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Math Help - Stuck on a question

  1. #1
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    Stuck on a question

    Hey, I have a trig question that I am stuck on and would appreciate any help, Thanks.
    Two sides of a parallelogram measure 7.0 cm and 9.0 cm. The longer
    diagonal is 12.0 cm long.
    a) calculate all the interior angles.
    b) How long is the other diagonal?

    The answer page got for a) 96 and 84 degrees. for b) 10.8cm

    I got the correct degrees, but I can't get 10.8cm.
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  2. #2
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    Use Law of Cosines with side lengths 7 and 9, and angle measure 84 deg.

    (Make sure your calculator is in degrees)
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  3. #3
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    I used the cosine law, maybe i'm doing it wrong. This is what I did.

    my sketch

    A------------------D For some reason I can't draw the pic , but D and C are connected as well as A and B. (parrallelogram)
    / /
    / /7cm
    / /
    B-----------------C
    9cm

    With a line from A to C which is 12.0cm.
    And a line from B to D which I need to figure out.

    I used Angle BDC

    c^2 = b^2 + d^2 - 2bd cosC
    144 = 130 - 126cosC
    130-144 = -126cosC
    -14 = -126cosC
    -14/-126 = cosC
    0.1111... = cosC
    c= cos^-1(0.1111...)
    c= 83 degrees

    but how can it be 83 degrees if its obtuse? what did i do wrong? did I draw it wrong?
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Detroit, MI
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    You've got a,b, and c. all you have to do is solve this for \arccos(\gamma)

    The fact that a parrallelogram is 360degrees should do the rest.
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  5. #5
    MHF Contributor
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    Cosine Rule

    Hello waxer1987
    Quote Originally Posted by waxer1987 View Post
    I used the cosine law, maybe i'm doing it wrong. This is what I did.

    my sketch

    A------------------D For some reason I can't draw the pic , but D and C are connected as well as A and B. (parrallelogram)
    / /
    / /7cm
    / /
    B-----------------C
    9cm

    With a line from A to C which is 12.0cm.
    And a line from B to D which I need to figure out.

    I used Angle BDC

    c^2 = b^2 + d^2 - 2bd cosC
    144 = 130 - 126cosC
    130-144 = -126cosC Should be 144 - 130 = -126 cos C
    -14 = -126cosC
    -14/-126 = cosC
    0.1111... = cosC
    c= cos^-1(0.1111...) So cos C = -0.1111...
    c= 83 degrees C = 96 degrees

    but how can it be 83 degrees if its obtuse? what did i do wrong? did I draw it wrong?
    I've corrected your mistake. So the acute angles are both 84^o (to the nearest degree).

    For the shorter diagonal, use the acute angle in the Cosine Rule:

    a^2 = 130 - 126 \cos 83.6^o


    = 116

    \Rightarrow a = 10.8

    Grandad
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