Some Applications of Trigonometry
CBSE Class 10 · Mathematics · Notes, formulas and practice questions
This chapter applies the trigonometric ratios to find heights of towers and buildings, distances to objects, and lengths of ladders or kite strings, using the angles of elevation and depression for the standard angles 30°, 45° and 60°.
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This chapter is about using the trigonometric ratios to find heights and distances that cannot be measured directly. When a straight line from the eye of an observer meets an object, that line is called the line of sight. The angle it makes with the horizontal is an angle of elevation (if the object is above) or depression (if below). We form a right triangle by drawing a vertical line from the object to the ground, then apply the ratios of 30°, 45° and 60° to solve.
The core skill is to identify the correct right triangle and choose the ratio that connects the given side and the required side. For example, tan is used when the two legs of the triangle are involved, while sin or cos is used when the hypotenuse (the line of sight, string or ladder) appears. Many errors come from mixing these up, or from placing the angle in the wrong corner of the triangle. Drawing a clean labelled diagram before starting the calculation prevents most of these mistakes.
Another common source of error is forgetting to include the observer's own height when the observation is made from a person standing on the ground. When the observer is on a building or cliff, the angle of depression from the top is equal to the angle of elevation from the object below, because the two horizontal lines are parallel. Recognising this equality lets you solve the problem using a single right-triangle ratio.
The chapter introduces no new trigonometry; the values needed come from the standard table for the angles 30°, 45° and 60°. The examination tests this as a word problem, usually involving a tower, building, kite, ladder or cliff. To score, first read the problem and mark the given angle and side, then choose the appropriate ratio. Practice with a variety of situations builds speed and confidence.
Key terms
- Line of sight
- The straight line drawn from the observer's eye to the object being viewed. In height and distance problems, it is always one side of the right triangle formed.
- Horizontal line
- A line through the observer's eye that is parallel to the ground. It is used as the reference for measuring the angle of elevation or depression.
- Angle of elevation
- When the object is above the observer, the angle between the horizontal line and the line of sight, measured upwards. For example, looking at the top of a tower from a point on the ground gives an angle of elevation.
- Angle of depression
- When the object is below the observer, the angle between the horizontal line and the line of sight, measured downwards. For example, looking from the top of a building at an object on the ground gives an angle of depression.
- Equality of elevation and depression
- For the same line of sight, the angle of depression from the upper point equals the angle of elevation from the lower point. This happens because the two horizontals are parallel and the line of sight acts as a transversal.
- Height-distance right triangle
- In a typical problem, the vertical height, the horizontal distance on the ground and the line of sight form a right-angled triangle. The right angle lies at the point where the perpendicular from the object touches the ground.
- Eye-level adjustment
- Some problems state the height of the observer’s eye above the ground. When they do, the triangle gives only the part of the tower above eye level, so the observer’s height must be added at the end. When no such height is given — which is the usual case — the angle is measured from the ground and nothing is added. If the observer is on a cliff or a building, that given height is normally the vertical side of the triangle itself.
- Choosing the correct ratio
- Use tan θ when the problem involves the two perpendicular sides (height and distance). Use sin θ when the height and the line of sight are involved, and use cos θ when the distance and the line of sight are involved. Marking the known and unknown sides on the diagram makes this choice easy.
Formula sheet
| What | Formula | Notes |
|---|---|---|
| Height from horizontal distance and angle | h = d × tan θ | h is the vertical height, d is the horizontal distance from the point below the object to the observer, and θ is the angle of elevation (or depression) at the observer. |
| Horizontal distance from height and angle | d = h / tan θ | d is the horizontal distance, h is the vertical height, and θ is the angle of elevation or depression involved in the problem. |
| Height from length of line of sight and angle | h = l × sin θ | h is the vertical height of the object, l is the length of the line of sight, string, or ladder, and θ is the angle it makes with the horizontal ground. |
| Horizontal distance from length of line of sight and angle | d = l × cos θ | d is the horizontal distance from the foot of the observer to the point directly below the object, l is the length of the line of sight or ladder, and θ is the angle with the ground. |
Practice questions with answers
1. From a point 15 m away from the foot of a tower, the angle of elevation of the top is 60°. Find the height of the tower (take √3 ≈ 1.732).
Let h be the tower height. In the right triangle, tan 60° = h/15. Since tan 60° = √3, h = 15√3 m ≈ 15 × 1.732 = 25.98 m. The tower is approximately 25.98 m high.
2. A kite is flying at a vertical height of 30 m. The string is taut and makes a 45° angle with the ground. Find the length of the string.
Let l be the length of the string. The height, string and the ground form a right triangle, so sin 45° = 30/l, giving l = 30/sin 45° = 30√2 m ≈ 42.42 m. The string is approximately 42.42 m long.
3. From the top of a building 10√3 m high, the angle of depression of an object on the ground is 30°. How far is the object from the foot of the building?
The angle of depression at the top equals the angle of elevation at the object, so tan 30° = (10√3)/d, where d is the required distance. Therefore d = (10√3)/tan 30° = 10√3 × √3 = 30 m. The object is 30 m away.
4. A ladder leaning against a wall makes an angle of 30° with the ground. The foot of the ladder is 2.5 m away from the wall. Find the length of the ladder (take √3 ≈ 1.732).
Let l be the length of the ladder. The distance 2.5 m is adjacent to the 30° angle, so cos 30° = 2.5/l. Hence l = 2.5 / cos 30° = 2.5 / (√3/2) = 5/√3 = 5√3/3 m ≈ 2.89 m. The ladder is about 2.89 m long.
5. When the sun's elevation is θ, a vertical pole casts a shadow that is 1/√3 times its height. Find θ.
Let the height be h and the shadow be h/√3. Then tan θ = height/shadow = h/(h/√3) = √3. Therefore θ = 60°, so the sun's altitude is 60°.
6. From the top of a cliff 120 m high, the angle of depression of a boat is 30°. Calculate the horizontal distance of the boat from the base of the cliff.
Using tan 30° = 120/d, where d is the horizontal distance, we get d = 120/tan 30° = 120√3 m. Taking √3 ≈ 1.732, d ≈ 207.8 m. The boat is about 207.8 m away from the cliff base.
7. Two men on opposite sides of a tower of height 50 m observe the top at angles of elevation 45° and 30°. Find the distance between the two men.
Let the distances from the tower to the men be x and y. From tan 45° = 50/x, we get x = 50 m. From tan 30° = 50/y, we get y = 50√3 m. The distance between the men is x + y = 50 + 50√3 = 50(1 + √3) m ≈ 136.6 m.
8. A kite is flying at a height of 90 m from the ground. If the string makes an angle of 60° with the ground, find the length of the string.
Let l be the string length. Since sin 60° = 90/l, l = 90/sin 60° = 90 × (2/√3) = 180/√3 = 60√3 m. Therefore the string length is 60√3 m, approximately 103.9 m.
9. Define the angle of depression of an object from an observer, with reference to the line of sight.
When the observer looks at an object lying below the horizontal line through his eye, the angle between that horizontal line and the line of sight is the angle of depression. It is measured downwards from the horizontal.
10. A man of height 1.6 m stands 12.5 m away from a tower. He sees the top of the tower at an angle of elevation of 45°. Find the height of the tower.
In the right triangle formed at eye level, the part of the tower above the eye is 12.5 × tan 45° = 12.5 m. Adding the man's height gives the full tower height = 12.5 + 1.6 = 14.1 m. The tower is 14.1 m high.
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