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Math Help - Trig functions question

  1. #1
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    Trig functions question

    Two discs,each of length 11.6 cm are laid on a table with their centers 14cm apart.
    find the area common to the two discs?
    i used the formula 1/2(r^2)theta to find the area of sector and then subtract the area of triangle from it 1/2(r^2)sin.theta
    the values i get are 60.14 for area of sector
    and 105.62 (i'm in radians)
    whats going wrong here.are my value right?
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  2. #2
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    Hello kandyfloss
    Quote Originally Posted by kandyfloss View Post
    Two discs,each of length 11.6 cm are laid on a table with their centers 14cm apart.
    find the area common to the two discs?
    i used the formula 1/2(r^2)theta to find the area of sector and then subtract the area of triangle from it 1/2(r^2)sin.theta
    the values i get are 60.14 for area of sector
    and 105.62 (i'm in radians)
    whats going wrong here.are my value right?
    Have a look at the diagram I've attached.

    I assume that it's the radius of the discs that is 11.6 cm. So in the \triangle ABO:
    \cos \theta = \frac{7}{11.6}

     \Rightarrow \theta = 0.9230 radians

    and the area of the overlap is 4 times the area of the shape bounded by the straight lines AB and BP and the arc AP; which is:
    4\times\big(\tfrac12\cdot11.6^2\cdot0.9230 - \tfrac12\cdot11.6\cdot7\cdot\sin(0.9230)\big) (that's the area of sector OAP minus the area of triangle OAB)
    =118.9 \text{ cm}^2, if my arithmetic is correct
    Grandad
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  3. #3
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    Yes 11.6 was the radius and the answer is right 118.9.
    i tried doing the same thing but by using the angle AOC (pls see the picture)
    as 90 deg..because it forms a square with equal radii's as its sides.
    why can i not use 1/2r^2theta now to find the area of sector and area 1/2r^2sintheta of one circle ,subtract them and multiply by two to find the total common area?
    Attached Thumbnails Attached Thumbnails Trig functions question-untitled.jpg  
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  4. #4
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    Hello kandyfloss
    Quote Originally Posted by kandyfloss View Post
    Yes 11.6 was the radius and the answer is right 118.9.
    i tried doing the same thing but by using the angle AOC (pls see the picture)
    as 90 deg..because it forms a square with equal radii's as its sides.
    why can i not use 1/2r^2theta now to find the area of sector and area 1/2r^2sintheta of one circle ,subtract them and multiply by two to find the total common area?
    But it isn't a square just because two adjacent sides are equal, is it? It's simply a rhombus.

    If \angle AOC = 90^o, then \theta would have to be 45^o. Which it isn't. It's \arccos (7/11.6)=52.88^o

    Grandad
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  5. #5
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    ok thank a lot
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