Question

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

Answer 1

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 )

       = $\sqrt{1-\frac{441}{841} }$

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

Answer 2

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 +    |