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Dmitriy789 [7]
3 years ago
13

Express sin A, cos A, and tan A as ratios ​

Mathematics
1 answer:
vlabodo [156]3 years ago
5 0
<h3>Answer: Choice B</h3><h3>sqrt(3)/2,    1/2,   sqrt(3)</h3>

================================================

Explanation:

Sine of an angle is the ratio of the opposite side over the hypotenuse. For reference angle A, the opposite side is BC = 6sqrt(3). The hypotenuse is the longest side AB = 12

Sin(angle) = opposite/hypotenuse

sin(A) = BC/AB

sin(A) = 6sqrt(3)/12

sin(A) = sqrt(3)/2

---------------

Cosine is the ratio of the adjacent and hypotenuse

cos(angle) = adjacent/hypotenuse

cos(A) = AC/AB

cos(A) = 6/12

cos(A) = 1/2

---------------

Tangent is the ratio of the opposite and adjacent

tan(angle) = opposite/adjacent

tan(A) = BC/AC

tan(A) = 6sqrt(3)/6

tan(A) = sqrt(3)

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Give the values of a, b, and c needed to write the equation's general form.<br>1/4x^2+5=0
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Let's solve your equation step-by-step.

\frac{1}{4} x^2+5=0

For this equation: a=0.25, b=0, c=5

0.25x^2 + 0x + 5 = 0

Step 1: Use quadratic formula with a=0.25, b=0, c=5.

x = \frac{-b ± \sqrt{b^2 - 4ac} }{2a}

x = \frac{-(0) ± \sqrt{(0)^2 -4 (0.25) (5)  } }{2(0.25)}

x = \frac{0 ± \sqrt{-5} }{0.5}

Answer : No Real Solutions.

❧⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯☙

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If this helped you, could you maybe give brainliest..?

Also Have a great day/night!

❀*May*❀

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Step-by-step explanation:

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3 years ago
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<img src="https://tex.z-dn.net/?f=9%5C%3A%20%20%5C%3A%20%20%5Cfrac%7B%20%5Csin%28%20%5Ctheta%29%20%7D%7B1%20%2B%20%20%5Ccos%28%2
PIT_PIT [208]

Option (b) is your correct answer.

Step-by-step explanation:

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

Given Trigonometric expression is

\rm :\longmapsto\:\dfrac{sin\theta }{1 + cos\theta }

So, on rationalizing the denominator, we get

\rm \:  =  \: \dfrac{sin\theta }{1 + cos\theta }  \times \dfrac{1 - cos\theta }{1 - cos\theta }

We know,

\purple{\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) =  {x}^{2} -  {y}^{2} \: }}}

So, using this, we get

\rm \:  =  \: \dfrac{sin\theta (1  -  cos\theta )}{1 -  {cos}^{2}\theta  }

We know,

\purple{\rm :\longmapsto\:\boxed{\tt{  {sin}^{2}x +  {cos}^{2}x = 1}}}

So, using this identity, we get

\rm \:  =  \: \dfrac{sin\theta (1 -  cos\theta )}{{sin}^{2}\theta  }

\rm \:  =  \: \dfrac{1 - cos\theta }{sin\theta }

<u>Hence, </u>

\\ \red{\rm\implies \:\boxed{\tt{ \rm \:\dfrac{sin\theta }{1 + cos\theta }   =  \: \dfrac{1 - cos\theta }{sin\theta } }}} \\

3 0
2 years ago
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