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Mars2501 [29]
3 years ago
8

Given that sine= 21/29, what is the value of cos 0, for 0° <0<90°? A -square root of 20/29 B -20/29 C 20/29 D square root

of 20/29
Mathematics
2 answers:
Allushta [10]3 years ago
8 0

Answer:

C

Step-by-step explanation:

Using the trigonometric identity

sin²x + cos²x = 1 ⇒ cosx = ± \sqrt{1-sin^2x}

Given

sinx = \frac{21}{29}, then

cosx = \sqrt{1-(\frac{21}{29})^2 } ( positive since 0 < x < 90 )

       = \sqrt{1-\frac{441}{841} }

       = \sqrt{\frac{400}{841} } = \frac{20}{29}

Stolb23 [73]3 years ago
4 0

Answer:

Step-by-step explanation:

sin=y/r

cos=x/r

sin=21/29

cos=x/29

x^2+y^2=r^2

x^2+21^2=29^2

x^2+441=841

x=sqrt(841-441)

x=20

cos=20/29

                       

  Only              |

   sin +             |     All Positive

---------------------|-------------------

           only      |    Only cos +

            tan +    |

                     

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Answer:

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Solution:

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Answer:

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