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Math Help - P is a point on a unit circle with coordinates (-0.37, 0.93)

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    P is a point on a unit circle with coordinates (-0.37, 0.93)

    If the angle between OP and the positive direction of the x axis is θdegrees find cosθ, tanθ and θdegrees.

    The "positive direction of x axis" throws me. Obviously new to trig.

    Please provide solution with steps.


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    Re: P is a point on a unit circle with coordinates (-0.37, 0.93)

    Quote Originally Posted by durrrr View Post
    If the angle between OP and the positive direction of the x axis is θdegrees find cosθ, tanθ and θdegrees.
    The "positive direction of x axis" throws me. Obviously new to trig.
    Please provide solution with steps.

    There is a problem with this post!
    The point (-.37,.93) is not on the unit circle !
    What are we to make of that?
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    Re: P is a point on a unit circle with coordinates (-0.37, 0.93)

    Well that is confusing. One of the previous questions relating to the question in my OP ask you to draw a circle with it's centre at the origin and radius of one unit. This is a unit circle, is it not?
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    Re: P is a point on a unit circle with coordinates (-0.37, 0.93)

    It isn't quite on the unit circle but it is very close, implying that the given point is an approximation to a point that is on the unit circle. Make sure your point is as exact as it needs to be for this context. (The magnitude of the point you provided is 1.0008995...)

    The "positive direction of x axis" throws me. Obviously new to trig.
    Notice that when you start out at the origin, the x-axis extends forever to both the positive and negative sides. Positive x-axis simply means you go to the right starting from the origin, along the positive numbers. That forms a line, and θ is the angle between this line and OP.
    Last edited by SworD; September 3rd 2012 at 08:55 PM.
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    Re: P is a point on a unit circle with coordinates (-0.37, 0.93)

    Unit circle or not, if \theta is the angle a line from (0, 0) to (x, y) makes with the positive x-axis, then dropping a perpendicular from (x, y) to the x-axis, gives a right triangle having that line as hypotenuse. Assuming that (x, y) is in the first quadrant (that both x and y are positive) the length of the near side, from (0, 0) to (x, 0), is x, the length of the "opposite side", from (x, 0) to (x, y), is y, and the length of the hypotenuse is \sqrt{x^2+ y^2}. You can construct all trig functions from their basic definition and that information.
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