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ABCD is a quadrilateral where AB is parallel to DC. If x = (4y)/3 and y = (3z)/8 find x, y, and z.

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\[\angle A+\angle D=180^\circ\\\angle B + \angle C=180^\circ\] Try to solve it now :)
I tried but I'm getting z as more than 90 degrees.

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Other answers:

z is an acute angle isn't it?
z + y = 180 and \[y = \frac{ 3z }{ 8 }\] \[z + \frac{ 3z }{ 8 } = 180\] \[\frac{ 8z + 3z }{ 8 } = 180\] \[\frac{ 11z }{ 8 } = 180\] \[z = \frac{ 180\times8 }{ 11 }\]
I don't think that's right.
Go ahead, you are right. Try to put z instead of y in your graph ;)
What do you mean 'in your graph'?
if y = 3/8 of z, it will be smaller than z. Are you sure you have y and z labelled correctly?
Yes. You think it could be a printing mistake?
No clue, but if z is acute, and y is 3/8 of z, then there's no way y can be obtuse.
Yeah, I never thought of that.
Sorry, graphic*. I mean that you have the wrong graphic or wrong data.
The answers given are 96, 96 and 84.
Try it the other way around. z = 3y/8
Then you get 11y/8 = 180
\(y = \frac{180 \times 8}{11}\)
That gives an angle of 131, meaning z would be 49. Doesn't work.
Well, it works, but not to the answers you're given. :)
I already tried all that.
Well, thanks for your help anyway. I'll ask my maths teacher.
No problem. :)

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