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    Not sure how to start don't really understand question...
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    (Original post by tsrforum)
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    not sure how to start don't really understand question...
    xb + by = al
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    (Original post by TeeEm)
    xb + by = al
    Still confused...


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    (Original post by TSRforum)
    Still confused...


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    It is meant to be in capitals
    AX + BY = AL
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    (Original post by TeeEm)
    It is meant to be in capitals
    AX + BY = AL
    How am I suppose to use that equation? 😣


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    (Original post by TSRforum)
    How am I suppose to use that equation? 😣


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    Yes

    mark the angles ... (more places where the angle is theta
    then get expressions for these lengths
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    I think your L is meant to be a Z, TeeEm.
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    Still don't understand, what do you mean by "mark the angles"?
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    AX + BY = AZ is what TeeEm meant.

    For example, if you consider the length adjacent to the angle \theta (i.e the width of the rectangle), you can use trig ratios to find expressions for AX and XB.

    That should give you an idea of what the question's asking for
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    (Original post by appliedmathlete)
    I think your L is meant to be a Z, TeeEm.
    (Original post by Indeterminate)
    AX + BY = AZ is what TeeEm meant.

    For example, if you consider the length adjacent to the angle \theta (i.e the width of the rectangle), you can use trig ratios to find expressions for AX and XB.

    That should give you an idea of what the question's asking for
    sorry the picture is cropped off on the top right and it looked like an L ...
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    It's asking "For what value of theta does the rectangle have the greatest perimeter (Part b), and what is this perimeter(Part a)".

    You can work out that;

    AX = 2 cos(θ)
    AC = sqrt(5)
    ∠ BAC = 45°
    ∠ CAZ = (45-θ)°
    AZ = sqrt(5)*cos(45-θ)

    Therefore the entire perimeter = 4*cos(θ) + 2*sqrt(5)*cos(45-θ)

    I have no idea how to simplify this to R*cos(θ-α) or how to solve it numerically but plotting it on a graph shows the maximum perimeter is approximately 7.39 when θ is approximately 0.55㎭

    PS. I've just started C2 trig identities so this could all be wrong.
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    (Original post by Indeterminate)
    AX + BY = AZ is what TeeEm meant.

    For example, if you consider the length adjacent to the angle \theta (i.e the width of the rectangle), you can use trig ratios to find expressions for AX and XB.

    That should give you an idea of what the question's asking for
    So 2cos(theta) = AX and XB = 2sin(theta) ?


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    (Original post by Notorious AHIG)
    It's asking "For what value of theta does the rectangle have the greatest perimeter (Part b), and what is this perimeter(Part a)".

    You can work out that;

    AX = 2 cos(θ)
    AC = sqrt(5)
    ∠ BAC = 45°
    ∠ CAZ = (45-θ)°
    AZ = sqrt(5)*cos(45-θ)

    Therefore the entire perimeter = 4*cos(θ) + 2*sqrt(5)*cos(45-θ)

    I have no idea how to simplify this to R*cos(θ-α) or how to solve it numerically but plotting it on a graph shows the maximum perimeter is approximately 5.77 when θ is approximately 0.74㎭

    PS. I've just started C2 trig identities so this could all be wrong.
    The equation expanded is Rcos(X)cos(a) + Rsin(X)sin(a)

    Rcos(a) = 4
    Rsin(a) = 2*sqrt(5)

    R = 6 a= 0.84


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    (Original post by TSRforum)
    So 2cos(theta) = AX and XB = 2sin(theta) ?


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    Yup. But remember that AX = YZ, so you've got 2 sides sorted out.

    Can you spot another theta?
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    (Original post by Indeterminate)
    Yup. But remember that AX = YZ, so you've got 2 sides sorted out.

    Can you spot another theta?
    What about BY? or do I just do AZ-XB to get BY?


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    (Original post by TSRforum)
    What about BY? or do I just do AZ-XB to get BY?


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    BY involves another angle! Think about geometric properties of angles.

    Then do the same stuff again and you'll be good to go!
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    (Original post by Indeterminate)
    BY involves another angle! Think about geometric properties of angles.

    Then do the same stuff again and you'll be good to go!
    You mean the Z/F rule? I don't see how I can use that here


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    (Original post by TSRforum)
    You mean the Z/F rule? I don't see how I can use that here


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    It might help to label the angles in triangle AXB (considering the sum is 180), and then use the fact that angles on a straight line add up to 180 as well
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    (Original post by Indeterminate)
    It might help to label the angles in triangle AXB (considering the sum is 180), and then use the fact that angles on a straight line add up to 180 as well
    I don't have enough data to do that how can I find ABX? I need to know at least 1 angle.
    Then all I need to do is 180-90-(angle ABX) to get CBY to then get BY


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    (Original post by TSRforum)
    I don't have enough data to do that how can I find ABX? I need to know at least 1 angle.
    Then all I need to do is 180-90-(angle ABX) to get CBY to then get BY


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    If your angle is \theta, and you know that one of the other two angles is a right-angle, then the third angle is

    180 - 90 -  \theta

    Think about it!
 
 
 
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