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Quantitative Aptitude

Heights & Distances

Probability28 / 51Quadratic Equations
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Concepts (2)

Angles of Elevation (up from horizontal) and Depression (down from horizontal) use trigonometry, mainly **tan**, to find heights and distances. Master 30°, 45°, 60° ratios for speed.

Core Formula

Angles of Elevation and Depression are fundamental concepts in Heights & Distances. They are always measured with respect to a horizontal line.

  • Angle of Elevation: The angle formed by the line of sight with the horizontal when an observer looks UP at an object.
  • Angle of Depression: The angle formed by the line of sight with the horizontal when an observer looks DOWN at an object.

Key Principle: The angle of elevation from point A to point B is equal to the angle of depression from point B to point A (due to alternate interior angles).

Primary Trigonometric Ratios for Right Triangles:

  • tan(θ) = Opposite / Adjacent (Most frequently used for height and distance problems)
  • sin(θ) = Opposite / Hypotenuse
  • cos(θ) = Adjacent / Hypotenuse

Worked Example 1 (Elevation)

Q: A vertical pole is 50m high. What is the angle of elevation of its top from a point on the ground 50√3m away from its base? A:

  1. Identify the knowns: Height (Opposite) = 50m, Distance (Adjacent) = 50√3m.
  2. Use the tan ratio: tan(θ) = Opposite / Adjacent
  3. Substitute values: tan(θ) = 50 / (50√3) = 1/√3
  4. Recall standard angles: We know tan(30°) = 1/√3.
  5. Therefore, the angle of elevation θ = 30°.

Worked Example 2 (Depression)

Q: From the top of a lighthouse 75m high, the angle of depression of a ship is 45°. Find the distance of the ship from the base of the lighthouse. A:

  1. Draw the scenario: The angle of depression from the lighthouse top to the ship is 45°. This means the angle of elevation from the ship to the lighthouse top is also 45° (alternate interior angles).
  2. Identify the knowns: Height (Opposite) = 75m, Angle = 45°.
  3. Let the distance of the ship from the base (Adjacent) be x.
  4. Use the tan ratio: tan(45°) = Opposite / Adjacent
  5. Substitute values: 1 = 75 / x (since tan(45°) = 1)
  6. Solve for x: x = 75m.
  7. The ship is 75m away from the base of the lighthouse.

Shortcuts & Tricks

  1. Standard Angle Ratios (Sides): Memorize these for quick calculations:
    • For θ = 30°: Sides are in ratio 1 : √3 : 2 (Opposite : Adjacent : Hypotenuse)
    • For θ = 45°: Sides are in ratio 1 : 1 : √2 (Opposite : Adjacent : Hypotenuse)
    • For θ = 60°: Sides are in ratio √3 : 1 : 2 (Opposite : Adjacent : Hypotenuse)
    • Example 1 (revisited with shortcut): Height = 50, Base = 50√3. The ratio is 1 : √3. This directly corresponds to a 30° angle.
    • Example 2 (revisited with shortcut): Height = 75, Angle = 45°. For 45°, Opposite : Adjacent is 1 : 1. So, if Opposite is 75, Adjacent is also 75.
  2. Complementary Angles Shortcut: If the angles of elevation of the top of a tower from two points at distances 'a' and 'b' from the base (in the same straight line and on the same side) are complementary (sum to 90°), then the height of the tower h = √(ab).

Common Mistakes

  1. Incorrect Angle Placement: Students often place the angle of depression inside the triangle at the top vertex, rather than between the line of sight and the horizontal. Always draw the horizontal line first.
  2. Misremembering Trig Values: Confusing tan(30°) with tan(60°) or sin with cos values for standard angles leads to incorrect answers.
  3. Calculation Errors with √3: Approximating √3 too early or making arithmetic mistakes when multiplying/dividing with surds.

Derivation (brief)

Heights and Distances problems are direct applications of basic trigonometry, specifically the relationships within right-angled triangles. When an observer looks at an object, their line of sight, the horizontal line from their eye level, and the vertical line from the object to the horizontal form a right-angled triangle. The angles of elevation or depression are always measured from the horizontal. The tan function is predominantly used because it directly links the 'opposite' side (height) to the 'adjacent' side (horizontal distance), which are the most common parameters in these problems. By knowing an angle and one side, the other side can be easily found.

Advanced Examples

Q: From a point on the ground, the angle of elevation of the top of a tower is 30°. On moving 40m closer to the tower, the angle of elevation becomes 60°. Find the height of the tower. A:

  1. Let the height of the tower be h and the initial distance from the tower be x.

  2. From the first position (angle 30°): tan(30°) = h / x => 1/√3 = h / x => x = h√3 (Equation 1)

  3. From the second position (angle 60°, 40m closer): tan(60°) = h / (x - 40) => √3 = h / (x - 40) => x - 40 = h/√3 (Equation 2)

  4. Substitute x from Equation 1 into Equation 2: h√3 - 40 = h/√3

  5. Multiply the entire equation by √3 to clear denominators: 3h - 40√3 = h

  6. Rearrange and solve for h: 2h = 40√3 => h = 20√3 m.

    Time-Saving Formula for this type: If angles are α and β (where β > α) and the distance moved is d, then the height h = d / (cot(α) - cot(β)). Applying this: h = 40 / (cot(30°) - cot(60°)) = 40 / (√3 - 1/√3) = 40 / ((3-1)/√3) = 40 / (2/√3) = 20√3 m. This is significantly faster.

Variation Types

  1. Two Objects/Two Angles: Problems involving two towers, or an object observed from two different points, often leading to two simultaneous equations from two right triangles.
  2. Observer's Height: When the observer is on a building or cliff, their height must be added to or subtracted from the calculated height if the angle is taken from their eye level and the total height of the observed object is required.
  3. Shadow Problems: The length of a shadow cast by an object is determined by the sun's angle of elevation. This is a direct application of tan(θ) = Height / Shadow Length.
  4. Angles from a Single Point: Observing the top and bottom of an object (e.g., a flagstaff on a building) from a single point, involving both elevation and depression or two elevations.

Time-Saving Methods

  1. Ratio Method Mastery: For 30°, 45°, 60° angles, internalize the side ratios (1:√3:2, 1:1:√2, √3:1:2). This allows you to scale sides directly without writing explicit tan equations, saving crucial seconds.
  2. Visual Sketching: Always draw a quick, clear diagram. This helps visualize the problem, correctly identify the opposite/adjacent sides, and place the angles accurately.
  3. Memorize Common Values: Know √3 ≈ 1.732 and 1/√3 ≈ 0.577 for quick approximations if needed, but prefer working with surds until the final step for accuracy.
  4. Pattern Recognition: Identify common problem patterns (like the 'moving observer' example above) and memorize their specific shortcut formulas. This converts a multi-step calculation into a single formula application.
  5. Avoid Premature Rounding: Work with exact values (like √3) as long as possible. Rounding too early can introduce errors, especially in competitive exams where options might be close.
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Master two-observer height/distance problems using quick formulas and trigonometric ratios. Speed up calculations for elevation/depression angles from same or opposite sides to ace SSC CGL.

Core Formula

Two-observer problems typically involve finding the height of an object or the distance between observers/objects using angles of elevation or depression.

Case 1: Observers on Opposite Sides of the Object If two observers are on opposite sides of a vertical object (like a tower or building) and observe its top with angles of elevation θ₁ and θ₂, and the distance between the observers is d, then the height h of the object is: h = d / (cotθ₁ + cotθ₂)

Case 2: Observers on the Same Side of the Object If two observers are on the same side of a vertical object and observe its top with angles of elevation θ₁ (nearer to the base) and θ₂ (farther from the base), and the distance between the observers is d (distance between their positions), then the height h of the object is: h = d / (cotθ₂ - cotθ₁) (Note: θ₂ is the angle from the farther point, θ₁ from the nearer point, so θ₂ < θ₁) Alternatively, using tangents: h = d * (tanθ₁ * tanθ₂) / (tanθ₁ - tanθ₂) (where θ₁ is the larger angle, θ₂ is the smaller angle)

Worked Example 1

Problem: Two points P and Q are on opposite sides of a tower. The angles of elevation of the top of the tower from P and Q are 30° and 45° respectively. If the distance between P and Q is 100m, find the height of the tower.

Solution:

  1. Identify the case: Observers on opposite sides.
  2. Given: d = 100m, θ₁ = 30°, θ₂ = 45°.
  3. Apply the formula: h = d / (cotθ₁ + cotθ₂)
  4. Substitute values: cot30° = √3, cot45° = 1 h = 100 / (√3 + 1)
  5. Rationalize the denominator: h = 100 / (√3 + 1) * (√3 - 1) / (√3 - 1) h = 100(√3 - 1) / (3 - 1) h = 100(√3 - 1) / 2 h = 50(√3 - 1) m (Approximate: h = 50(1.732 - 1) = 50 * 0.732 = 36.6 m)

Worked Example 2

Problem: From two points on the ground, 50m apart, and on the same side of a building, the angles of elevation of the top of the building are 60° and 30°. Find the height of the building.

Solution:

  1. Identify the case: Observers on the same side.
  2. Given: d = 50m, θ₁ = 60° (nearer point), θ₂ = 30° (farther point).
  3. Apply the formula: h = d / (cotθ₂ - cotθ₁)
  4. Substitute values: cot30° = √3, cot60° = 1/√3 h = 50 / (√3 - 1/√3)
  5. Simplify the denominator: √3 - 1/√3 = (3 - 1) / √3 = 2/√3
  6. Calculate h: h = 50 / (2/√3) = 50 * √3 / 2 h = 25√3 m (Approximate: h = 25 * 1.732 = 43.3 m)

Shortcuts & Tricks

  1. Standard Triangle Ratios: For angles like 30°, 45°, 60°, always visualize the right-angled triangles and use side ratios (e.g., for 30-60-90, sides are x : x√3 : 2x; for 45-45-90, sides are x : x : x√2). This often avoids complex cot or tan calculations.
  2. Direct Tangent Formula (Same Side): For observers on the same side, if d is the distance between them, and angles are θ₁ (larger, nearer) and θ₂ (smaller, farther), the height h = d * (tanθ₁ * tanθ₂) / (tanθ₁ - tanθ₂). This is faster than cot if you're more comfortable with tan.
  3. Complementary Angles: If the angles of elevation of the top of a tower from two points at distances a and b from its base (on the same straight line and on the same side) are complementary (θ and 90°-θ), then the height of the tower h = √(ab). (This is a very specific but common shortcut).
  4. Memorize common values: √2 ≈ 1.414, √3 ≈ 1.732. Use these for quick approximations if options are far apart.

Common Mistakes

  1. Confusing Same vs. Opposite Side Formulas: Incorrectly using + instead of - or vice-versa in the denominator. Always draw a diagram to clarify the situation.
  2. Incorrect Angle Assignment: For same-side problems, ensure θ₁ (nearer, larger angle) and θ₂ (farther, smaller angle) are correctly placed in the formula h = d / (cotθ₂ - cotθ₁). The smaller angle's cotangent should be subtracted from the larger angle's cotangent (or vice-versa, ensuring the result is positive).
  3. Calculation Errors: Errors in rationalizing denominators or simplifying expressions involving √3 and 1/√3 are common. Practice these calculations thoroughly.
  4. Mixing Elevation and Depression: While the core geometry remains similar, misinterpreting the angle (e.g., using angle with vertical instead of horizontal) can lead to errors. Always assume angles are with the horizontal unless specified.

Derivation (brief)

Let's derive the formula for observers on opposite sides. Consider a tower of height h. Let the two observers be at points A and B, d meters apart. Let the angles of elevation from A and B be θ₁ and θ₂ respectively. Let the distance from the base of the tower to A be x₁ and to B be x₂.

From ΔTAC (where T is top, C is base, A is observer 1): tanθ₁ = h / x₁ => x₁ = h / tanθ₁ = h cotθ₁

From ΔTBC (where T is top, C is base, B is observer 2): tanθ₂ = h / x₂ => x₂ = h / tanθ₂ = h cotθ₂

Since A and B are on opposite sides, the total distance d = x₁ + x₂. Substitute x₁ and x₂: d = h cotθ₁ + h cotθ₂ d = h (cotθ₁ + cotθ₂) Therefore, h = d / (cotθ₁ + cotθ₂)

The derivation for same side follows a similar logic, but d = |x₁ - x₂|, leading to h = d / |cotθ₁ - cotθ₂|.

Advanced Examples

Problem: A man on a cliff observes a boat at an angle of depression of 30°. After sailing 50m towards the cliff, the angle of depression becomes 45°. Find the height of the cliff.

Solution:

  1. Draw a diagram. Let the height of the cliff be h. Let the initial position of the boat be B1 and the final position be B2. The distance B1B2 = 50m.
  2. Angles of depression from the top of the cliff (T) are 30° and 45°. By alternate interior angles, the angles of elevation from the boat to the top of the cliff are also 30° (from B1) and 45° (from B2).
  3. This is a 'same side' observer problem (the boat is moving on the same side towards the cliff).
  4. Given: d = 50m (distance between the two observation points of the boat).
  5. θ₁ = 45° (nearer angle), θ₂ = 30° (farther angle).
  6. Apply the formula: h = d / (cotθ₂ - cotθ₁)
  7. Substitute values: cot30° = √3, cot45° = 1 h = 50 / (√3 - 1)
  8. Rationalize the denominator: h = 50 / (√3 - 1) * (√3 + 1) / (√3 + 1) h = 50(√3 + 1) / (3 - 1) h = 50(√3 + 1) / 2 h = 25(√3 + 1) m (Approximate: h = 25(1.732 + 1) = 25 * 2.732 = 68.3 m)

Variation Types

  1. Two Towers: Finding the distance between two towers or the height of one tower given the other, and angles of elevation/depression from their tops/bases.
  2. Observer at a Height: When the observer is not on the ground but at a certain height (e.g., from a window of a building, observing another building or object).
  3. Angle of Depression: Problems involving angles of depression are essentially similar to elevation problems; just remember that the angle of depression from point A to point B is equal to the angle of elevation from point B to point A (alternate interior angles).
  4. Moving Object/Observer: Problems where an object or observer moves, changing the angle of elevation/depression over a known distance or time.

Time-Saving Methods

  1. Draw Diagrams: Always start with a clear, labeled diagram. This helps in correctly identifying angles, distances, and the specific case (same side/opposite side).
  2. Ratio Method for Specific Angles: For 30°, 45°, 60° problems, use the side ratios directly. For example, if angle is 45°, height = base. If angle is 30°, base = height * √3. If angle is 60°, base = height / √3. Combine these for two-observer problems.
    • Example: For the cliff problem (45° and 30°): Let cliff height be h. For 45°, distance from base to boat is h. For 30°, total distance from base to farther boat position is h√3. The difference is h√3 - h = h(√3 - 1). This difference is given as 50m. So, h(√3 - 1) = 50 => h = 50 / (√3 - 1) = 25(√3 + 1). This is much faster if you're adept at ratios.
  3. Avoid Unnecessary Rationalization: Only rationalize at the very end if the options are numerical. If options are in terms of √3, leave it as is.
  4. Practice Mental Math: Quickly calculate cot or tan values and simplify expressions involving √3 and 1/√3 in your head.
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