math
posted by Amina .
(3/5, 2) is a point on the terminal side of theta, find the value of the six trig functions

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°