Trigonometry

Trigonometry

🟢 Lite

Key Rule / Formula

sin θ = Opposite/Hypotenuse. cos θ = Adjacent/Hypotenuse. tan θ = sin/cos = Opposite/Adjacent. Key identity: sin²θ + cos²θ = 1.

Memory Trick

SOH-CAH-TOA: Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent. Complementary angle: sin θ = cos(90°−θ).

1-Sentence Summary

Trigonometry in SSC tests angle measurement, identity-based simplifications, and practical heights-and-distances problems using right triangle trigonometry.

Quick Example

Q: Find sin 30° + cos 60°. A: ½ + ½ = 1 — sin 30° = ½, cos 60° = ½.

🟡 Standard

Concept

Trigonometry in SSC Tier 2 has two components: algebraic manipulation using trigonometric identities, and practical application in heights-and-distances word problems. The algebraic side tests your ability to simplify expressions using standard identities, while the geometry side tests your ability to model real situations as right triangles.

Standard Values to Memorise:

Angle	0°	30°	45°	60°	90°
sin	0	1/2	1/√2	√3/2	1
cos	1	√3/2	1/√2	1/2	0
tan	0	1/√3	1	√3	∞

Key Identities:

• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ (for θ ≠ 90°)
• 1 + cot²θ = cosec²θ (for θ ≠ 0°)
• sin(A+B), sin(A−B), cos(A+B), cos(A−B) expansion formulas
• Max/min values: For a sin θ + b cos θ, range is [−√(a²+b²), √(a²+b²)]

Heights and Distances:

• Angle of elevation: when looking UP from horizontal.
• Angle of depression: when looking DOWN from a height — equals angle of elevation from the lower point to the higher point.
• Always draw the right triangle — mark known sides, mark unknown (what you’re solving for).

Key Points

• tan θ = sin θ / cos θ. cot θ = cos θ / sin θ = 1/tan θ.
• sec θ = 1/cos θ. cosec θ = 1/sin θ.
• Complementary angles: sin θ = cos(90°−θ), tan θ = cot(90°−θ).
• For heights and distances, if the angle and one side are known, use the appropriate trig ratio to find the other side.
• When two angles of elevation are given from two different points to the same object, use tan values to set up two equations.

Worked Example

Q: The angle of elevation of the top of a tower from a point 30m away from its base is 30°. Find the height of the tower. Approach: tan 30° = Height / 30 → 1/√3 = h/30 → h = 30/√3 = 10√3 m. Answer: 10√3 m

SSC Pattern / Tips

• In heights-and-distances, always check whether the point is on level ground with the base or above/below it.
• When an object is viewed from two different points, use both angles to create two equations.
• For identity simplification, try to convert everything to sin and cos first — it’s often the most general approach.
• Maximum/minimum value of expressions like sin θ + cos θ = √2 × sin(θ+45°), so max = √2.

🔴 Extended

Full Concept

Trigonometry in SSC CGL Tier 2 goes far beyond the basic sin-cos-tan definitions. You need mastery over three interconnected domains: trigonometric identities and equations, heights and distances (applications), and angular relationships.

The six trigonometric ratios — sine, cosine, tangent, cosecant, secant, and cotangent — are defined as ratios of sides in a right-angled triangle. The fundamental identity sin²θ + cos²θ = 1 serves as the bedrock. From this flows a family of identities: 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ. SSC frequently tests transformations using these — converting sin to cos, swapping denominators, and reducing complex expressions to simple values like 1 or 0.

Angles in trigonometry problems span three systems: degrees, radians, and the special SSC-favorite — angles measured in degrees but treated with sexagesimal arithmetic. The standard angles 0°, 30°, 45°, 60°, 90° and their complementary pairs form the core. You must have instant recall for sin 15°, sin 75°, tan 15°, tan 75° and similar values because paper-carrying time is precious. Compound angle formulas (sin(A+B), cos(A+B), etc.) appear less frequently but when they do, they separate high-scorers from average attempts.

Heights and distances constitute about 30-35% of Tier 2 trigonometry questions. The approach is always the same: draw a clean figure, identify the reference angle (angle of elevation or depression), label the known side, apply the appropriate trigonometric ratio. The angle of elevation from the horizontal line of sight to the object equals the angle of depression from the object to the observer (alternate angles). Questions often chain two or three observations — a common pattern is finding the height of a tower given angles of elevation from two different points and the distance between them.

Maximum and minimum values of trigonometric expressions require understanding the range of each function. sinθ and cosθ range from -1 to +1, while tanθ has no finite maximum or minimum. Questions ask for the greatest or least value of expressions like sinθ + cosθ (which ranges from -√2 to √2) or sinθ · cosθ (maximum 1/2 at 45°).

SSC CGL Deep Analysis

Trigonometry consistently contributes 8-10 questions out of 100 in Tier 2 Quantitative Abilities, making it one of the highest-yield topics. Questions break down into three difficulty bands: easy (direct identity application, standard angle values — roughly 40%), moderate (compound expressions, word problems with heights/distances — roughly 45%), and hard (maximum/minimum value problems, trigonometric equations, chained observations — roughly 15%).

The 2019-2024 trend shows a shift toward applied trigonometry. Pure identity simplification questions are declining; simultaneous heights-and-distances problems with two observation points are increasing. SSC also started embedding trigonometry inside geometry questions — using angle bisector theorems or properties of cyclic quadrilaterals alongside trig ratios.

Common question stems: “The angle of elevation of a tower from point A is 30°…” (moderate difficulty), “If sinθ + cosθ = √2, then find θ” (moderate to high), “Maximum value of (sinθ + cosθ) is…” (moderate), “From a point on the ground 20m from the base of a tower, the angle of elevation is 60°. The height of the tower is…” (easy).

The cot(90° - θ) = tanθ identity is frequently tested in the “complementary angles” question type. Similarly, sin(90° - θ) = cosθ appears so often that failing to recognise it wastes 2-3 minutes per question.

High-Scoring Strategy

Step 1: Memorise all standard angle values for sin, cos, tan, cosec, sec, cot from 0° to 90° on a single table. Write this table 5 times by hand until it becomes reflex — no thinking, just recall.

Step 2: For identity problems, always look for opportunities to convert everything to sin and cos. This simplifies the expression and reveals patterns. For example, (1 + tanθ)/(1 + cotθ) simplifies to tanθ because 1 + cotθ = (sinθ + cosθ)/sinθ and 1 + tanθ = (sinθ + cosθ)/cosθ, giving (cosθ/sinθ) = tanθ.

Step 3: In heights and distances, always draw the figure even if the question doesn’t explicitly ask for one. Label the known distance (base), the unknown (height), and the angle. The three ratios (sin, cos, tan) will immediately tell you which one to use. If adjacent and opposite are given → tan; if hypotenuse and opposite → sin; if hypotenuse and adjacent → cos.

Step 4: For maximum/minimum problems, express the given function as R sin(θ + α) or R cos(θ - α) where R = √(a² + b²). This transforms trigonometric expressions into simple range problems — the maximum is +R and minimum is -R.

Step 5: Answer elimination works powerfully here. If asked for tan 15° and options include √3 - 1, √3 + 1, √3/2, 2 - √3 — you know 2 - √3 is correct. Train yourself to spot which option is clearly wrong before computing.

SSC-Level Practice

Q1: From a point on the ground, the angle of elevation of the top of a tower 60m tall is 30°. From another point on the ground 40√3 m away from the first point towards the tower, the angle of elevation is 60°. Find the height of the tower.

Answer: 60 m — Working: Let h be the height and x the horizontal distance from the nearer (second) point to the base. At the second point, h/x = tan 60° = √3, so x = h/√3. At the first point, which is 40√3 m farther back, h/(x + 40√3) = tan 30° = 1/√3. Substituting: h/[(h/√3) + 40√3] = 1/√3 → h√3 = h + 120 → 2h = 120 → h = 60 m. ✓

Q2: If 0° < θ < 90° and sin 7θ = cos 5θ, find the value of 3θ.

Answer: 135° — Working: sin 7θ = cos 5θ = sin(90° - 5θ). Using sin x = sin y ⇒ x = 180° - y, take 7θ = 180° - (90° - 5θ) = 90° + 5θ, so 2θ = 90° → θ = 45°. Then 3θ = 135°. ✓

Common Traps

• Forgetting the complementary angle identity: sin(90° - θ) = cosθ and cot(90° - θ) = tanθ are extremely common in SSC — always look for complementary pairs in the problem before starting calculation.
• Wrong angle selection in heights & distances: Mixing up which side is opposite and which is adjacent, especially in problems with two observation points. Always mark the angle first, then identify the side opposite it.
• Assuming acute angles: Many students implicitly assume θ is acute even when not stated. Trigonometric ratios for obtuse angles (90° < θ < 180°) behave differently — sin is positive but cos and tan are negative. This catches roughly 1 in 4 students on trick questions.

Content adapted based on your selected roadmap duration.