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