Chapter 9: Some Applications of Trigonometry
1. Heights and Distances
This chapter applies trigonometric ratios to find the heights of objects (towers, buildings, trees, poles) and the distances between points without actually measuring them. The line of sight, the horizontal, and the vertical object form a right triangle, and the ratios sin, cos and tan connect its sides with the angle of observation.
2. Basic Terms
- Line of sight: The line drawn from the eye of an observer to the point on the object viewed by the observer.
- Horizontal level: The horizontal line through the eye of the observer.
- Angle of elevation: The angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level, i.e. when we raise our head to look at the object.
- Angle of depression: The angle formed by the line of sight with the horizontal when the point being viewed is below the horizontal level, i.e. when we lower our head to look at the object.
Key fact: The angle of depression of a point B as seen from a point A is equal to the angle of elevation of A as seen from B (they are alternate interior angles between the two parallel horizontal lines).
The angle of depression is measured from the horizontal at the observer's eye, NOT from the vertical wall/tower. Drawing the horizontal line at the top of the tower and marking the depression angle there, then transferring it to the ground point as an equal alternate angle, is the single most important step in depression problems.
3. Which Ratio to Use?
| Known / Required sides | Ratio to use |
|---|---|
| Opposite (height) and Adjacent (distance) | tan θ = Opposite/Adjacent |
| Opposite (height) and Hypotenuse (slant length: ladder, string, slide) | sin θ = Opposite/Hypotenuse |
| Adjacent (distance) and Hypotenuse | cos θ = Adjacent/Hypotenuse |
In most problems the height and horizontal distance are involved, so tan θ is used most often. Values needed (only 30°, 45°, 60° are in the syllabus):
tan 30° = 1/√3 | tan 45° = 1 | tan 60° = √3 | sin 30° = 1/2, sin 60° = √3/2, sin 45° = 1/√2 | cos 30° = √3/2, cos 60° = 1/2, cos 45° = 1/√2
4. Standard Method for Height-Distance Problems
- Read and draw: Draw a neat figure showing the object as a vertical line, the ground as horizontal, and right angles marked. A correct figure carries marks by itself.
- Mark angles: Put the angle of elevation at the observer on the ground, or the angle of depression at the top (then transfer it down as an alternate angle).
- Account for observer's height: If the observer's eye is at height h (e.g. a 1.5 m tall boy), measure the angles from eye level and add the observer's height at the end.
- Choose the ratio: Pick tan/sin/cos so that the equation contains exactly one unknown.
- Form equations: One equation per right triangle. Two angles (two positions or two objects) give two equations — solve simultaneously.
- Solve and state: Simplify surds (use √3 ≈ 1.73 only if the question says so), and write the final answer with units and a statement.
5. Standard Situations (Board Patterns)
- Single tower, single angle: tan θ = h/d; find h or d directly. (1–2 marks)
- Ladder / string / slide (hypotenuse given): use sin or cos. Example: a ladder making 60° with the ground reaching a window uses sin 60° = height/ladder. (2 marks)
- Two angles from the same point (e.g. bottom and top of a tower on a hill, or building + flagstaff): two tan equations with common base distance; subtract to get the extra height. (3 marks)
- Two positions of the observer (walking towards the tower changes elevation from 30° to 60°): tan 30° = h/(d + x) and tan 60° = h/d; solve the pair. (3–5 marks)
- Angles of depression of two objects (two ships/cars seen from a lighthouse/tower on the same side or opposite sides): distance between objects = difference or sum of the two base distances. (3–5 marks)
- Object on both sides (a tree/pole between two points, or angles of elevation from opposite banks of a river): the two base segments add up to the total width. (3 marks)
- Broken tree: the broken part becomes the hypotenuse; total original height = standing part + broken (slant) part = d tan θ + d sec θ where d is the distance of the top from the foot. (3 marks)
- Moving object (balloon/aeroplane) at constant height: the height stays fixed; two elevations give two horizontal distances; distance travelled = difference of the distances. (5 marks / case study)
6. Worked Micro-Example
The angle of elevation of the top of a tower from a point 30 m away from its foot is 30°. Height?
tan 30° = h/30 ⇒ h = 30 × 1/√3 = 30/√3 = 10√3 m ≈ 17.32 m
Rationalise at the end: 30/√3 = 30√3/3 = 10√3. Leaving the answer as 30/√3 may lose the final accuracy mark. Also, never forget to add back the observer's eye height when it is given.
7. Exam Checklist
- Figure drawn and labelled with right angle, angles, known lengths — compulsory.
- Angle of depression transferred correctly using alternate angles, with the reason written.
- All angles used are 30°, 45° or 60° only, as per the syllabus — if your equation produces any other angle, re-check the figure.
- Final statement written in words: "Therefore, the height of the tower is 10√3 m."
Deleted in rationalised syllabus (2026-27): No major topic of this chapter is deleted. Note, however, that problems are restricted to simple situations involving at most two right triangles, and angles of elevation/depression of 30°, 45° and 60° only.