Trigonometry: Notes और MCQ Practice

Explanation

आसान भाषा में, Trigonometry angles aur unke side ratios ko study karta hai. Right‑angled triangle mein sin, cos, tan ka use hota hai aur ye ratios har angle ke liye constant hote hain.

Important Formulas / Rules

• sin θ = opposite ÷ hypotenuse
• cos θ = adjacent ÷ hypotenuse
• tan θ = opposite ÷ adjacent
• sin²θ + cos²θ = 1
• tanθ = sinθ ÷ cosθ
• sin(90°‑θ) = cosθ, cos(90°‑θ) = sinθ, tan(90°‑θ) = cotθ

Quick Tricks

1. 45°‑45‑90 triangle: sides = 1 : 1 : √2 → sin45 = cos45 = 1/√2.
2. 30°‑60‑90 triangle: sides = 1 : √3 : 2 → sin30 = 1/2, cos30 = √3/2, tan30 = 1/√3.

Common Mistakes

• Mixing up opposite and adjacent for a given angle → result wrong tan/cos.
• Forgetting that sin²+cos²=1 only for same angle, not for different angles.

Trigonometry

1. Concept

Trigonometry deals with the relationship between the angles and sides of a right‑angled triangle. In SSC CGL only the basic ratios (sine, cosine, tangent and their reciprocals) and the ability to apply them to find missing sides or angles are required. The key idea is:

• For a right‑angled triangle with angle θ, opposite side O, adjacent side A, and hypotenuse H:
• sin θ = O ÷ H
• cos θ = A ÷ H
• tan θ = O ÷ A
• cosec θ = H ÷ O
• sec θ = H ÷ A
• cot θ = A ÷ O

All angles are measured in degrees for the exam.

2. Important formulas / rules / facts

• Pythagorean theorem: O² + A² = H²
• Fundamental identities:
• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ
• Complementary angle relations (θ + 90° = 90°):
• sin (90°‑θ) = cos θ
• cos (90°‑θ) = sin θ
• tan (90°‑θ) = cot θ
• Special angles (values for 0°, 30°, 45°, 60°, 90°):
• sin 0°=0, sin 30°=½, sin 45°=√2⁄2, sin 60°=√3⁄2, sin 90°=1
• cos 0°=1, cos 30°=√3⁄2, cos 45°=√2⁄2, cos 60°=½, cos 90°=0
• tan 0°=0, tan 30°=1⁄√3, tan 45°=1, tan 60°=√3, tan 90° undefined
• Area of a right triangle: (1⁄2) × O × A
• Height from the right angle to the hypotenuse: h = (O × A) ÷ H

3. Solved examples (exam style)

Example 1

Question: In a right‑angled triangle, the hypotenuse is 13 cm and one side is 5 cm. Find the other side and the acute angles.

Solution:

1. Use Pythagoras: other side² = 13² – 5² = 169 – 25 = 144 → other side = 12 cm.
2. sin θ = opposite ÷ hypotenuse = 5 ÷ 13 → θ = sin⁻¹(5/13) ≈ 22.6°.
3. The other acute angle = 90° – 22.6° ≈ 67.4°.

Answer: Other side = 12 cm, angles ≈ 22.6° and 67.4°.

Example 2

Question: A ladder 15 m long leans against a wall making an angle of 60° with the ground. Find the height reached on the wall and the distance of the foot from the wall.

Solution:

• Height (opposite) = H × sin 60° = 15 × √3⁄2 = 15√3⁄2 ≈ 12.99 m.
• Base distance (adjacent) = H × cos 60° = 15 × ½ = 7.5 m.

Answer: Height ≈ 13 m, foot distance = 7.5 m.

Example 3

Question: If tan θ = 3⁄4 and θ is an acute angle, find sin θ and cos θ.

Solution:

• Let opposite = 3k, adjacent = 4k → hypotenuse = √(3k)² + (4k)² = 5k.
• sin θ = opposite ÷ hypotenuse = 3k ÷ 5k = 3⁄5.
• cos θ = adjacent ÷ hypotenuse = 4k ÷ 5k = 4⁄5.

Answer: sin θ = 3⁄5, cos θ = 4⁄5.

Example 4

Question: Two trees are 30 m apart. From a point between them, the angles of elevation to the tops are 30° and 45° respectively. Find the height of the taller tree.

Solution:

1. Let the point be x m from the shorter tree; then distance to taller tree = 30‑x.
2. For the shorter tree: tan 30° = height ÷ x → height = x × 1⁄√3.
3. For the taller tree: tan 45° = height ÷ (30‑x) → height = 30‑x.
4. Equate heights: x ÷ √3 = 30‑x → x(1 + 1/√3) = 30 → x = 30 ÷ (1 + 1/√3) ≈ 13.2 m.
5. Height = 30‑x ≈ 16.8 m.

Answer: Height of taller tree ≈ 16.8 m.

Example 5

Question: Find the value of sin (2θ) if sin θ = 5⁄13 and θ is acute.

Solution:

• cos θ = √(1‑sin²θ) = √(1‑25⁄169) = √(144⁄169) = 12⁄13.
• sin 2θ = 2 sin θ cos θ = 2 × (5⁄13) × (12⁄13) = 120⁄169.

Answer: sin 2θ = 120⁄169.

4. Shortcuts and tricks

• Remember the 3‑4‑5 triangle: sides 3,4,5 give sin θ = 3⁄5, cos θ = 4⁄5, tan θ = 3⁄4. Useful for quick mental calculations.
• Use complementary angles to convert a difficult tan or cot to a known sin or cos (e.g., tan 75° = cot 15°).
• When tan θ is given, compute sin and cos by forming a right triangle with opposite = tan θ × 1 and adjacent = 1; then hypotenuse = √(1 + tan²θ).
• Area method: For height from right angle to hypotenuse, use h = (product of the two legs) ÷ hypotenuse – avoids extra steps.
• Multiplication by √2 or √3: Memorise sin 45° = √2⁄2 and sin 30° = ½, cos 30° = √3⁄2 to speed up.

5. Common mistakes to avoid

• Mixing up opposite and adjacent when the angle is not the one at the base; always draw a quick sketch.
• Using tan⁻¹ directly on a fraction without checking the quadrant – for SSC CGL all angles are acute, but still verify.
• Forgetting that tan 90° is undefined; never write a numeric value.
• Applying Pythagoras to non‑right triangles – only right‑angled triangles qualify.
• Incorrect simplification of radicals – keep √2⁄2 as is; do not convert to 0.707 unless asked for decimal.