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Math Help - polar

  1. #1
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    polar

    13: THANKS HWHELPER

    17: Point by point, a dilation transforms the circle x^2 - 6x + y^2 - 8y = -24 onto the circle x^2 - 14x + y^2 - 4y = -44. Find the center and the magnification factor of this transformation.

    Also:

    19: Use the unit circle to find sin 240 and cos 240? -- (How do I use a unit circle to tell that?! I don't need an answer, I just need someone to point me in the right direction =/ I know what a unit circle is, I just have no idea how I can use that to find sin 240 and cos 240. I know that if I graph (cos theta, sin theta) I get a unit circle. Do I somehow use that to help me?)

    20: Given that cos 80 = .0173648, explain how to find cos100, cos 260, cos 280, and sin 190 without using a calculator.
    Last edited by pyrosilver; September 17th 2008 at 04:49 PM.
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  2. #2
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    13. If r = 12 inches, what is c = circumference? What is 9/c?
    area = 9/c * area of circle, same for angle.
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  3. #3
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    THANK YOU!
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  4. #4
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    Hello, pyrosilver!

    13) A sector of a circle is enclosed by two 12 inch radii and a 9 inch arc.
    What is the area of this sector, to the nearest tenth of a square inch?
    What is the central angle of the sector, to the nearest tenth of a degree?

    The arc length formula is: . s \:=\:r\theta
    . . where r is the radius, and \theta is the central angle in radians.

    We are given: . s = 9,\;r = 12
    Hence: . 9 \:=\:12\theta \quad\Rightarrow\quad <br />
\theta \:=\:\frac{9}{12}\:=\:0.75 \:\approx\:\boxed{43.0^o}

    The area formula is: . A \:=\;\frac{1}{2}r^2\theta

    Therefore: . A \;=\;\frac{1}{2}(12^2)(0.75) \;=\;\boxed{54\text{ in}^2}




    17) Point by point, a dilation transforms the circle x^2 - 6x + y^2 - 8y \:=\: -24
    onto the circle x^2 - 14x + y^2 - 4y \:=\: -44

    Find the center and the magnification factor of this transformation.
    You're expected to know how to complete-the-square.


    The original circle is: . (x-3)^2 + (y-4)^2 \:=\:1

    The new circle is: . (x-7)^2 + (y-2)^2 \:=\:9


    The center moved from (3, 4) to (7, 2); it was magnified by a factor of 3.

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  5. #5
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    Thank you so much for the help! I never learned how to complete the square, but I'll go look up how to do it so I can fully understand how to answer it.

    I added a couple more I was stumped no, btw. Thanks for helping me out, guys.
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    I ADDED NEW QUESTIONS GUYS!!
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  7. #7
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    Quote Originally Posted by pyrosilver View Post
    20: Given that cos 80 = .0173648, explain how to find cos100, cos 260, cos 280, and sin 190 without using a calculator.
    Basic things you need to know:
    1. what is cos 90? cos 180? cos 0? cos(-x) given cos(x)?
    2. cos(x-y) = cos(x) cos(y) + sin(x) sin(y)
    3. sin(x-y) = sin(x)cos(y) - cos(x)sin(y)

    100 = 180 - 80
    260 = 180 + 80
    280 = 360 - 80
    190 = 270 - 80 and 270 = 360 - 90 = 0 - 90

    Does that make it easier?

    Oh and since you cannot use a calculator, you probably cannot use sin 80 = sqrt(1 - cos^2 80).
    But I think none of these actually involve knowing the value of sin 80 since it is always paired with sin 180 or sin 360.
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  8. #8
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    Hello,

    19: Use the unit circle to find sin 240 and cos 240? -- (How do I use a unit circle to tell that?! I don't need an answer, I just need someone to point me in the right direction =/ I know what a unit circle is, I just have no idea how I can use that to find sin 240 and cos 240. I know that if I graph (cos theta, sin theta) I get a unit circle. Do I somehow use that to help me?)
    Note that 240=180+60

    So you find the point that is 60 degrees after 180. See what its coordinates are. If you want to compare them to given values, get the point at 60 degrees, which is the symmetric of 240 with respect to the center of the circle.
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