Height and Distance: Notes और MCQ Practice

आसान भाषा में:

Height and Distance problems में दो objects की relative vertical motion को compare किया जाता है. जब ek object dusre से upar या neeche hota है, तब हम relative speed = speed1 - speed2 use karte हैं.

Important Formulas / Rules

• Relative Speed = Speed₁ − Speed₂ (same direction) or Speed₁ + Speed₂ (opposite direction)
• Time = Distance ÷ Relative Speed
• Height = (Speed²) ÷ (2 × g) (g = 10 m/s² for quick calc)
• Distance covered in t seconds = Initial Speed × t + ½ × a × t²

Quick Tricks

1. अगर dono speeds same direction hain, to relative speed ko subtract karo, nahi to add karo.
2. g ko 10 rakh ke calculations 2‑digit tak round off kar lo – bahut fast answer milta hai.

Common Mistakes

1. Relative speed ko add karna jab objects same direction में हों.
2. Height formula में 2×g को miss कर देना या g को 9.8 use करना, जिससे answer complex हो जाता है.

Height and Distance

1. Concept

Height and Distance problems involve finding the height of an object (tower, tree, building, etc.) or the horizontal distance between two points using the principles of right‑angled triangles. The key idea is that the line of sight, the ground, and the vertical height form a right triangle. By using trigonometric ratios (tan θ = opposite ÷ adjacent) or the properties of similar triangles, we can relate the known angle and a known side to the unknown side.

2. Important formulas / rules / facts

• Basic trigonometric ratio: tan θ = height ÷ distance → height = distance × tan θ, distance = height ÷ tan θ
• Angle of elevation: angle measured upward from the horizontal line of sight.
• Angle of depression: angle measured downward from the horizontal line of sight (use the same tan formula; just consider the vertical drop as the height).
• Similar triangles: If two right triangles share an angle, the ratios of corresponding sides are equal.
• Conversion: 1° = (π/180) rad (only needed if a calculator works in radians).
• When two angles are given: Use the fact that the sum of angles in a triangle is 180°; often you will have complementary angles (θ and 90°‑θ) which give tan θ = cot (90°‑θ).

3. Solved examples (exam style)

Example 1

A pole stands on a flat ground. From a point 30 m away, the angle of elevation to the top of the pole is 45°. Find the height of the pole.

Solution

1. Draw the right triangle: distance = 30 m (adjacent), height = ? (opposite), angle = 45°.
2. Use tan 45° = 1.
3. height = distance × tan 45° = 30 m × 1 = 30 m.

Example 2

From the top of a 50 m high building, the angle of depression to a point on the ground is 30°. Find the horizontal distance of the point from the base of the building.

Solution

1. Angle of depression = angle of elevation from the ground, so tan 30° = opposite ÷ adjacent = 50 ÷ distance.
2. tan 30° = 1/√3 ≈ 0.577.
3. distance = 50 ÷ tan 30° = 50 ÷ 0.577 ≈ 86.6 m.

Example 3

A man standing 2 m tall is 10 m away from a tree. He looks at the top of the tree and the angle of elevation is 30°. Find the height of the tree.

Solution

1. Height of tree = height of man + extra height.
2. tan 30° = extra height ÷ 10 m → extra height = 10 × tan 30° = 10 × 0.577 = 5.77 m.
3. Total height = 2 m + 5.77 m ≈ 7.77 m.

Example 4

Two poles of equal height stand 80 m apart. From a point between them, the angles of elevation to the tops are 30° and 45° respectively. Find the height of each pole.

Solution

1. Let the point be x m from the pole with 30° elevation; then distance to the other pole = 80 - x.
2. Using tan 30° = height ÷ x → height = x × tan 30° = 0.577x.
3. Using tan 45° = height ÷ (80‑x) → height = (80‑x) × 1 = 80‑x.
4. Equate the two expressions: 0.577x = 80‑x → 1.577x = 80 → x ≈ 50.75 m.
5. Height = 80‑x ≈ 80‑50.75 = 29.25 m (or using 0.577x gives the same).

4. Shortcuts and tricks

• Memorise key tan values: tan 30° = 1/√3 ≈ 0.577, tan 45° = 1, tan 60° = √3 ≈ 1.732. This speeds up calculations without a calculator.
• Use complementary angles: tan θ = 1 ÷ tan (90°‑θ). Helpful when the given angle is 60° (use tan 30°).
• When two angles are given from the same point (as in Example 4), set up two equations for the same height and solve for the unknown distance first.
• If the height of the observer is not negligible, subtract it from the final height (as in Example 3).
• Quick estimation: For small angles (≤ 10°), tan θ ≈ θ in radians. This can give a fast approximate answer when exact precision is not required.

5. Common mistakes to avoid

• Mixing up angle of elevation and depression – always treat them with the same tan formula; only the “opposite” side changes sign conceptually.
• Using sin or cos instead of tan – tan relates opposite (height) to adjacent (distance) directly; sin and cos involve the hypotenuse and lead to extra steps.
• Forgetting to subtract the observer’s height when the problem states the person’s height.
• Incorrect unit handling – keep all distances in the same unit (metres, kilometres, etc.) before applying the formula.
• Rounding too early – keep intermediate results to at least three decimal places; round only in the final answer to avoid cumulative error.