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Original post by tsrforum
not sure how to start don't really understand question...
not sure how to start don't really understand question...
xb + by = al
Original post by TeeEm
xb + by = al
Still confused...
Posted from TSR Mobile
Original post by TSRforum
It is meant to be in capitals
AX + BY = AL
Original post by TeeEm
It is meant to be in capitals
AX + BY = AL
AX + BY = AL
How am I suppose to use that equation? 😣
Posted from TSR Mobile
Original post by TSRforum
Yes
mark the angles ... (more places where the angle is theta
then get expressions for these lengths
I think your L is meant to be a Z, TeeEm.
AX + BY = AZ is what @TeeEm meant.
For example, if you consider the length adjacent to the angle (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
For example, if you consider the length adjacent to the angle (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
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 (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
For example, if you consider the length adjacent to the angle (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 ...
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
.
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
.(edited 8 years ago)
Original post by Indeterminate
AX + BY = AZ is what @TeeEm meant.
For example, if you consider the length adjacent to the angle (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
For example, if you consider the length adjacent to the angle (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) ?
Posted from TSR Mobile
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.
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
Posted from TSR Mobile
Original post by TSRforum
Yup. But remember that AX = YZ, so you've got 2 sides sorted out.
Can you spot another theta?
Original post by Indeterminate
Yup. But remember that AX = YZ, so you've got 2 sides sorted out.
Can you spot another theta?
Can you spot another theta?
What about BY? or do I just do AZ-XB to get BY?
Posted from TSR Mobile
Original post by TSRforum
BY involves another angle! Think about geometric properties of angles.
Then do the same stuff again and you'll be good to go!
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!
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
Posted from TSR Mobile
Original post by TSRforum
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
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
Posted from TSR Mobile
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
Posted from TSR Mobile
Then all I need to do is 180-90-(angle ABX) to get CBY to then get BY
Posted from TSR Mobile
If your angle is , and you know that one of the other two angles is a right-angle, then the third angle is
Think about it!
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