Heights & Distances
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Solved Examples
Study MaterialSolved Examples – Heights & Distances
Solved examples help students understand the practical application of trigonometric ratios, angle of elevation, angle of depression, and right-angled triangle concepts in competitive examinations. These examples are designed from basic to advanced level and cover important questions frequently asked in SSC, Banking, Railway, Insurance, Defence, CAT, NDA, CDS, and various aptitude examinations.
Topics Covered in Solved Examples
- Angle of Elevation Problems
- Angle of Depression Problems
- Tower and Building Questions
- Ladder Problems
- Shadow Problems
- Observer Height Applications
- Distance and Height Calculations
- Advanced Trigonometry Applications
Example 1: Basic Height Problem
Question: From a point 20 m away from the foot of a tower, the angle of elevation of the top of the tower is 45°. Find the height of the tower.
Solution:
Using:
tan θ = Height / Distance
tan 45° = Height / 20
1 = Height / 20
Height = 20 m
Answer: 20 m
Example 2: Angle of Elevation Problem
Question: The angle of elevation of the top of a building from a point 30 m away is 60°. Find the height of the building.
Solution:
tan 60° = Height / 30
√3 = Height / 30
Height = 30√3 m
≈ 30 × 1.732
≈ 51.96 m
Answer: 51.96 m
Example 3: Finding Distance
Question: The angle of elevation of the top of a pole is 30°. If the height of the pole is 10 m, find the distance of the observer from the pole.
Solution:
tan 30° = 10 / Distance
1/√3 = 10 / Distance
Distance = 10√3 m
≈ 17.32 m
Answer: 17.32 m
Example 4: Ladder Problem
Question: A ladder 13 m long rests against a wall. If the foot of the ladder is 5 m away from the wall, find the height reached by the ladder.
Solution:
Using Pythagoras theorem:
Height² = 13² − 5²
= 169 − 25
= 144
Height = 12 m
Answer: 12 m
Example 5: Shadow Problem
Question: A pole casts a shadow 20 m long when the angle of elevation of the sun is 45°. Find the height of the pole.
Solution:
tan 45° = Height / 20
1 = Height / 20
Height = 20 m
Answer: 20 m
Example 6: Angle of Depression Problem
Question: From the top of a building 50 m high, the angle of depression of a car on the ground is 45°. Find the distance of the car from the building.
Solution:
Angle of depression = Angle of elevation
tan 45° = 50 / Distance
1 = 50 / Distance
Distance = 50 m
Answer: 50 m
Example 7: Two Angles Problem
Question: The angle of elevation of the top of a tower from two points 20 m apart are 30° and 60°. Find the height of the tower.
Solution:
Let distance from nearer point = x m
Then distance from farther point = (x + 20) m
Using tan 60°:
√3 = Height / x
Height = x√3
Using tan 30°:
1/√3 = Height / (x + 20)
Height = (x + 20)/√3
Equating:
x√3 = (x + 20)/√3
3x = x + 20
2x = 20
x = 10
Height = 10√3 m
≈ 17.32 m
Answer: 17.32 m
Example 8: Observer Height Problem
Question: A man standing 1.5 m tall observes the top of a building at an angle of elevation 45°. If he is 25 m away from the building, find the height of the building.
Solution:
tan 45° = Height above observer / 25
1 = Height above observer / 25
Height above observer = 25 m
Total building height:
= 25 + 1.5
= 26.5 m
Answer: 26.5 m
Example 9: Airplane Problem
Question: An airplane is flying at a height of 3000 m. The angle of elevation from a point on the ground is 30°. Find the horizontal distance of the airplane from the observer.
Solution:
tan 30° = 3000 / Distance
1/√3 = 3000 / Distance
Distance = 3000√3 m
≈ 5196 m
Answer: 5196 m
Example 10: Tree Problem
Question: The angle of elevation of the top of a tree is 60°. If the observer is 15 m away from the tree, find the height of the tree.
Solution:
tan 60° = Height / 15
√3 = Height / 15
Height = 15√3
≈ 25.98 m
Answer: 25.98 m
Example 11: Bridge Problem
Question: From the top of a bridge 40 m high, the angle of depression of a boat is 30°. Find the distance of the boat from the bridge.
Solution:
tan 30° = 40 / Distance
1/√3 = 40 / Distance
Distance = 40√3 m
≈ 69.28 m
Answer: 69.28 m
Example 12: Pole and Wire Problem
Question: A wire attached to the top of a pole makes an angle of 60° with the ground. If the wire is 20 m long, find the height of the pole.
Solution:
sin 60° = Height / 20
√3/2 = Height / 20
Height = 20 × √3/2
= 10√3 m
≈ 17.32 m
Answer: 17.32 m
Example 13: Building and Shadow Problem
Question: A building casts a shadow 50 m long when the angle of elevation of the sun is 60°. Find the height of the building.
Solution:
tan 60° = Height / 50
√3 = Height / 50
Height = 50√3 m
≈ 86.6 m
Answer: 86.6 m
Example 14: Two Buildings Problem
Question: From the top of a 30 m building, the angle of elevation of another building is 45°. If the buildings are 20 m apart, find the height of the second building.
Solution:
tan 45° = Height Difference / 20
1 = Height Difference / 20
Height Difference = 20 m
Height of second building:
= 30 + 20
= 50 m
Answer: 50 m
Example 15: Mountain Problem
Question: The angle of elevation of the top of a mountain from a point is 30°. After moving 100 m nearer, the angle becomes 60°. Find the height of the mountain.
Solution:
Let distance from nearer point = x m
Then distance from farther point = (x + 100) m
Using tan 60°:
Height = x√3
Using tan 30°:
Height = (x + 100)/√3
Equating:
x√3 = (x + 100)/√3
3x = x + 100
2x = 100
x = 50
Height = 50√3 m
≈ 86.6 m
Answer: 86.6 m
Important Exam Tips
- Draw diagrams carefully before solving.
- Use tan θ in most problems.
- Memorize important trigonometric values.
- Remember Pythagoras theorem.
- Practice observer height problems regularly.
- Improve simplification speed.
- Practice previous year aptitude questions.
Common Mistakes to Avoid
- Confusing angle of elevation and depression.
- Using wrong trigonometric ratio.
- Ignoring observer height.
- Calculation mistakes in square roots.
- Using incorrect triangle sides.
Practicing solved examples regularly improves conceptual clarity, logical thinking, and calculation speed in solving Heights & Distances aptitude questions in competitive examinations.