Question

How do you sketch the angle whose terminal side in standard position passes through `(-7,6sqrt2)` and how do you find sin and cos?

Answer 1

`sin t = (6sqrt2)/11`
`cos t = - 7/11`

Explanation:

Consider right triangle MOm, with
Om = x = - 7
mM = y = `6sqrt2`.
The hypotenuse `OM = sqrt(49 + 72) = sqrt121 = 11`
`sin t = (6sqrt2)/11`
`cos t = - 7/11`