Introduction to Trigonometry — Telangana (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Introduction to Trigonometry" — important questions with detailed answers, download PDF for board exam preparation.

Introduction to Trigonometry — Telangana (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Introduction to Trigonometry" — important questions with detailed answers, download PDF for board exam preparation.

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Introduction to Trigonometry — Textbook Questions Answered (9)

Q1: Define the six trigonometric ratios for a right triangle. State the trigonometric values for the standard angles 0°, 30°, 45°, 60°, and 90°.

Trigonometric Ratios (for a right triangle with reference angle θ): Let a right triangle have: - Opposite side (opposite to angle θ) = opposite - Adjacent side (adjacent to angle θ) = adjacent - Hypotenuse = hypotenuse The six ratios are: 1. sin θ = opposite / hypotenuse 2. cos θ = adjacent / hypotenuse 3. tan θ = opposite / adjacent 4. cot θ = adjacent / opposite = 1/tan θ 5. sec θ = hypotenuse / adjacent = 1/cos θ 6. cosec θ = hypotenuse / opposite = 1/sin θ MNEMONIC: SOH-CAH-TOA Sin=Opposite/Hypotenuse, Cos=Adjacent/Hypotenuse, Tan=Opposite/Adjacent ═══════════════════════════════════════════════════════════════ TRIGONOMETRIC VALUES FOR STANDARD ANGLES ═══════════════════════════════════════════════════════════════ Angle θ | 0° | 30° | 45° | 60° | 90° -----------|-------|-------|-------|-------|-------- sin θ | 0 | 1/2 | 1/√2 | √3/2 | 1 cos θ | 1 | √3/2 | 1/√2 | 1/2 | 0 tan θ | 0 | 1/√3 | 1 | √3 | ∞(undef) cot θ | ∞ | √3 | 1 | 1/√3 | 0 sec θ | 1 | 2/√3 | √2 | 2 | ∞(undef) cosec θ | ∞ | 2 | √2 | 2/√3 | 1 Note: 1/√2 = √2/2, and 1/√3 = √3/3 (rationalized forms) Key Observations: - As angle increases from 0° to 90°, sin increases from 0 to 1 - As angle increases from 0° to 90°, cos decreases from 1 to 0 - sin²θ + cos²θ = 1 (Fundamental Identity)

Q2: In a right triangle ABC with right angle at B, AB = 5 cm and AC = 13 cm. Find all six trigonometric ratios of angle A.

Given: Right triangle ABC with ∠B = 90° AB = 5 cm (side adjacent to angle A) AC = 13 cm (hypotenuse) Find: All six trigonometric ratios of angle A Step 1: Find BC (the side opposite to angle A) using Pythagorean theorem. AC² = AB² + BC² 13² = 5² + BC² 169 = 25 + BC² BC² = 144 BC = 12 cm Step 2: Identify the sides relative to angle A. Opposite side (opposite to ∠A) = BC = 12 cm Adjacent side (adjacent to ∠A) = AB = 5 cm Hypotenuse = AC = 13 cm Step 3: Calculate sin A. sin A = opposite / hypotenuse = 12/13 Step 4: Calculate cos A. cos A = adjacent / hypotenuse = 5/13 Step 5: Calculate tan A. tan A = opposite / adjacent = 12/5 Step 6: Calculate cot A. cot A = adjacent / opposite = 5/12 Step 7: Calculate sec A. sec A = hypotenuse / adjacent = 13/5 Step 8: Calculate cosec A. cosec A = hypotenuse / opposite = 13/12 Answer: sin A = 12/13 cos A = 5/13 tan A = 12/5 cot A = 5/12 sec A = 13/5 cosec A = 13/12

Q3: Prove the identity: sin²θ + cos²θ = 1

Prove: sin²θ + cos²θ = 1 Proof: Consider a right triangle with: - Angle θ at one vertex - Opposite side = a - Adjacent side = b - Hypotenuse = c Step 1: Write the definitions of sin θ and cos θ. sin θ = a/c cos θ = b/c Step 2: Square both ratios. sin²θ = a²/c² cos²θ = b²/c² Step 3: Add the squared ratios. sin²θ + cos²θ = a²/c² + b²/c² sin²θ + cos²θ = (a² + b²)/c² Step 4: Apply the Pythagorean theorem. In a right triangle: a² + b² = c² Substitute: sin²θ + cos²θ = c²/c² sin²θ + cos²θ = 1 Hence proved. ✓ This is called the Pythagorean identity and is one of the most important trigonometric identities. Other useful identities derived from this: 1. 1 + tan²θ = sec²θ 2. 1 + cot²θ = cosec²θ

Q4: Simplify: (sin 30° + cos 60°)/(tan 45°) + sec 60° - cosec 30°

Given: Simplify (sin 30° + cos 60°)/(tan 45°) + sec 60° - cosec 30° Step 1: Recall the values of trigonometric functions for standard angles. sin 30° = 1/2 cos 60° = 1/2 tan 45° = 1 sec 60° = 2 cosec 30° = 2 Step 2: Substitute these values. = (1/2 + 1/2)/(1) + 2 - 2 Step 3: Simplify the first term. = (1)/(1) + 2 - 2 = 1 + 2 - 2 = 1 Answer: The simplified expression equals 1

Q5: If sin θ = 3/5 and θ is acute, find cos θ and tan θ.

Given: sin θ = 3/5, θ is acute (0° < θ < 90°) Find: cos θ and tan θ Step 1: Use the Pythagorean identity sin²θ + cos²θ = 1. sin²θ + cos²θ = 1 (3/5)² + cos²θ = 1 9/25 + cos²θ = 1 cos²θ = 1 - 9/25 cos²θ = 25/25 - 9/25 cos²θ = 16/25 Step 2: Take the square root. cos θ = ±4/5 Since θ is acute, cos θ > 0. cos θ = 4/5 Step 3: Find tan θ. tan θ = sin θ / cos θ tan θ = (3/5) / (4/5) tan θ = (3/5) × (5/4) tan θ = 3/4 Answer: cos θ = 4/5 = 0.8 tan θ = 3/4 = 0.75 Verification: sin²θ + cos²θ = (3/5)² + (4/5)² = 9/25 + 16/25 = 25/25 = 1 ✓

Q6: If tan θ = 5/12 and θ is in the first quadrant, find sin θ and cos θ.

Given: tan θ = 5/12, θ is in the first quadrant Find: sin θ and cos θ Step 1: Understand what tan θ = 5/12 means. tan θ = opposite / adjacent = 5/12 This means in a right triangle: Opposite side = 5 Adjacent side = 12 Step 2: Find the hypotenuse using Pythagorean theorem. hypotenuse² = opposite² + adjacent² hypotenuse² = 5² + 12² hypotenuse² = 25 + 144 hypotenuse² = 169 hypotenuse = 13 Step 3: Find sin θ. sin θ = opposite / hypotenuse = 5/13 Step 4: Find cos θ. cos θ = adjacent / hypotenuse = 12/13 Answer: sin θ = 5/13 cos θ = 12/13 Verification: sin²θ + cos²θ = (5/13)² + (12/13)² = 25/169 + 144/169 = 169/169 = 1 ✓ tan θ = sin θ / cos θ = (5/13) / (12/13) = 5/12 ✓

Q7: Prove that: (1 - sin²θ) × sec²θ = 1

Prove: (1 - sin²θ) × sec²θ = 1 Proof: Method 1: Using Pythagorean Identity Step 1: Simplify the first bracket using sin²θ + cos²θ = 1. 1 - sin²θ = cos²θ Step 2: Substitute this into the expression. (1 - sin²θ) × sec²θ = cos²θ × sec²θ Step 3: Recall that sec θ = 1/cos θ, so sec²θ = 1/cos²θ. = cos²θ × (1/cos²θ) = cos²θ / cos²θ = 1 Hence proved. ✓ Method 2: Direct substitution Step 1: Start with the left side. (1 - sin²θ) × sec²θ Step 2: Use sec²θ = 1/cos²θ. = (1 - sin²θ) × (1/cos²θ) = (1 - sin²θ) / cos²θ Step 3: Use sin²θ + cos²θ = 1, so 1 - sin²θ = cos²θ. = cos²θ / cos²θ = 1 Hence proved. ✓

Q8: Evaluate: 3sin 30° + 4cos 60° - 2tan 45° + sec 45°

Given: Evaluate 3sin 30° + 4cos 60° - 2tan 45° + sec 45° Step 1: Recall the standard angle values. sin 30° = 1/2 cos 60° = 1/2 tan 45° = 1 sec 45° = 1/cos 45° = 1/(1/√2) = √2 Step 2: Substitute these values into the expression. = 3(1/2) + 4(1/2) - 2(1) + √2 Step 3: Calculate each term. = 3/2 + 4/2 - 2 + √2 = 3/2 + 2 - 2 + √2 = 3/2 + √2 Step 4: Convert to decimal if needed. = 1.5 + 1.414... = 2.914... (approximately) Answer: 3/2 + √2 or (3 + 2√2)/2

Q9: In a right triangle, one angle is 30°. If the side opposite to the 30° angle is 5 cm, find the hypotenuse and the other leg.

Given: Right triangle with one angle = 30° Side opposite to 30° = 5 cm Find: Hypotenuse and the other leg Setup: In a right triangle, we have: - Right angle (90°) - Angle of 30° - Third angle = 180° - 90° - 30° = 60° Step 1: Find the hypotenuse using sin 30°. sin 30° = opposite / hypotenuse 1/2 = 5 / hypotenuse hypotenuse = 5 × 2 hypotenuse = 10 cm Step 2: Find the other leg (adjacent to 30°) using tan 30°. tan 30° = opposite / adjacent 1/√3 = 5 / adjacent adjacent = 5 × √3 adjacent = 5√3 cm Alternative method for Step 2: Using Pythagorean theorem: hypotenuse² = opposite² + adjacent² 10² = 5² + adjacent² 100 = 25 + adjacent² adjacent² = 75 adjacent = √75 = √(25 × 3) = 5√3 cm Answer: Hypotenuse = 10 cm Other leg (adjacent to 30°) = 5√3 cm ≈ 8.66 cm Verification: sin 30° = 5/10 = 1/2 ✓ tan 30° = 5/(5√3) = 1/√3 ✓ 5² + (5√3)² = 25 + 75 = 100 = 10² ✓

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