CBSE Class 10 Mathematics Assignment – Trigonometry
Name: ____________________________ Class: 10 Subject: Mathematics
Topic: Trigonometry
Date: ____________________________
Assignment: Trigonometry – Questions with Answers
1. Q1. Define trigonometric ratios for an acute angle of a right-angled triangle.
A1. In a right-angled triangle ABC with right angle at B:
sin θ = Perpendicular / Hypotenuse
cos θ = Base / Hypotenuse
tan θ = Perpendicular / Base
cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ
2. Q2. Find the value of sin 30°, cos 45°, and tan 60°.
A2. sin 30° = 1/2, cos 45° = 1/√2, tan 60° = √3
3. Q3. If sin A = 3/5, find cos A and tan A.
A3. sin A = 3/5 ⇒ Perpendicular = 3, Hypotenuse = 5
Base = √(5² - 3²) = 4
So, cos A = 4/5 and tan A = 3/4
4. Q4. Prove that sin²θ + cos²θ = 1.
A4. In a right triangle, sin θ = P/H and cos θ = B/H.
Then sin²θ + cos²θ = (P²/H²) + (B²/H²) = (P² + B²)/H² = H²/H² = 1
5. Q5. If tan A = 1, find the value of sin A + cos A.
A5. tan A = 1 ⇒ A = 45°
sin 45° + cos 45° = (1/√2) + (1/√2) = √2/1 = √2
6. Q6. Evaluate: sin 60° × cos 30° + cos 60° × sin 30°
A6. sin 60° × cos 30° + cos 60° × sin 30° = (√3/2 × √3/2) + (1/2 × 1/2) = 3/4 + 1/4 = 1
Hence, sin 60° × cos 30° + cos 60° × sin 30° = sin(60° + 30°) = sin 90° = 1
7. Q7. If cot θ = 7/24, find sin θ and cos θ.
A7. cot θ = 7/24 ⇒ tan θ = 24/7
Let opposite = 24, adjacent = 7 ⇒ hypotenuse = √(24² + 7²) = 25
sin θ = 24/25, cos θ = 7/25
8. Q8. Find the value of sin 2A, if sin A = 3/5 and A is an acute angle.
Name: ____________________________ Class: 10 Subject: Mathematics
Topic: Trigonometry
Date: ____________________________
Assignment: Trigonometry – Questions with Answers
1. Q1. Define trigonometric ratios for an acute angle of a right-angled triangle.
A1. In a right-angled triangle ABC with right angle at B:
sin θ = Perpendicular / Hypotenuse
cos θ = Base / Hypotenuse
tan θ = Perpendicular / Base
cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ
2. Q2. Find the value of sin 30°, cos 45°, and tan 60°.
A2. sin 30° = 1/2, cos 45° = 1/√2, tan 60° = √3
3. Q3. If sin A = 3/5, find cos A and tan A.
A3. sin A = 3/5 ⇒ Perpendicular = 3, Hypotenuse = 5
Base = √(5² - 3²) = 4
So, cos A = 4/5 and tan A = 3/4
4. Q4. Prove that sin²θ + cos²θ = 1.
A4. In a right triangle, sin θ = P/H and cos θ = B/H.
Then sin²θ + cos²θ = (P²/H²) + (B²/H²) = (P² + B²)/H² = H²/H² = 1
5. Q5. If tan A = 1, find the value of sin A + cos A.
A5. tan A = 1 ⇒ A = 45°
sin 45° + cos 45° = (1/√2) + (1/√2) = √2/1 = √2
6. Q6. Evaluate: sin 60° × cos 30° + cos 60° × sin 30°
A6. sin 60° × cos 30° + cos 60° × sin 30° = (√3/2 × √3/2) + (1/2 × 1/2) = 3/4 + 1/4 = 1
Hence, sin 60° × cos 30° + cos 60° × sin 30° = sin(60° + 30°) = sin 90° = 1
7. Q7. If cot θ = 7/24, find sin θ and cos θ.
A7. cot θ = 7/24 ⇒ tan θ = 24/7
Let opposite = 24, adjacent = 7 ⇒ hypotenuse = √(24² + 7²) = 25
sin θ = 24/25, cos θ = 7/25
8. Q8. Find the value of sin 2A, if sin A = 3/5 and A is an acute angle.
