pratu043
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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pratu043
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dw:1353909972372:dw

darthjavier
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\[\angle A+\angle D=180^\circ\\\angle B + \angle C=180^\circ\]
Try to solve it now :)

pratu043
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I tried but I'm getting z as more than 90 degrees.

pratu043
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z is an acute angle isn't it?

pratu043
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z + y = 180
and
\[y = \frac{ 3z }{ 8 }\]
\[z + \frac{ 3z }{ 8 } = 180\]
\[\frac{ 8z + 3z }{ 8 } = 180\]
\[\frac{ 11z }{ 8 } = 180\]
\[z = \frac{ 180\times8 }{ 11 }\]

pratu043
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I don't think that's right.

darthjavier
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Go ahead, you are right. Try to put z instead of y in your graph ;)

pratu043
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What do you mean 'in your graph'?

geoffb
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if y = 3/8 of z, it will be smaller than z. Are you sure you have y and z labelled correctly?

pratu043
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Yes. You think it could be a printing mistake?

geoffb
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No clue, but if z is acute, and y is 3/8 of z, then there's no way y can be obtuse.

pratu043
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Yeah, I never thought of that.

darthjavier
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Sorry, graphic*. I mean that you have the wrong graphic or wrong data.

pratu043
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The answers given are 96, 96 and 84.

geoffb
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Try it the other way around. z = 3y/8

geoffb
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Then you get 11y/8 = 180

geoffb
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\(y = \frac{180 \times 8}{11}\)

geoffb
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That gives an angle of 131, meaning z would be 49. Doesn't work.

geoffb
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Well, it works, but not to the answers you're given. :)

pratu043
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yes.

pratu043
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I already tried all that.

pratu043
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Well, thanks for your help anyway. I'll ask my maths teacher.

geoffb
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No problem. :)