Question

Given that sin 0 = 21/29, what is the value of cose, for 0° <0<90°? please help

Answer 1

3rd option

Explanation:

Using the identity

sin²x + cos²x = 1 ( subtract sin²x from both sides )

cos²x = 1 - sin²x ( take the square root of both sides )

cosx = ±
`√(1-sin^2x)`

Given

sinθ =
`(21)/(29)` and 0 < θ < 90 , then

cosθ

=
`\sqrt{1-((21)/(29))^2 }`

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

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