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Math Help - Finding the radius, solving, and finding theta?

  1. #1
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    Finding the radius, solving, and finding theta?

    #1 In a circle, an arc of length 30 cm contains a central angle of 120 degrees. Determine the radius of this circle.

    Work: 30/(2(pi)r) = (3pi/4)/2pi
    r= 40/pi... But this isn't right.

    Answer: 45/pi

    #2 Solve: 5(sin(x))^2 = cos(x)

    Work: Unsure how to start this and finish it. I also tried graphing it and finding the intersections, but I got HUGE numbers (3 digit range).

    Answer: 0.44, 5.84

    #3 The point (5, -6) is on the terminal arm of the standard position angle theta. Determine the smallest positive measure of theta in radians.


    Work: graphed coordinates (5, -6), but don't know how to continue.

    Answer: 5.41
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  2. #2
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    Quote Originally Posted by AlphaRock View Post
    #1 In a circle, an arc of length 30 cm contains a central angle of 120 degrees. Determine the radius of this circle.

    Work: 30/(2(pi)r) = (3pi/4)/2pi
    r= 40/pi... But this isn't right.

    Answer: 45/pi

    #2 Solve: 5(sin(x))^2 = cos(x)

    Work: Unsure how to start this and finish it. I also tried graphing it and finding the intersections, but I got HUGE numbers (3 digit range).

    Answer: 0.44, 5.84

    #3 The point (5, -6) is on the terminal arm of the standard position angle theta. Determine the smallest positive measure of theta in radians.


    Work: graphed coordinates (5, -6), but don't know how to continue.

    Answer: 5.41
    first question

    the angle is 120 in rad it equal \frac{2\pi}{3} this is the angle facing the arc

    the arc length given by

    \theta (r) theta is the angle facing the arc and r is the radius of the circle so you have theta and the arc length r is unknown

    r(\theta) = arclength you solve the equation


    second question 5sin^2(x) = cos(x)

    5(1-cos^2(x) ) = cos(x)

    5-5cos^2(x) = cos(x) \Rightarrow 5cos^2(x) - cos(x) -5 let t=cos(x)

    5t^2 - t - 5 = 0 solve for t by the formula

    \frac{-b\pm \sqrt{b^2-4(a)(b)}}{2a} since a,b and c are

    ax^2 + bx + c =0 after finding t value sub it this t=cosx

    the last question the head of the angle in the origin or what ?
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  3. #3
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    Quote Originally Posted by AlphaRock View Post
    #3 The point (5, -6) is on the terminal arm of the standard position angle theta. Determine the smallest positive measure of theta in radians.

    Work: graphed coordinates (5, -6), but don't know how to continue.

    Answer: 5.41
    Draw a right triangle in the coordinate plane (see Trig Function Point Definitions ) The point is in Quadrant IV, so x = 5 and y = -6. Use the tangent definition:
    \begin{aligned}<br />
\tan \theta &= \frac{y}{x} \\<br />
\tan \theta &= \text{-}\frac{6}{5}<br />
\end{aligned}
    Solve for \theta by taking the inverse tangent of both sides. Remember, though, that inverse tangent only gives angles in \text{-}\frac{\pi}{2} < \theta < \frac{\pi}{2}. Our point is in Quadrant IV, so we're okay.
    \theta = \tan^{-1}\left(\text{-}\frac{6}{5}\right) \approx \text{-}50.19^{\circ} + 360^{\circ} \approx 309.81^{\circ}


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