Trigonometry
SSC-CGL Exams
1. Overview
Trigonometry is heavily tested in SSC CGL for:
Trigonometric ratios
Identities and simplifications
Heights & distances problems
Focus on shortcuts, formulas, and visualization for fast problem-solving.
2. Trigonometric Ratios
For a right-angled triangle with angle θ:
| Ratio | Formula |
|---|---|
| sin θ | opposite/hypotenuse |
| cos θ | adjacent/hypotenuse |
| tan θ | opposite/adjacent |
| cot θ | 1/tan θ = adjacent/opposite |
| sec θ | 1/cos θ = hypotenuse/adjacent |
| cosec θ | 1/sin θ = hypotenuse/opposite |
Example 1
Right triangle, opposite = 3, adjacent = 4, hypotenuse = 5 → find sin θ, cos θ, tan θ
sin θ = 3/5, cos θ = 4/5, tan θ = 3/4
3. Important Identities
Pythagorean Identities
sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
1 + cot²θ = csc²θ
Reciprocal Identities
sin θ = 1/csc θ, cos θ = 1/sec θ, tan θ = 1/cot θ
Co-function Identities
sin(90−θ) = cos θ, cos(90−θ) = sin θ, tan(90−θ) = cot θ
Example 2
If sin θ = 3/5, find cos θ
cos θ = √(1 − sin²θ) = √(1 − 9/25) = √(16/25) = 4/5
4. Heights & Distances
SSC focuses on practical problems involving:
Angles of elevation/depression
Heights of objects
Distance between points
Formulas (Right Triangle approach)
Height = Distance × tan θ
Distance = Height / tan θ
Use sin θ, cos θ for hypotenuse / vertical / horizontal distances
Short Tricks
| Angle | Approx tan θ | Approx sin θ | Approx cos θ |
|---|---|---|---|
| 30° | 1/√3 ≈ 0.577 | 1/2 | √3/2 |
| 45° | 1 | √2/2 | √2/2 |
| 60° | √3 ≈ 1.732 | √3/2 | 1/2 |
Use tan θ = perpendicular/base for SSC "heights & distances" problems.
Example 3
Tree casts shadow 10 m long, angle of elevation = 30° → height of tree?
tan 30 = height/10 ⇒ 1/√3 = height/10 ⇒ height = 10/√3 ≈ 5.77 m
Example 4
Tower height = 20 m, distance from foot = 10 m → angle of elevation?
tan θ = height/distance = 20/10 = 2 ⇒ θ = tan⁻¹(2) ≈ 63.43°
5. SSC Short Tricks / Tips
Always draw the triangle — visualize shadow, height, distance.
Memorize tan 30°, 45°, 60° — most SSC problems use these angles.
Use 1-√3-2 rule: 30° → tan 30 = 1/√3, 60° → tan 60 = √3
Small angles approximation: For tiny angles, tan θ ≈ sin θ ≈ θ (in radians)
Angle of depression = angle of elevation (alternate angle property)
Combined problems: Two towers, bridges → break into two right triangles
6. Practice Section
Q1. Right triangle: opposite = 7, adjacent = 24 → sin θ, cos θ, tan θ
View Answer
Hypotenuse = √(7²+24²) = √(49+576)=√625=25
sin θ = 7/25, cos θ = 24/25, tan θ = 7/24
7/25, 24/25, 7/24
Q2. Angle of elevation = 45°, distance = 10 m → height?
View Answer
tan 45 = h/10 ⇒ h=10 m
10 m
Q3. Tree height = 15 m, shadow = 15√3 m → angle of elevation?
View Answer
tan θ = 15/(15√3) = 1/√3 ⇒ θ=30°
30°
Q4. sin θ = 3/5 → cos θ?
View Answer
cos θ = √(1−9/25) = √16/25 = 4/5
4/5
Q5. Tower A height = 20 m, shadow = 20 m; Tower B height = ? shadow = 10 m → tan comparison?
View Answer
Tower A: tan θ = 20/20 =1; Tower B: tan θ = h/10
If same angle θ → 1 = h/10 ⇒ h=10 m
10 m
Q6. Find sin²θ + cos²θ if sin θ = 4/5
View Answer
cos θ = √(1−16/25)=3/5 ⇒ sin²θ + cos²θ = 16/25 + 9/25 = 1
1
7. Quick Recap Table
| Topic | Key Formula / Property | SSC Tip |
|---|---|---|
| Ratios | sin θ = opp/hyp, cos θ = adj/hyp, tan θ = opp/adj | Memorize basic 30°,45°,60° |
| Identities | sin²+cos²=1, 1+tan²=sec², 1+cot²=csc² | Use for simplification |
| Heights & Distances | tan θ = height/distance | Draw triangle & apply |
| Angle of Elevation/Depression | Equal (alternate angles) | Draw lines from eye-level |
| Common shortcut | 1-√3-2 rule | Apply tan 30°, 45°, 60° directly |
You've completed Article 12: Trigonometry!
Courage Tip: For SSC, draw a simple triangle, label all sides, use tan for height, distance, angle problems. Shortcuts like 30°-45°-60° save time in exam conditions.
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