We are given: - The ladder makes an angle of 60° with the ground. - The distance from the foot of the ladder to the wall is 2 m. Using the cosine function: cosθ = (Adjacent side) / (Hypotenuse), where θ = 60°, Adjacent side = 2 m, and Hypotenuse = Length of the ladder (L). Substitute the values: cosRead more
We are given:
– The ladder makes an angle of 60° with the ground.
– The distance from the foot of the ladder to the wall is 2 m.
Using the cosine function:
cosθ = (Adjacent side) / (Hypotenuse),
where θ = 60°, Adjacent side = 2 m, and Hypotenuse = Length of the ladder (L).
Substitute the values:
cos60° = 2 / L.
From trigonometric values, cos60° = 1/2:
1/2 = 2 / L.
Solve for L:
L = 2 × 2 = 4.
This question related to Chapter 9 Mathematics Class 10th NCERT. From the Chapter 9 Applications of Trigonometry. Give answer according to your understanding.
We are given: - The angle of elevation of the sun (θ) = 60°, - Length of the shadow = 30 m. Using the tangent function: tanθ = (Height of the tower) / (Length of the shadow). Substitute the values: tan60° = Height / 30. From trigonometric values, tan60° = √3: √3 = Height / 30. Solve for Height: HeigRead more
We are given:
– The angle of elevation of the sun (θ) = 60°,
– Length of the shadow = 30 m.
Using the tangent function:
tanθ = (Height of the tower) / (Length of the shadow).
Solve for Height:
Height = 30 × √3 = 30√3 m.
This question related to Chapter 9 Mathematics Class 10th NCERT. From the Chapter 9 Applications of Trigonometry. Give answer according to your understanding.
We are given the equations: a cosθ + b sinθ = 4 — (1) a sinθ - b cosθ = 3 — (2) We need to find the value of a² + b². Step 1: Square both equations Square both sides of equation (1): (a cosθ + b sinθ)² = 4² Expand the left-hand side: a² cos²θ + 2ab cosθ sinθ + b² sin²θ = 16 — (3) Square both sides oRead more
We are given the equations:
a cosθ + b sinθ = 4 — (1)
a sinθ – b cosθ = 3 — (2)
We need to find the value of a² + b².
Step 1: Square both equations
Square both sides of equation (1):
(a cosθ + b sinθ)² = 4²
Expand the left-hand side:
a² cos²θ + 2ab cosθ sinθ + b² sin²θ = 16 — (3)
Square both sides of equation (2):
(a sinθ – b cosθ)² = 3²
Expand the left-hand side:
a² sin²θ – 2ab sinθ cosθ + b² cos²θ = 9 — (4)
Step 2: Add equations (3) and (4)
Add the expanded forms of equations (3) and (4):
(a² cos²θ + 2ab cosθ sinθ + b² sin²θ) + (a² sin²θ – 2ab sinθ cosθ + b² cos²θ) = 16 + 9
Simplify the terms:
– The terms involving 2ab cosθ sinθ cancel out.
– Combine the remaining terms:
a² (cos²θ + sin²θ) + b² (sin²θ + cos²θ) = 25
Step 3: Use the Pythagorean identity
From the Pythagorean identity, we know:
cos²θ + sin²θ = 1.
Substitute this into the equation:
a² (1) + b² (1) = 25
Simplify:
a² + b² = 25.
The question is connected to Chapter 8, “Introduction to Trigonometry,” in the Class 10th NCERT Mathematics book. Respond based on your comprehension of the chapter.
Given: sinθ - cosθ = 0. 1. Solve for θ: sinθ = cosθ ⇒ tanθ = 1 ⇒ θ = 45°. 2. For θ = 45°: sinθ = cosθ = 1/√2. 3. Calculate sin⁴θ + cos⁴θ: sin⁴θ + cos⁴θ = (sin²θ)² + (cos²θ)² = (1/2)² + (1/2)² = 1/4 + 1/4 = 1/2. The question is based on Chapter 8 of the Class 10th NCERT Mathematics textbook, titled "Read more
Given: sinθ – cosθ = 0.
1. Solve for θ:
sinθ = cosθ ⇒ tanθ = 1 ⇒ θ = 45°.
2. For θ = 45°:
sinθ = cosθ = 1/√2.
3. Calculate sin⁴θ + cos⁴θ:
sin⁴θ + cos⁴θ = (sin²θ)² + (cos²θ)²
= (1/2)² + (1/2)²
= 1/4 + 1/4
= 1/2.
The question is based on Chapter 8 of the Class 10th NCERT Mathematics textbook, titled “Introduction to Trigonometry.” Provide your response in line with the concepts covered in this chapter.
4. Result: 2sinθcosθ / sinθcosθ = 2.
This question related to Chapter 8 Mathematics Class 10th NCERT. From the Chapter 8 Introduction to Trigonometry. Give answer according to your understanding.
A ladder makes an angle of 60° with the ground when placed against a wall. If the foot of ladder is 2 m away from the wall, then the length of the ladder ( in metres) is
We are given: - The ladder makes an angle of 60° with the ground. - The distance from the foot of the ladder to the wall is 2 m. Using the cosine function: cosθ = (Adjacent side) / (Hypotenuse), where θ = 60°, Adjacent side = 2 m, and Hypotenuse = Length of the ladder (L). Substitute the values: cosRead more
We are given:
– The ladder makes an angle of 60° with the ground.
– The distance from the foot of the ladder to the wall is 2 m.
Using the cosine function:
cosθ = (Adjacent side) / (Hypotenuse),
where θ = 60°, Adjacent side = 2 m, and Hypotenuse = Length of the ladder (L).
Substitute the values:
cos60° = 2 / L.
From trigonometric values, cos60° = 1/2:
1/2 = 2 / L.
Solve for L:
L = 2 × 2 = 4.
This question related to Chapter 9 Mathematics Class 10th NCERT. From the Chapter 9 Applications of Trigonometry. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
If the altitude of the sum is at 60°, then the height of the vertical tower that will cast a shadow of length 30m is
We are given: - The angle of elevation of the sun (θ) = 60°, - Length of the shadow = 30 m. Using the tangent function: tanθ = (Height of the tower) / (Length of the shadow). Substitute the values: tan60° = Height / 30. From trigonometric values, tan60° = √3: √3 = Height / 30. Solve for Height: HeigRead more
We are given:
– The angle of elevation of the sun (θ) = 60°,
– Length of the shadow = 30 m.
Using the tangent function:
tanθ = (Height of the tower) / (Length of the shadow).
Substitute the values:
tan60° = Height / 30.
From trigonometric values, tan60° = √3:
√3 = Height / 30.
Solve for Height:
Height = 30 × √3 = 30√3 m.
This question related to Chapter 9 Mathematics Class 10th NCERT. From the Chapter 9 Applications of Trigonometry. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
If a cos θ + b sinθ = 4 and a sin θ – b cos θ = 3, then a² + b² =
We are given the equations: a cosθ + b sinθ = 4 — (1) a sinθ - b cosθ = 3 — (2) We need to find the value of a² + b². Step 1: Square both equations Square both sides of equation (1): (a cosθ + b sinθ)² = 4² Expand the left-hand side: a² cos²θ + 2ab cosθ sinθ + b² sin²θ = 16 — (3) Square both sides oRead more
We are given the equations:
a cosθ + b sinθ = 4 — (1)
a sinθ – b cosθ = 3 — (2)
We need to find the value of a² + b².
Step 1: Square both equations
Square both sides of equation (1):
(a cosθ + b sinθ)² = 4²
Expand the left-hand side:
a² cos²θ + 2ab cosθ sinθ + b² sin²θ = 16 — (3)
Square both sides of equation (2):
(a sinθ – b cosθ)² = 3²
Expand the left-hand side:
a² sin²θ – 2ab sinθ cosθ + b² cos²θ = 9 — (4)
Step 2: Add equations (3) and (4)
Add the expanded forms of equations (3) and (4):
(a² cos²θ + 2ab cosθ sinθ + b² sin²θ) + (a² sin²θ – 2ab sinθ cosθ + b² cos²θ) = 16 + 9
Simplify the terms:
– The terms involving 2ab cosθ sinθ cancel out.
– Combine the remaining terms:
a² (cos²θ + sin²θ) + b² (sin²θ + cos²θ) = 25
Step 3: Use the Pythagorean identity
From the Pythagorean identity, we know:
cos²θ + sin²θ = 1.
Substitute this into the equation:
a² (1) + b² (1) = 25
Simplify:
a² + b² = 25.
The question is connected to Chapter 8, “Introduction to Trigonometry,” in the Class 10th NCERT Mathematics book. Respond based on your comprehension of the chapter.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
If sin θ – cos θ = 0, then the value of sin⁴ θ + cos⁴ θ is
Given: sinθ - cosθ = 0. 1. Solve for θ: sinθ = cosθ ⇒ tanθ = 1 ⇒ θ = 45°. 2. For θ = 45°: sinθ = cosθ = 1/√2. 3. Calculate sin⁴θ + cos⁴θ: sin⁴θ + cos⁴θ = (sin²θ)² + (cos²θ)² = (1/2)² + (1/2)² = 1/4 + 1/4 = 1/2. The question is based on Chapter 8 of the Class 10th NCERT Mathematics textbook, titled "Read more
Given: sinθ – cosθ = 0.
1. Solve for θ:
sinθ = cosθ ⇒ tanθ = 1 ⇒ θ = 45°.
2. For θ = 45°:
sinθ = cosθ = 1/√2.
3. Calculate sin⁴θ + cos⁴θ:
sin⁴θ + cos⁴θ = (sin²θ)² + (cos²θ)²
= (1/2)² + (1/2)²
= 1/4 + 1/4
= 1/2.
The question is based on Chapter 8 of the Class 10th NCERT Mathematics textbook, titled “Introduction to Trigonometry.” Provide your response in line with the concepts covered in this chapter.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
The value of (1 + cot θ – cosec θ) (1 + tan θ + sec θ) is
Given: (1 + cotθ - cosecθ)(1 + tanθ + secθ). 1. Rewrite in terms of sinθ and cosθ: (1 + cosθ/sinθ - 1/sinθ)(1 + sinθ/cosθ + 1/cosθ). 2. Simplify each factor: First factor: (sinθ + cosθ - 1)/sinθ. Second factor: (cosθ + sinθ + 1)/cosθ. 3. Multiply and simplify: Numerator: (sinθ + cosθ)² - 1 = 2sinθcoRead more
Given: (1 + cotθ – cosecθ)(1 + tanθ + secθ).
1. Rewrite in terms of sinθ and cosθ:
(1 + cosθ/sinθ – 1/sinθ)(1 + sinθ/cosθ + 1/cosθ).
2. Simplify each factor:
First factor: (sinθ + cosθ – 1)/sinθ.
Second factor: (cosθ + sinθ + 1)/cosθ.
3. Multiply and simplify:
Numerator: (sinθ + cosθ)² – 1 = 2sinθcosθ.
Denominator: sinθcosθ.
4. Result: 2sinθcosθ / sinθcosθ = 2.
This question related to Chapter 8 Mathematics Class 10th NCERT. From the Chapter 8 Introduction to Trigonometry. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/