Heights and Distances
SSC GD Exam Preparation - Quantitative Aptitude
Heights and Distances problems are based on trigonometry, mainly using angles of elevation, depression, and right-angled triangles. These questions frequently appear in SSC, Railway, and Police exams.
🔹 1. Key Concepts
1. Angle of Elevation
The angle formed above the horizontal when an observer looks upwards at an object.
Looking UP → Angle ABOVE horizontal
2. Angle of Depression
The angle formed below the horizontal when an observer looks downwards at an object.
Looking DOWN → Angle BELOW horizontal
3. Right-Angled Triangle in Heights & Distances
Hypotenuse
Line of sight
Opposite side
Height difference
Adjacent side
Horizontal distance
🔹 2. Trigonometric Ratios Used
| Ratio | Formula | Application |
|---|---|---|
| sin θ | Opposite / Hypotenuse | When line of sight is known |
| cos θ | Adjacent / Hypotenuse | When horizontal distance is part of line of sight |
| tan θ | Opposite / Adjacent | Most common - height/distance problems |
Most problems use tan θ = height / distance
🔹 3. Important Formulas
1. Height of Object
Height = Distance × tan θ
2. Distance from Object
Distance = Height / tan θ
3. Two Points Formula
When angles of elevation from two points are given:
Distance = Height difference / (tan θ₁ - tan θ₂)
🔹 4. Quick Tips for Solving
Draw a diagram for visualization
Identify right-angled triangles
Label height, distance, angles clearly
Use tan θ for most calculations
Check units (meters, km) carefully
Remember observer height when given
🧠 Practice Section: 15 Questions (With Answers)
Each Q followed by its Answer (SSC style). Click on "View Answer" to check your understanding.
Q1. The angle of elevation of the top of a building from a point 50 m away is 30°. Find the height of the building.
View Answer
Height = 50 × tan 30° = 50 × 1/√3 ≈ 28.87 m
Q2. A tower is 40 m high. Find the angle of elevation from a point 30 m away.
View Answer
θ = tan⁻¹(height / distance) = tan⁻¹(40/30) = tan⁻¹(4/3) ≈ 53.13°
Q3. The top of a pole is seen at an angle of 45° from a distance of 20 m. Find the height.
View Answer
Height = 20 × tan 45° = 20 × 1 = 20 m
Q4. A man 1.8 m tall sees the top of a building at 60°. Find the building's height if he is 10 m away from the base.
View Answer
Height of building = 10 × tan 60° + 1.8 = 10 × √3 + 1.8 ≈ 18.48 m
Q5. The angle of elevation of a tower from two points 30 m apart on the same line is 30° and 60°. Find the height of the tower.
View Answer
Height = 30 × tan 30° × tan 60° / (tan 60° – tan 30°) = 30 × (1/√3 × √3) / (√3 – 1/√3) ≈ 15 m
Q6. The angle of elevation of a tree from a point 50 m away is 45°. Find the height of the tree.
View Answer
Height = 50 × tan 45° = 50 m
Q7. A person is standing 10 m from a pole. The angle of elevation of the top is 30°. Find the height of the pole.
View Answer
Height = 10 × tan 30° ≈ 5.77 m
Q8. From the top of a 20 m high building, the angle of depression of a car is 30°. Find the distance of the car from the building.
View Answer
Distance = 20 / tan 30° = 20 × √3 ≈ 34.64 m
Q9. The angle of elevation of the top of a building from a point 40 m away is 60°. Find the height.
View Answer
Height = 40 × tan 60° = 40 × √3 ≈ 69.28 m
Q10. A balloon rises vertically and the angle of elevation changes from 30° to 60° over 50 m distance. Find the height of the balloon.
View Answer
Height = 50 × (tan 60° × tan 30°) / (tan 60° – tan 30°) ≈ 21.65 m
Q11. A tree's top is seen at an angle of 30° from 15 m away. Find its height.
View Answer
Height = 15 × tan 30° ≈ 8.66 m
Q12. From a point 20 m away from a building, the angle of elevation is 45°. Height of the building?
View Answer
Height = 20 × tan 45° = 20 m
Q13. A man is 50 m from a building. Angle of elevation to the top is 60°. Find height.
View Answer
Height = 50 × tan 60° = 50 × √3 ≈ 86.60 m
Q14. A 1.5 m tall person sees the top of a building at 30° from 20 m away. Height of building?
View Answer
Height = 20 × tan 30° + 1.5 ≈ 13.08 m
Q15. Angle of depression of a boat from a 25 m high cliff is 45°. Find distance of the boat.
View Answer
Distance = 25 / tan 45° = 25 m
✅ SSC GD Exam Strategy
🟢 Always draw a rough diagram to visualize the problem
🟢 Remember the difference between elevation and depression
🟢 Use tan θ = height/distance for most problems
🟢 Don't forget to add observer height when applicable
🟢 Time yourself - aim for 45-60 seconds per height-distance problem
You've completed Heights and Distances!
SSC GD Tip: Mastering heights and distances problems requires strong visualization skills and quick application of trigonometric ratios. Remember that angle of elevation is measured upward from horizontal, while angle of depression is measured downward. Always identify the right triangle in the problem and use the appropriate trigonometric ratio. Practice drawing quick diagrams and labeling the known values. Regular practice with previous year SSC GD questions will build your speed and accuracy in solving these problems efficiently.
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