How to Solve: Height and Distance (Angles of Elevation/Depression)

How to Solve: Height and Distance (Angles of Elevation/Depression)

For SSC, Bank, Railway Exams

Introduction

"Mastering angles of elevation and depression can get you 3-5 marks in SSC, Bank, or Railway exams—enough to push you into the next cutoff bracket. These questions test your ability to apply trigonometry to real-world problems like finding the height of a building or the distance of a ship from shore. Let’s break it down step by step so you never lose marks on this topic again."

What You Need To Know First

Before diving in, ensure you understand: 1. Basic trigonometry ratios (sin, cos, tan) – You must know these by heart. 2. Right-angled triangles – How to identify the hypotenuse, opposite, and adjacent sides. 3. Complementary angles – Angles that add up to 90° (e.g., 30° and 60°).

If any of these are unclear, review them first—this guide assumes you’re comfortable with them.

Key Vocabulary

Formulas To Know

Step-by-Step Method

Step 1: Draw a Clear Diagram

• Sketch the scenario as a right-angled triangle.
• Label:
• The observer’s position (point A).
• The object (point B, e.g., top of a building).
• The horizontal line (ground or water level).
• The angle of elevation/depression (θ).
• All given values (heights, distances, angles).

Step 2: Identify the Right-Angled Triangle

• The horizontal line and the line of sight form the two sides.
• The vertical height (or depth) is the third side.
• Mark the right angle (90°) where the horizontal and vertical lines meet.

Step 3: Label the Sides

• Opposite side: The side opposite the angle θ (usually the height).
• Adjacent side: The side next to the angle θ (usually the horizontal distance).
• Hypotenuse: The line of sight (longest side).

Step 4: Choose the Correct Trigonometric Ratio

• If you have:
• Opposite and Adjacent → Use tan θ.
• Opposite and Hypotenuse → Use sin θ.
• Adjacent and Hypotenuse → Use cos θ.
• Most problems use tan θ because they give angles and ask for heights/distances.

Step 5: Plug in the Values and Solve

• Write the equation (e.g., ( \tan θ = \frac{\text{Opposite}}{\text{Adjacent}} )).
• Substitute the known values.
• Solve for the unknown (cross-multiply if needed).

Step 6: Adjust for Observer’s Height (If Given)

• If the observer is not at ground level (e.g., standing on a platform), add/subtract their height from the final answer.
• Example: If the observer is 1.5m tall and the calculated height is 10m, the total height = 10m + 1.5m = 11.5m.

Step 7: Check Units and Answer the Question

• Ensure all units are consistent (meters, degrees, etc.).
• Reread the question – Are you asked for the total height or just the height above the observer?

Worked Examples

Example 1 – Basic (Angle of Elevation)

Question: From a point 50m away from the base of a tower, the angle of elevation to the top is 30°. Find the height of the tower.

Step-by-Step Solution:

1. Draw the diagram:
2. Point A (observer) → 50m from tower (point B).
3. Angle of elevation = 30°.

4. Tower height = BC (unknown).

5. Identify the triangle:

6. Right-angled at B (base of tower).
7. Opposite = BC (height).
8. Adjacent = AB = 50m.

9. Angle θ = 30°.

10. Choose the ratio:

11. We have opposite (BC) and adjacent (AB) → Use tan 30°.

12. Write the equation:
    [
    \tan 30° = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{50}
    ]

13. Substitute tan 30°:
    [
    \frac{1}{\sqrt{3}} = \frac{BC}{50}
    ]

14. Solve for BC:
    [
    BC = 50 \times \frac{1}{\sqrt{3}} = \frac{50}{\sqrt{3}} \approx 28.87 \text{ m}
    ]
    (Rationalise if needed: ( \frac{50\sqrt{3}}{3} ))

15. Final answer:
    The height of the tower is ( \frac{50}{\sqrt{3}} ) m (or ~28.87m).

What we did and why: - We used tan θ because we had the adjacent side (distance) and needed the opposite side (height). - The observer was at ground level, so no height adjustment was needed.

Example 2 – Medium (Angle of Depression + Observer’s Height)

Question: From the top of a 20m high building, the angle of depression to a car on the road is 45°. Find the distance of the car from the base of the building.

Step-by-Step Solution:

1. Draw the diagram:
2. Building height = 20m (AB).
3. Observer at A, car at C.

4. Angle of depression = 45° → Angle of elevation from C to A = 45° (alternate angles).

5. Identify the triangle:

6. Right-angled at B (base of building).
7. Opposite = AB = 20m.
8. Adjacent = BC (distance, unknown).

9. Angle θ = 45°.

10. Choose the ratio:

11. We have opposite (AB) and need adjacent (BC) → Use tan 45°.

12. Write the equation:
    [
    \tan 45° = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{20}{BC}
    ]

13. Substitute tan 45°:
    [
    1 = \frac{20}{BC}
    ]

14. Solve for BC:
    [
    BC = 20 \text{ m}
    ]

15. Final answer:
    The car is 20m away from the building.

What we did and why: - The angle of depression from the top equals the angle of elevation from the car (alternate angles). - We used tan 45° = 1 to simplify the calculation.

Example 3 – Exam-Style (Disguised Problem)

Question: A man standing on a cliff 100m high observes two boats in the same direction. The angles of depression to the boats are 30° and 45°. Find the distance between the two boats.

Step-by-Step Solution:

1. Draw the diagram:
2. Cliff height = AB = 100m.
3. Boats at C and D.

4. Angles of depression: 30° (to C) and 45° (to D).

5. Convert to angles of elevation:

6. Angle of elevation from C to A = 30°.

7. Angle of elevation from D to A = 45°.

8. Find distance to Boat D (45°):

9. In triangle ABD:
   [
   \tan 45° = \frac{AB}{BD} \implies 1 = \frac{100}{BD} \implies BD = 100 \text{ m}
   ]

10. Find distance to Boat C (30°):

11. In triangle ABC:
    [
    \tan 30° = \frac{AB}{BC} \implies \frac{1}{\sqrt{3}} = \frac{100}{BC} \implies BC = 100\sqrt{3} \text{ m}
    ]

12. Calculate distance between boats (CD):
    [
    CD = BC - BD = 100\sqrt{3} - 100 = 100(\sqrt{3} - 1) \text{ m}
    ]
    (Approx. 73.2m)

13. Final answer:
    The distance between the boats is ( 100(\sqrt{3} - 1) ) m.

What we did and why: - We treated the two boats separately, using tan θ for each. - The distance between boats is the difference in their distances from the cliff.

Common Mistakes

Exam Traps

1-Minute Recap (Night Before the Exam)

"Listen up—this is all you need to remember for height and distance problems: 1. Draw a diagram—always. Label the observer, object, and angles. 2. Angle of elevation = looking up. Angle of depression = looking down (but use the angle of elevation in your triangle). 3. tan θ = opposite/adjacent—this is your best friend. Use it when you have the angle and need height/distance. 4. Adjust for observer’s height—if the person is standing on a platform, add their height to the answer. 5. Check units—meters, degrees, no mixing. 6. Practice 2-3 problems—focus on SSC/Bank past papers. You’ve got this!