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3 years ago

A square has an area of 4 ft^2 what is the lenght of each side

Mathematics

1 answer:

sergiy2304 [10]3 years ago

5 0

area of square = 4ft²

a s area of square = length×lenght

let lenght =x

so area = x²

x²= 4ft²

taking square root

x=2 ftSo lebght of each side is 2 ft.

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Prove the following

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Answer:

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

<h2 /><h2><u>Consider</u></h2>

$\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \dfrac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \}cos(23π+x)cos(2π+x)$

$\rm \: \cos \bigg( \dfrac{3\pi}{2} + x \bigg) = sinx$

$\rm \: {cos \: (2\pi + x) }$

$\rm \: \cot \bigg( \dfrac{3\pi}{2} - x \bigg) \: = \: tanx$

$\rm \: cot(2\pi + x) \: = \: cotx$

So, on substituting all these values, we get

$\rm \: = \: sinx \: cosx \: (tanx \: + \: cotx)$

$\rm \: = \: sinx \: cosx \: \bigg(\dfrac{sinx}{cosx} + \dfrac{cosx}{sinx}$

$\rm \: = \: sinx \: cosx \: \bigg(\dfrac{ {sin}^{2}x + {cos}^{2}x}{cosx \: sinx}$

$\rm \: = \: 1=1$

<h2>Hence,</h2>

$\boxed{\tt{ \cos \bigg( \frac{3\pi}{2} + x \bigg) \cos \: (2\pi + x) \bigg \{ \cot \bigg( \frac{3\pi}{2} - x \bigg) + cot(2\pi + x) \bigg \} = 1}}$

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<h2>ADDITIONAL INFORMATION :-</h2>

Sign of Trigonometric ratios in Quadrants

sin (90°-θ) = cos θ
cos (90°-θ) = sin θ
tan (90°-θ) = cot θ
csc (90°-θ) = sec θ
sec (90°-θ) = csc θ
cot (90°-θ) = tan θ
sin (90°+θ) = cos θ
cos (90°+θ) = -sin θ
tan (90°+θ) = -cot θ
csc (90°+θ) = sec θ
sec (90°+θ) = -csc θ
cot (90°+θ) = -tan θ
sin (180°-θ) = sin θ
cos (180°-θ) = -cos θ
tan (180°-θ) = -tan θ
csc (180°-θ) = csc θ
sec (180°-θ) = -sec θ
cot (180°-θ) = -cot θ
sin (180°+θ) = -sin θ
cos (180°+θ) = -cos θ
tan (180°+θ) = tan θ
csc (180°+θ) = -csc θ
sec (180°+θ) = -sec θ
cot (180°+θ) = cot θ
sin (270°-θ) = -cos θ
cos (270°-θ) = -sin θ
tan (270°-θ) = cot θ
csc (270°-θ) = -sec θ
sec (270°-θ) = -csc θ
cot (270°-θ) = tan θ
sin (270°+θ) = -cos θ
cos (270°+θ) = sin θ
tan (270°+θ) = -cot θ
csc (270°+θ) = -sec θ
sec (270°+θ) = cos θ
cot (270°+θ) = -tan θ

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