Cheat Sheet — Some Applications of Trigonometry

📐 Key Definitions

Line of Sight — line from observer's eye to the object
Angle of Elevation — angle above horizontal (looking UP)
Angle of Depression — angle below horizontal (looking DOWN)
Horizontal — reference line parallel to ground at eye level

💡 Angles are ALWAYS measured from the horizontal, never from the vertical!

⚡ Core Formulas

Finding height: h = d × tan θ
Finding distance: d = h / tan θ = h × cot θ
Line of sight length: L = h / sin θ

θ = angle of elevation or depression; h = height; d = horizontal distance

📊 Quick Reference Table

Angle θ	tan θ	If h = 100 m, then d = ?	If d = 100 m, then h = ?	Key Relationship
30°	1/√3 ≈ 0.577	100√3 ≈ 173.2 m	100/√3 ≈ 57.7 m	d = √3 × h
45°	1	100 m	100 m	d = h (equal!)
60°	√3 ≈ 1.732	100/√3 ≈ 57.7 m	100√3 ≈ 173.2 m	h = √3 × d

💡 At 30°, distance is √3 times the height. At 60°, height is √3 times the distance. At 45°, they're equal.

🔺 Two-Angle Formula (Same Side)

h = d · (tan α · tan β) / (tan β − tan α)

d = distance between two points; α = far angle (smaller); β = near angle (larger)

Setup: Let height = h, distance from nearer point to base = x

tan β = h/x → x = h/tan β
tan α = h/(x+d) → x+d = h/tan α
Subtract and solve for h

🔺 Two-Angle Formula (Opposite Sides)

h = d / (cot α + cot β)

Points on opposite sides of the tower; d = total distance between them

Setup: d₁ + d₂ = d

d₁ = h·cot α, d₂ = h·cot β
h(cot α + cot β) = d
Solve for h

🏗️ Building + Tower Problem

From the top of a building of height a, angle of elevation of tower top = α, angle of depression of tower base = β.

Distance to tower: d = a / tan β
Tower height: H = a + d·tan α = a(1 + tan α / tan β)

💡 The building height gives you the depression equation; use it to find d first.

⚠️ Common Mistakes

❌ Measuring angle from the vertical instead of horizontal
❌ Forgetting to add observer's height to calculated height
❌ Confusing "angle at base" with "angle of elevation"
❌ Using sin/cos when tan is needed (most problems use tan!)
❌ Same side vs opposite sides — using wrong formula
❌ Not drawing the diagram (costs 1 mark!)

📝 Problem-Solving Steps

Draw a neat labeled diagram (mark 90°, θ, sides)
Identify the right triangle(s)
Label known values and let unknown = h or x
Choose correct trig ratio (usually tan θ)
Solve the equation(s)
Verify — does the answer make sense?

🔢 Useful Values

√3 ≈ 1.732 | 1/√3 ≈ 0.577
√2 ≈ 1.414 | 1/√2 ≈ 0.707
tan 30° = 1/√3 | tan 45° = 1 | tan 60° = √3

Rationalisation:

h/√3 → h√3/3 (multiply by √3/√3)
CBSE prefers 20√3 m over 20/√3 × √3/√3

🎯 CBSE Exam Pattern

1–2 questions from this chapter (total 5–6 marks)
Usually one 5-mark long answer (two-triangle problem)
Diagram carries 1 mark — never skip it!
Most common angles: 30°, 45°, 60°
Leave answer in surd form unless told otherwise
Key phrase: "find height" → use tan θ = h/d

⚠️ This is one of the highest-scoring chapters — problems are formulaic once you master the setup!

✅ Worked Example (5-Mark Pattern)

Q: The angles of elevation of the top of a tower from two points at distances a and b from the base (same side) are complementary. Prove that the height of the tower is √(ab).

Let height = h, angles = θ and (90°−θ)

From distance a: tan θ = h/a → h = a·tan θ ... (i)
From distance b: tan(90°−θ) = h/b → cot θ = h/b → h = b·cot θ = b/tan θ ... (ii)

Multiply (i) × (ii): h² = a·tan θ × b/tan θ = ab
∴ h = √(ab) ■

💡 "Complementary angles" is a huge hint — it means their product eliminates tan θ!