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Thread: trigo 6

  1. #1
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    trigo 6

    A rectangular lamina ABCD has sides AB=CD=2a and BC=AD=a . The midpoint of CD is M and the midpoint of BM is N . THe lamina is folded along BM such that the planes BMC and BMDA are perpendicular . For this folded lamina , find in terms of surds if necessary .

    (a) AN
    (b) AC
    (c) angle ACB

    This is what i did :

    (a) BM = $\displaystyle a\sqrt{2} $, $\displaystyle BN =\frac{a\sqrt{2}}{2}$ , AB=2a

    AM=AB=2a , cos angle MBC $\displaystyle = \frac{\sqrt{2}}{2} \Rightarrow $
    angle MBC=45 degree , therefore ABN= 45 degree

    $\displaystyle
    AN^2=(2a)^2+(\frac{a\sqrt{2}}{2})^2-2(2a)(a\sqrt{2}{2})cos 45
    $

    $\displaystyle
    AN=\frac{\sqrt{10}}{2}a
    $

    (b) AMC=90 degree

    MC=a , AM = 2a

    $\displaystyle AC=\sqrt{(2a)^2+a^2}=a\sqrt{5}$

    (c) ABC=90

    tan angle ACB = 2a/a

    ACB=arctan 2

    = 63 degree 26 mins


    CAn somepne pls check my working . Thanks a lot !!
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  2. #2
    MHF Contributor Amer's Avatar
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    Jordan
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    Quote Originally Posted by thereddevils View Post
    A rectangular lamina ABCD has sides AB=CD=2a and BC=AD=a . The midpoint of CD is M and the midpoint of BM is N . THe lamina is folded along BM such that the planes BMC and BMDA are perpendicular . For this folded lamina , find in terms of surds if necessary .

    (a) AN
    (b) AC
    (c) angle ACB

    This is what i did :

    (a) BM = $\displaystyle a\sqrt{2} $, $\displaystyle BN =\frac{a\sqrt{2}}{2}$ , AB=2a

    AM=AB=2a , cos angle MBC $\displaystyle = \frac{\sqrt{2}}{2} \Rightarrow $
    angle MBC=45 degree , therefore ABN= 45 degree

    $\displaystyle
    AN^2=(2a)^2+(\frac{a\sqrt{2}}{2})^2-2(2a)(a\sqrt{2}{2})cos 45
    $


    $\displaystyle
    AN=\frac{\sqrt{10}}{2}a
    $

    (b) AMC=90 degree

    MC=a , AM = 2a

    $\displaystyle AC=\sqrt{(2a)^2+a^2}=a\sqrt{5}$

    (c) ABC=90

    tan angle ACB = 2a/a

    ACB=arctan 2

    = 63 degree 26 mins


    CAn somepne pls check my working . Thanks a lot !!

    $\displaystyle
    AN^2=(2a)^2+(\frac{a\sqrt{2}}{2})^2-2(2a)(a\sqrt{2}{2})cos 45
    $
    should be

    $\displaystyle AN^2 = (2a)^2 + \left(\frac{a\sqrt{2}}{2}\right)^2 - 2(2a)\left(\frac{a\sqrt{2}}{2}\right)\cos 45^o$


    $\displaystyle AN^2 = (2a)^2 + \left(\frac{a\sqrt{2}}{2}\right)^2 - 2(2a)\left(\frac{a\sqrt{2}}{2}\right)\left(\frac{1 }{\sqrt{2}}\right)$


    $\displaystyle AN^2 = (2a)^2 + \left(\frac{a}{\sqrt{2}}\right)^2 - 2(2a)\left(\frac{a}{\sqrt{2}}\right)\left(\frac{1} {\sqrt{2}}\right)$

    $\displaystyle AN^2 = 4a^2 + \left(\frac{a^2}{2}\right) - 2a^2$

    $\displaystyle AN^2=2a^2+\frac{a^2}{2}$

    $\displaystyle AN^2=\frac{5a^2}{2}\Rightarrow AN=a\sqrt{\frac{5}{2}}$


    b) and c) is correct
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  3. #3
    Senior Member
    Joined
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    Quote Originally Posted by Amer View Post
    $\displaystyle
    AN^2=(2a)^2+(\frac{a\sqrt{2}}{2})^2-2(2a)(a\sqrt{2}{2})cos 45
    $
    should be

    $\displaystyle AN^2 = (2a)^2 + \left(\frac{a\sqrt{2}}{2}\right)^2 - 2(2a)\left(\frac{a\sqrt{2}}{2}\right)\cos 45^o$


    $\displaystyle AN^2 = (2a)^2 + \left(\frac{a\sqrt{2}}{2}\right)^2 - 2(2a)\left(\frac{a\sqrt{2}}{2}\right)\left(\frac{1 }{\sqrt{2}}\right)$


    $\displaystyle AN^2 = (2a)^2 + \left(\frac{a}{\sqrt{2}}\right)^2 - 2(2a)\left(\frac{a}{\sqrt{2}}\right)\left(\frac{1} {\sqrt{2}}\right)$

    $\displaystyle AN^2 = 4a^2 + \left(\frac{a^2}{2}\right) - 2a^2$

    $\displaystyle AN^2=2a^2+\frac{a^2}{2}$

    $\displaystyle AN^2=\frac{5a^2}{2}\Rightarrow AN=a\sqrt{\frac{5}{2}}$


    b) and c) is correct
    THANKS so i think the answer given for b and c in my book is wrong .
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