.
- Using tan where the hypotenuse is involved (should be sin or cos).
- Forgetting to add the observer's height when required.
Some Applications of Trigonometry — MCQ Quiz
10 questions with instant feedback. Use number keys 1–4 to answer.
The angle of elevation is measured from the horizontal to the line of sight when the object is:
Practice questions
Short answer
Define angle of depression.
The angle below the horizontal from the observer's eye to an object that is lower than the observer.
A tower is 20 m high and a point is 20 m away. What is the angle of elevation of the top?
tan θ = 20/20 = 1, so θ = 45°.
Which ratio uses the hypotenuse and the opposite side?
sine (sin).
Long answer
From a point on the ground, the angle of elevation of the top of a 60 m tower is 60°. Find the distance of the point from the foot of the tower.
tan 60° = 60/d ⇒ √3 = 60/d ⇒ d = 60/√3 = 20√3 m ≈ 34.6 m.
An observer 1.5 m tall is 28.5 m from a tower. The angle of elevation of the top of the tower from his eyes is 45°. Find the height of the tower.
From the eye level, height above the eyes = 28.5 × tan 45° = 28.5 m. Adding the observer's height: total tower height = 28.5 + 1.5 = 30 m.
HOTS (Higher Order Thinking)
The angles of elevation of the top of a tower from two points at distances 4 m and 9 m from the base (on the same side) are complementary. Find the height of the tower.
Let the angles be θ and 90° − θ, and height h. Then tan θ = h/9 and tan(90° − θ) = cot θ = h/4. Multiplying: (h/9)(h/4) = tan θ · cot θ = 1, so h² = 36 ⇒ h = 6 m.
Why do we often get 'nice' answers like 10√3 in these problems?
Because the questions use standard angles (30°, 45°, 60°) whose trig values are simple surds, the arithmetic stays clean.
Quick revision
Revision notes
- Line of sight: eye to object; elevation = up, depression = down.
- tan(angle) = height ÷ distance.
- Use sin/cos when the hypotenuse (ladder, rope, string) is involved.
- Angle of depression = angle of elevation (alternate angles).
Key takeaways
- A correct, labelled diagram is half the solution.
- Pick the ratio that connects the side you know to the side you want.
- Watch for the observer's own height in some problems.
Frequently asked questions
What is the difference between elevation and depression?
Elevation is the angle looking up at a higher object; depression is the angle looking down at a lower object — both measured from the horizontal.
Which trig ratio should I use?
tan for height and ground distance, sin for hypotenuse and opposite, cos for hypotenuse and adjacent.
Do I need to memorise new formulas for this chapter?
No — you reuse the standard trig ratios; the skill is translating the word problem into a right triangle.