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Math Help - Trig - Finding Radius

  1. #1
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    Trig - Finding Radius Q

    I'll let the question speak for itself. I'll give the proceeding question in order to clarify the second question.

    You have a cone shaped bag. At the bottom of the bag is an orange with a radius of 2 inches. On top of the orange is a melon with a radius of 6 inches. It touches the orange and fits snugly in the bag, touching it in a ring around the orange. Its top is at the same level as the top of the bag. All of this is illustrated in this crude figure:


    The height of the cone is [18] inches, and its radius is [10.39230485] inches.

    ****Here is the next Question****

    This is like the preceding problem, except that the radii of the orange and the melon are general. Call the radius of the orange r, the radius of the melon R, and assume r < R. Your answer will be a mathematical expression involving r and R.

    I've drawn up a basic diagram of how this looks below...



    I know that h = 2R^2/R-r, but previously in the other question I used the angles and numbers and the laws of sine and cosine to get the answer. So, I'm a bit stumped on what to do next.

    Any thoughts or comments would be greatly appreciated.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by
    ****Here is the next Question****

    This is like the preceding problem, except that the radii of the orange and the melon are general. Call the radius of the orange r, the radius of the melon R, and assume r < R. Your answer will be a mathematical expression involving r and R.

    I've drawn up a basic diagram of how this looks below...

    [IMG
    http://images5.theimagehosting.com/Cone2Radius.JPG[/IMG]

    I know that h = 2R^2/R-r, but previously in the other question I used the angles and numbers and the laws of sine and cosine to get the answer. So, I'm a bit stumped on what to do next.

    Any thoughts or comments would be greatly appreciated.
    Umm, I have answered the same first question yesterday. Another person, if you are not him, posted it yesterday or the day before.
    Go to the Trigonometry forum/page to see my solution there.

    I did not use Law of sines there. And I will not use that here either. I will just follow my solution there.

    Let y = line segment from bottom of orange to bottom of cone.
    And theta = half of the angle of the "V" or bottom of the cone.
    And X = radius of the cone.
    And H = height of the cone.

    In the small right triangle formed by the radius of the orange, a portion of the wall of the cone, and a portion of H,
    sin(theta) = r / (r+y) ----------(1)

    In the medium right triangle formed by the radius of the melon, a portion of the wall of the cone, and a portion of H,
    sin(theta) = R / (R +2r +y) ----------(2)

    sin(theta) = sin(theta),
    r/(r+y) = R/(R+2r+y)
    Cross multiply,
    r*(R+2r+y) = (r+y)*R
    rR +2r^2 +ry = rR +Ry
    2r^2 = Ry -ry
    y = (2r^2)/(R-r) ----------(3)

    H = 2R +2r +y
    H = 2R +2r +(2r^2)/(R-r)
    H = 2(R+r) +2(r^2)/(R-r)
    H = 2[(R+r)(R-r) +r^2]/(R-r)
    H = 2[R^2 -r^2 +r^2]/(R-r)
    H = 2[R^2]/(R-r)
    H = 2R^2 / (R-r) ---------------answer.

    -------------
    For the cone's radius X:

    In the big right triangle formed by X, the whole wall of cone, and H,
    tan(theta) = X/H
    X = H*tan(theta)
    X = [(2R^2)/(R-r)]tan(theta) ------------------(4)

    We find tan(theta) by way of sin(theta)

    sin(theta) = r/(r+y) ---------(1)
    sin(theta) = r/[r +(2r^2)/(R-r)]
    sin(theta) = r/[r(R-r) +2r^2]/(R-r)
    sin(theta) = r/[rR -r^2 +2r^2]/(R-r)
    sin(theta) = r/[rR +r^2]/(R-r)
    sin(theta) = r/[r(R+r)/(R-r)]
    sin(theta) = (R-r)/(R+r) -----------***

    Using the Pythagorean trig identity
    sin^2(theta) +cos^2(theta) = 1, --------------***
    cos^2(theta) = 1 -sin^2(theta)
    cos^2(theta) = 1 -[(R-r)/(R+r)]^2
    Using (a^2 -b^2) = (a+b)(a-b),
    cos^2(theta) = {1 +(R-r)/(R+r)}*{1 -(R-r)/(R+r)}
    cos^2(theta) = {[1(R+r) +(R-r)]/(R+r)}*{[1(R+r) -(R-r)]/(R+r)}
    cos^2(theta) = {2R/(R+r)}*{2r/(R+r)}
    cos^2(theta) = (4rR)/(R+r)^2
    Take the square roots of both sides,
    cos(theta) = 2sqrt(rR) / (R+r) ------------***

    tan(theta) = sin(theta) / cos(theta)
    tan(theta) = {(R-r)/(R+r)} / {2sqrt(rR) /(R+r)}
    tan(theta) = (R-r) / 2sqrt(rR) ----------------------***

    Substitute that into (4),
    X = [(2R^2)/(R-r)]tan(theta) ------------------(4)
    X = [(2R^2)/(R-r)]*[(R-r)/ 2sqrt(rR)]
    X = (2R^2) / 2sqrt(rR)
    X = R^2 / sqrt(rR) ---------------answer.
    Or, simplifying that furthermore,
    X = R*sqrt(R) / sqrt(r)
    X = R*sqrt(R/r) ------------------answer.
    Or,
    X = R*sqrt(R)sqrt(r) / r
    X = (R/r)sqrt(rR) ----------------answer.
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