1. Chapter at a Glance
How tall is the Qutub Minar? How wide is a river you cannot cross? How high is a kite on a string? Surveyors and astronomers have answered such questions for centuries using trigonometry: measure one distance and one angle, and calculate the rest. In this chapter we learn the language of angles of elevation and depression and solve heights and distances problems — a favourite 3- to 5-mark board question.
- Line of sight, angle of elevation, angle of depression.
- Drawing correct diagrams for heights and distances problems.
- Using tan (most often), sin and cos to find unknown heights and distances.
- Problems with two angles (moving observer, two objects, top and bottom of a tower).
2. Key Terms
- Line of sight: the line from the observer's eye to the object.
- Angle of elevation: angle between the horizontal and the line of sight when the object is above the horizontal (you raise your head).
- Angle of depression: angle between the horizontal and the line of sight when the object is below the horizontal (you lower your head).
The angle of depression from A to B equals the angle of elevation from B to A, because the two horizontals are parallel (alternate angles).
3. Method
- Draw a neat diagram; mark the right angle, the known side and the angle(s).
- Choose the ratio linking the known side and the unknown side: tan for height ↔ horizontal distance, sin or cos when the slant length (string, ladder) is involved.
- Solve, rationalise if needed, and give units. Use √3 ≈ 1.732 when asked.
| Angle | tan | Height h for distance d |
|---|---|---|
| 30° | 1/√3 | d/√3 |
| 45° | 1 | d |
| 60° | √3 | d√3 |
4. One-angle Problems
A tower stands vertically on the ground. From a point 15 m from its foot, the angle of elevation of the top is 60°. Find the height.
Solution: tan 60° = h/15 → h = 15√3 ≈ 25.98 m.
A kite is flying at a height of 60 m. Its string makes 60° with the ground. Find the length of the string (assume it is straight).
Solution: sin 60° = 60/L → L = 120/√3 = 40√3 ≈ 69.28 m.
From the top of a 75 m lighthouse, the angle of depression of a ship is 30°. How far is the ship from the base?
Solution: tan 30° = 75/d → d = 75√3 ≈ 129.9 m.
5. Two-angle Problems
A man on the ground sees the top of a building at 30°. After walking 50 m towards it, the angle becomes 60°. Find the height.
Solution: Let height h and final distance x. tan 60° = h/x → h = x√3. tan 30° = h/(x + 50) → x + 50 = h√3 = 3x → x = 25. h = 25√3 ≈ 43.3 m.
From a point on the ground, the angles of elevation of the bottom and top of a flagstaff on top of a 20 m building are 45° and 60°. Find the flagstaff's length.
Solution: tan 45° = 20/d → d = 20. tan 60° = (20 + f)/20 → 20 + f = 20√3 → f = 20(√3 − 1) ≈ 14.64 m.
Two equal poles stand on either side of an 80 m wide road. From a point between them, the angles of elevation of their tops are 60° and 30°. Find the height of the poles and the distances of the point from them.
Solution: h = x√3 and h = (80 − x)/√3 → 3x = 80 − x → x = 20. h = 20√3 ≈ 34.64 m; distances 20 m and 60 m.
6. Solved Problems (Board Pattern)
A tree breaks and its top touches the ground making 30° with it, 8 m from the foot. Find the original height.
Solution: Standing part = 8 tan 30° = 8/√3. Broken part = 8/cos 30° = 16/√3. Height = 24/√3 = 8√3 ≈ 13.86 m.
From the top of a tower, the angle of depression of a car changes from 30° to 60° in 6 seconds as it approaches at uniform speed. How much more time will it take to reach the tower?
Solution: Distances: h√3 and h/√3. Distance covered in 6 s = h√3 − h/√3 = 2h/√3. Remaining h/√3 is half of that → 3 seconds.
From the top of a 7 m building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the tower's height.
Solution: Horizontal distance = 7 (tan 45°). Extra height above the building = 7 tan 60° = 7√3. Tower = 7(1 + √3) ≈ 19.12 m.
7. Master Formula Table
| Given | Use | Formula |
|---|---|---|
| Horizontal distance, angle | tan | h = d tan θ |
| Height, angle | tan | d = h / tan θ |
| Slant length (string/ladder), angle | sin | h = L sin θ |
| Slant length, angle | cos | d = L cos θ |
| Elevation = depression | Alternate angles | ∠ of elevation from B = ∠ of depression from A |
| Approximations | — | √3 ≈ 1.732, √2 ≈ 1.414 |
8. Master Summary Table
| Problem type | Setup | Key step |
|---|---|---|
| Single angle | One right triangle | One tan/sin equation |
| Moving observer | Two triangles, shared height | Express h twice; equate |
| Object on top of another | Bottom and top angles | Find d first |
| Two objects, one point | Distances add up | x + (w − x) |
| From a height | Elevation + depression | Split the tower at eye level |
9. Mind Map
- Elevation (up)
- Depression (down)
- tan: h ↔ d
- sin/cos: slant
- Moving observer
- Flagstaff, poles
- Mark right angle
- Label knowns
10. Tips and Tricks
- 45° means height = distance — a quick win.
- Elevations 30° and 60° from the same line: the farther distance is 3 times the nearer one.
- Keep answers in surd form until the last step, then use √3 ≈ 1.732.
- Marking the angle of depression at the observer on the ground instead of from the horizontal at the top.
- Using sin when the hypotenuse is not involved.
- Forgetting to add the observer's eye height when it is given.
11. Important One-liners
- Angle of elevation is measured upward from the horizontal.
- Angle of depression is measured downward from the horizontal.
- The angle of elevation of A from B equals the angle of depression of B from A.
- As an observer moves closer to a tower, the angle of elevation increases.
- If the angle of elevation is 45°, height equals horizontal distance.
12. Exam-pattern Questions with Detailed Answers
The ratio of the length of a pole to its shadow is 1 : √3. The sun's angle of elevation is:
- 30°
- 45°
- 60°
- 90°
Show answer
tan θ = 1/√3 → θ = 30°.
Answer: (A)
A ladder 10 m long makes 60° with the ground when it reaches a window. How high is the window?
Show answer
10 sin 60° = 5√3 ≈ 8.66 m.
Answer: 5√3 m
The angle of elevation of the top of a 30 m tower from a point on the ground is 45°. How far is the point from the foot?
Show answer
d = 30 (tan 45° = 1).
Answer: 30 m
A 1.5 m tall boy stands at some distance from a 30 m building. The angle of elevation of the top from his eyes rises from 30° to 60° as he walks towards it. Find the distance walked.
Show answer
Height above eye = 28.5. Distances: 28.5√3 and 28.5/√3. Walked = 28.5(√3 − 1/√3) = 28.5 × 2/√3 = 19√3 ≈ 32.9 m.
Answer: 19√3 m
From the top of a 100 m cliff, the angles of depression of two ships on opposite sides are 30° and 45°. Find the distance between the ships.
Show answer
Distances: 100√3 and 100. Total = 100(√3 + 1) ≈ 273.2 m.
Answer: 100(√3 + 1) m
A statue 1.6 m tall stands on a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and of the top of the pedestal is 45°. Find the pedestal's height.
Show answer
Let pedestal = p = d (45°). tan 60° = (p + 1.6)/p → p√3 = p + 1.6 → p = 1.6/(√3 − 1) = 0.8(√3 + 1) ≈ 2.19 m.
Answer: 0.8(√3 + 1) m
The shadow of a tower is 40 m longer when the sun's altitude is 30° than when it is 60°. Find the tower's height.
Show answer
h√3 − h/√3 = 40 → 2h/√3 = 40 → h = 20√3 ≈ 34.64 m.
Answer: 20√3 m
A kite string is 100 m long and makes 30° with the horizontal. Find the height of the kite.
Show answer
100 sin 30° = 50 m.
Answer: 50 m
The angles of elevation of the top of a tower from two points 4 m and 9 m from its base, on the same line, are complementary. Prove that the tower is 6 m tall.
Show answer
tan θ = h/4 and tan(90° − θ) = cot θ = h/9. Multiplying: tan θ × cot θ = 1 = h²/36 → h = 6 m. ∎
Answer: h = 6 m
A bird sits on top of an 80 m tree. From a point on the ground, its angle of elevation is 45°. It flies horizontally away, and after 2 seconds the angle is 30°. Find the bird's speed.
Show answer
Initial distance 80; later 80√3. Flown = 80(√3 − 1) ≈ 58.56 m in 2 s → ≈ 29.28 m/s.
Answer: 40(√3 − 1) ≈ 29.28 m/s
From a point P on the ground, the angle of elevation of the top of a 10 m building is 30°. A flag on the building has its top at an elevation of 45° from P. Find the flag's length and the distance of P from the building.
Show answer
d = 10√3 ≈ 17.32 m. Flag top height = d tan 45° = 10√3. Flag = 10√3 − 10 = 10(√3 − 1) ≈ 7.32 m.
Answer: Flag ≈ 7.32 m, distance ≈ 17.32 m
From a window 60 m above the ground, the angles of elevation and depression of the top and foot of another house are 60° and 45°. Find the height of the other house.
Show answer
Distance = 60 (depression 45°). Above window: 60 tan 60° = 60√3. Height = 60 + 60√3 = 60(1 + √3) ≈ 163.92 m.
Answer: 60(1 + √3) m
13. Quick Revision Summary
- Elevation: look up from horizontal; depression: look down from horizontal; they are equal for two points.
- Use tan for height–distance; sin/cos when a slant length is involved.
- Two-angle problems: write h in two ways and equate.
- 45° ⇒ h = d; 30°/60° pairs give factor 3.
- Draw the diagram first — it decides the method.