In a heights and distances problem, mathematics is usually the easy part. One right-angled triangle, one trigonometric ratio, and a short calculation. What goes wrong is the picture. A diagram with the angle in the wrong place, or the wrong side labelled, sends every later line in a perfectly logical direction towards a wrong answer.
This guide starts with the vocabulary, then gives a reliable method for drawing the figure, four worked problems, and a table of the diagram mistakes that cost the most marks. If you need the trigonometry itself first, read these notes on the introduction to trigonometry, the trigonometric ratios and the values for the specific angles, and for this chapter in full, the notes on some applications of trigonometry. The thinking is close to that in similar triangles in real life, where shadows and proportions find heights without climbing.
The Vocabulary That Draws the Picture
- Line of sight: the straight line from the observer’s eye to the object being looked at.
- Horizontal level: the level line through the observer’s eye. Every angle in these problems is measured from this line.
- Angle of elevation: the angle between the horizontal and the line of sight when the object is above the observer. The observer looks up.
- Angle of depression: the angle between the horizontal and the line of sight when the object is below the observer. The observer looks down.
Two facts remove most of the confusion. First, both angles are measured from the horizontal, never from the vertical. Second, the angle of depression of an object from the top of a tower equals the angle of elevation of the top of the tower from that object, because they are alternate angles between parallel lines.
Values You Need at Your Fingertips
| Angle | sin | cos | tan |
| 30° | 1/2 | √3/2 | 1/√3 |
| 45° | 1/√2 | 1/√2 | 1 |
| 60° | √3/2 | 1/2 | √3 |
In height problems, tan is used most, because it links the height (opposite side) and the horizontal distance (adjacent side), which are the two things the problems usually involve.
A Method for Drawing the Figure
- Read the problem twice, and underline the given angles and lengths.
- Draw the ground as a horizontal line, and the tower or building as a vertical line, with a right angle marked where they meet.
- Mark the observer as a point on the ground (or at the right height if the observer is on a building).
- Draw the line of sight, and then the horizontal line through the observer’s eye.
- Mark the angle at the observer’s end, between the horizontal and the line of sight, and write its value.
- Label what is unknown with a letter, and what is given with its number.
- Pick the right-angled triangle, and choose the ratio that connects the known and unknown sides.
Four Worked Problems
Problem one: a single angle
From a point on the ground 30 m from the foot of a tower, the angle of elevation of the top is 30°. Find the height of the tower.
tan 30° = height ÷ 30, so 1/√3 = h/30 and h = 30/√3 = 10√3 m, which is about 17.3 m.
Problem two: an observer with height
A boy 1.5 m tall stands 30 m from a tower. The angle of elevation of the top of the tower from his eyes is 30°. Find the height of the tower.
The triangle begins at the boy’s eye level, so the height found by the angle is measured above 1.5 m. Height above eye level = 30 tan 30° = 10√3 ≈ 17.32 m. The tower’s height is 17.32 + 1.5 = 18.82 m. Forgetting the 1.5 m is one of the most common slips.
Problem three: an angle of depression
From the top of a tower 50 m high, the angle of depression of a car on the ground is 30°. How far is the car from the foot of the tower?
The angle of depression equals the angle of elevation of the tower’s top from the car, so the angle at the car is 30°. tan 30° = 50 ÷ d, so d = 50 ÷ (1/√3) = 50√3 m, about 86.6 m.
Problem four: two positions
The angle of elevation of the top of a tower from a point on the ground is 30°. After walking 20 m towards the tower, the angle becomes 60°. Find the height of the tower.
Let the height be h, and let the distance from the nearer point to the foot be x. From the nearer point, tan 60° = h/x, so x = h/√3. From the farther point, tan 30° = h/(x + 20), so x + 20 = h√3. Substitute: h/√3 + 20 = h√3, so 20 = h(√3 − 1/√3) = h × 2/√3. Therefore h = 20√3/2 = 10√3 m, about 17.3 m.
Check. x = h/√3 = 10 m, and x + 20 = 30 m. tan 30° = 10√3/30 = √3/3 = 1/√3. ✓
Diagram Mistakes and Their Fixes
| Mistake | What it looks like | The fix |
| Measuring the angle from the vertical | Putting the angle of depression between the line of sight and the tower | Draw the horizontal first, and measure the angle from it |
| Marking the angle at the wrong end | The angle of elevation drawn at the top of the tower | The angle of elevation is at the observer |
| Missing the right angle | A tower that leans | Mark the right angle at the base |
| Ignoring the observer’s height | Using the ground as the starting level | Add the observer’s height to the result |
| Not using alternate angles | Treating the depression and elevation as different | They are equal as alternate angles |
| Wrong ratio | Using sin where tan is needed | Identify opposite and adjacent first |
| Reading the diagram’s scale | Drawing it so crowded that sides are unclear | Draw it big, and label neatly |
| Mixing two triangles | Using one angle with the wrong base | Write a separate equation for each triangle |
| Dropping units | The answer given as a bare number | Write “m” or “km” |
| Not simplifying surds | Leaving 30/√3 | Rationalise: 30/√3 = 10√3 |
Two More Situations You Will See
- Two buildings: the angle of elevation of the top of one building from the other’s top or base. Draw both buildings, mark a horizontal line from the shorter building’s top, and the difference in heights becomes the opposite side of a triangle.
- A ladder or a kite string: both are the hypotenuse of a right triangle, so you often need sin or cos, since the hypotenuse is involved.
Identities for the Next Step
Occasionally, problems use trigonometric identities to simplify or verify an expression. The next guide, on trigonometric identities and a memory framework, shows how to learn them without rote memorising.
Applications of Trigonometry in Your Syllabus
Some Applications of Trigonometry is a chapter of the Class 10 Maths book in both the CBSE and the Karnataka SSLC textbooks. Practise with the chapter MCQs, the important questions and the chapter notes. The ideas extend into Class 11 trigonometric functions, as in these notes on trigonometric functions, which matter for JEE Main and KCET. Confirm the current portions with your own board.
Frequently Asked Questions
Q1. What is the angle of elevation?
The angle between the horizontal and the line of sight when the object being viewed is above the observer’s eye level.
Q2. What is the angle of depression?
The angle between the horizontal and the line of sight when the object is below the observer’s eye level.
Q3. How are the angle of depression and the angle of elevation related?
The angle of depression of an object from a point equals the angle of elevation of that point from the object, because they are alternate angles between parallel horizontal lines.
Q4. Which trigonometric ratio is used most in height problems?
tan, because it connects the height (opposite side) with the horizontal distance (adjacent side).
Q5. How do I handle the height of the observer?
The angle is measured from the observer’s eye, so the height found by trigonometry is above eye level. Add the observer’s height to get the total height.
Draw First, Calculate Second
A heights and distances problem is solved the moment the diagram is right. Draw the horizontal, mark the angle at the observer, label the triangle, and the calculation is a single line of trigonometry.
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