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

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    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°

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