At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

|dw:1353909972372:dw|

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

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

z is an acute angle isn't it?

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. :)

yes.

I already tried all that.

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

No problem. :)