(-3/5, 2) is a point on the terminal side of theta, find the value of the six trig functions

Answer 1

r^2 = (-3/5)^2 + 2^2 = 109/25
r = √109/5

since (3/5,2) is in quadrant II

sinØ = 2/(√109/5) = 10/√109
cscØ = √109/10

cosØ = (-3/5) /(√109/5) = -3/√109
secØ = -√109/3

tanØ = 2/(-3/5) = -10/3
cotØ = -3/10

btw Ø = appr 106.7°