If x is an angle in standard position with point (6,-8) on its terminal side, find sin x. ?

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  • 4 months ago

    Point (6, -8) is in quadrant IV where sinθ < 0

    Also, we have a 6, 8, 10 triangle using Pythagoras' theorem.

    so, sinθ = -8/10 => -0.8

    :)>

  • They really should have used a different symbol for the angle, rather than x, but no matter.  Let's call it t, shall we?

    sin(t) = y / sqrt(x^2 + y^2)

    That's all there is to it

    y = -8

    x = 6

    sqrt(x^2 + y^2) = sqrt(36 + 64) = sqrt(100) = 10

    sin(t) = -8 / 10 = -4/5

    -4/5 is the answer.

    See, if I had given you the definition for the sine of an angle and I had used "x" to denote the angle, then this is what you would have gotten:

    sin(x) = y / sqrt(x^2 + y^2)

    And that's just needlessly confusing.  You should bring this up with your teacher.

    -4/5 = -0.8

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