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

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 + |

Answer 2

C

Explanation:

Using the trigonometric identity

sin²x + cos²x = 1 ⇒ cosx = ±
`√(1-sin^2x)`

Given

sinx =
`(21)/(29)`, then

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

=
`\sqrt{1-(441)/(841) }`

=
`\sqrt{(400)/(841) }` =
`(20)/(29)`