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If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to
We are given the equation: x tan 45° cos 60° = sin 60° cot 60°. Step 1: Substitute the trigonometric values Using standard trigonometric values: - tan 45° = 1, - cos 60° = 1/2, - sin 60° = √3/2, - cot 60° = 1/√3. Substitute these values into the equation: x (1) (1/2) = (√3/2) (1/√3). Step 2: SimplifRead more
We are given the equation:
x tan 45° cos 60° = sin 60° cot 60°.
Step 1: Substitute the trigonometric values
Using standard trigonometric values:
– tan 45° = 1,
– cos 60° = 1/2,
– sin 60° = √3/2,
– cot 60° = 1/√3.
Substitute these values into the equation:
x (1) (1/2) = (√3/2) (1/√3).
Step 2: Simplify both sides
Simplify the left-hand side:
x (1/2) = x/2.
Simplify the right-hand side:
(√3/2) (1/√3) = (√3 / √3) / 2 = 1/2.
Thus, the equation becomes:
x/2 = 1/2.
Step 3: Solve for x
Multiply through by 2 to isolate x:
x = 1.
Step 4: Final Answer
The value of x is 1.
The correct answer is:
a) 1
This question pertains to Chapter 8 of the Class 10th NCERT Mathematics textbook, which introduces the topic of Trigonometry. Provide the answer based on your understanding of the chapter.
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Work done in carrying an electron from AxoB lying on an equipotential surface of one volt potential is
The correct answer is (d) Zero. On an equipotential surface, all points have the same electric potential. Since work done (W = qΔV) depends on the potential difference (ΔV) and ΔV = 0 on an equipotential surface, no work is required to move an electron between points. For more visit here: https://wwRead more
The correct answer is (d) Zero.
On an equipotential surface, all points have the same electric potential. Since work done (W = qΔV) depends on the potential difference (ΔV) and ΔV = 0 on an equipotential surface, no work is required to move an electron between points.
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A charge of 10 p. C lies at the centre of a square. Work done in carrying a charge of 2 p C from one comer of square to the diagonally opposite comer is
The correct answer is (c) Zero. The work done in moving a charge in an electric field is given by W = qΔV. Since the charge is at the center of the square, all corners are at the same potential. The potential difference (ΔV) between diagonally opposite corners is zero, making the work done also zeroRead more
The correct answer is (c) Zero.
The work done in moving a charge in an electric field is given by W = qΔV. Since the charge is at the center of the square, all corners are at the same potential. The potential difference (ΔV) between diagonally opposite corners is zero, making the work done also zero, regardless of the path taken.
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N/C is the SI unit of
(c) electric intensity The SI unit of electric field intensity (also called electric field strength) is newtons per coulomb (N/C). This represents the force per unit charge exerted on a small positive test charge placed in the field. For more visit here: https://www.tiwariacademy.com/ncert-solutionsRead more
(c) electric intensity
The SI unit of electric field intensity (also called electric field strength) is newtons per coulomb (N/C). This represents the force per unit charge exerted on a small positive test charge placed in the field.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-2/
If 8 tan x = 15, then sin x – cos x is equal to
We are given: 8 tan x = 15. Step 1: Solve for tan x Rearrange the equation to solve for tan x: tan x = 15/8. Step 2: Express sin x and cos x in terms of tan x Using the identity tan x = sin x / cos x, we can write: sin x = 15k and cos x = 8k, where k is a positive constant such that sin²x + cos²x =Read more
We are given:
8 tan x = 15.
Step 1: Solve for tan x
Rearrange the equation to solve for tan x:
tan x = 15/8.
Step 2: Express sin x and cos x in terms of tan x
Using the identity tan x = sin x / cos x, we can write:
sin x = 15k and cos x = 8k,
where k is a positive constant such that sin²x + cos²x = 1 (Pythagorean identity).
Substitute sin x = 15k and cos x = 8k into the identity:
(15k)² + (8k)² = 1
225k² + 64k² = 1
289k² = 1
k² = 1/289
k = √(1/289)
k = 1/17.
Thus:
sin x = 15k = 15/17,
cos x = 8k = 8/17.
Step 3: Find sin x – cos x
Now, calculate sin x – cos x:
sin x – cos x = (15/17) – (8/17)
= (15 – 8)/17
= 7/17.
Step 4: Final Answer
The value of sin x – cos x is 7/17.
The correct answer is:
d) 7/17
This question related to Chapter 8 Mathematics Class 10th NCERT. From the Chapter 8 Introduction to Trigonometry. Give answer according to your understanding.
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See lesshttps://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-8/