the sine of an angle is 0.6. the constraint is the angle lies in quadrant II.

Question

how do I figure out what x is to solve the six functions?

Answer 1

sin Ø = .6 = 6/10 = 3/5
you should recognize the famous 3-4-5 right-angled triangle.
so in quadrant II,
construct your right-angled triangle in quadrant II

sinØ = opposite/hypotenuse = y/r
y = 3, r = 5
and in II, x = -4

so ...
sinØ = 3/5 , csc Ø = 5/3
cosØ = -4/5 , sec Ø = -5/4
tan Ø = - 3/4 , cot Ø = - 4/3

Answer 2

oh! thank you!