## pratu043 2 years ago ABCD is a quadrilateral where AB is parallel to DC. If x = (4y)/3 and y = (3z)/8 find x, y, and z.

1. pratu043

|dw:1353909972372:dw|

2. darthjavier

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

3. pratu043

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

4. pratu043

z is an acute angle isn't it?

5. pratu043

z + y = 180 and $y = \frac{ 3z }{ 8 }$ $z + \frac{ 3z }{ 8 } = 180$ $\frac{ 8z + 3z }{ 8 } = 180$ $\frac{ 11z }{ 8 } = 180$ $z = \frac{ 180\times8 }{ 11 }$

6. pratu043

I don't think that's right.

7. darthjavier

8. pratu043

What do you mean 'in your graph'?

9. geoffb

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

10. pratu043

Yes. You think it could be a printing mistake?

11. geoffb

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

12. pratu043

Yeah, I never thought of that.

13. darthjavier

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

14. pratu043

The answers given are 96, 96 and 84.

15. geoffb

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

16. geoffb

Then you get 11y/8 = 180

17. geoffb

$$y = \frac{180 \times 8}{11}$$

18. geoffb

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

19. geoffb

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

20. pratu043

yes.

21. pratu043

22. pratu043