trig: given m<x=30, m<z=60, yz=6 and xz=18

[pattie33]

given m<x=30, m<z=60, yz=6 and xz=18 to be honest i dont really understand this question and its really confusing me because i dont even know how to start this off to solve it

 

[o_O]

What are you looking for? Also, if x = 30 and z = 60, how does xz = 18? Post the problem in its entirety and maybe we can work something out.

 

[pattie33]

thank you for trying to help but i have the answer and i had to find out the segments that equaled 9 and the answer was YZ

 

[Denis]

pattie, nobody understands what you're saying:
post the original question word for word, IN FULL.

 

[stapel]

i had to find out the segments that equaled 9

You had to find the segments of what?

It sounds like you're working from a picture...? One that we cannot see...? :shock:

Eliz.

 

[soroban]

Hello, Pattie!

given m<x=30, m<z=60, yz=6 and xz=18

I've seen this type of notation.
Evidently, it involves angles and sides.
But the information is inconsistent . . . Is there a typo?


This is what I would guess the problem says:

We have a triangle \(\displaystyle XYZ\)
. . with: \(\displaystyle \:\angle X \,=\,30^o,\;\angle Z\,=\,60^o,\;YZ\,=\,6,\;XZ\,=\,18\)

                        Z
                        *
                     *  *
             18   *  60°*
               *        * 6
            *           *
         *  30°         *
    X * * * * * * * * * * Y

But the measurements are inconsistent.

If \(\displaystyle \angle X\,=\,30^o,\;\angle Z\,=\,60^o\), then \(\displaystyle \angle Y \,=\,90^o\)

We have a 30-60 right triangle.
The sides must be in the ratio: \(\displaystyle \,1\,:\,\sqrt{3}\,:\,2\)

So that if \(\displaystyle YZ\,=\,6\), then \(\displaystyle XZ\) must be 12.

Or if \(\displaystyle XZ\,=\,18\), then \(\displaystyle YZ\) must be 9.


Just read the response . . .
Evidently, we are NOT given \(\displaystyle YZ\).
. . We are asked to find it.