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    I've been looking at this question for so long i've lot hope. It shouldnt be that hard.. but i think cause i've been trying it for so long that im missing something.

    Help would be appreciated

    A road 12m wide runs through a tunnel 18m high. The cross section of the tunnel is a major segment of a circle as in diagram (see attatchment) C2 Question Diagram.doc

    a) Calculate the radius BC of the circle
    b) Find <BCD
    c)Find the cross sectional area of the tunnel

    Even help with a) would be great
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    Anyone?
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    A 12cm wide road?!
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    Whoops... should be metres!
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    a) BC is a radius so BC = 18/2 = 9
    b) Cosine rule because BC = CD
    c) Isn't there a formula 1/2 r^2 (A - sin A) where A is angle BCD? Check your textbook.
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    for part a) it's not 9 cause 18 isn't the diametre of the circle. It only goes up to the stright part.
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    Your diagram is not very clear, but assuming that I understood it, here is one method for a)
    consider the height of your triangle BCD (it cuts BD in half because your triangle is isocele)
    call this point A
    Therefore, using pythagora's theorem, you have BC²=CA²+BA²
    Then, you want to have CA
    so, CA=18-r, where r is the radius (look at the diagram)
    therefore, BC²=6²+(18-r)²
    so r²=36+18²-36r+r²
    <=> 360-36r=0
    <=> r=10
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    (b)
    sin(BCD)=sin(2BCA)=2sin(BCA)cos( BCA)
    =2*(AB/BC)*(AC/BC)
    =(2AB*AC)/(BC²)
    =2*6*8/10²
    =24/25

    Similarly, cos(BCD)=2cos²(BCA)-1 (duplication formulae)
    so cos(BCD)=2(AC/BC)²-1=7/25

    Now, you may use a calculator, (if you cannot use the reciprocal functions)
    but here is a way to do it approximately
    sin(x)=24/25 almost 1 and cos>0 so you'd be looking in the first quadrant, close to pi/2 (around 5pi/12) I do not know however how precisely you need it

    (the exact value still being arccos(0,28) and arcsine(0,96))
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    Sorry about the diagram. Thanks very much That helps
 
 
 
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Updated: January 26, 2010

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