im stuck in this problem..please help

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im stuck in this problem..please help

Mathematics
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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.

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|dw:1327972990128:dw| find the range of values for z
can u please redraw the figure??
ok

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|dw:1327973646692:dw| there?..please help?
that 4z-12 is an angle?
that's how it's drawn in my book..more like this..|dw:1327974111848:dw|
but it's same thing right?
well monster, can u tell if these triangles are right angle?
none of them are?
that's really funny, are you sure u have put every information?
i forgot the 16 on the first triangle..
|dw:1327974424468:dw|
so sorry, but i have no idea right now, hope that any other member can help, i'd like to see the solution
any ideas??
I see three unknowns, z, a (the shorter length) and b (the longer length). I find only two equations, namely by the cosine rule of each triangle. Also, z appears in only one place, so that's "fishy". Can you dig out some more information from the question? For example, copy the question verbatim?
Omar- Thank's Anyways Math Mate- i don't know that's all it says on my book..
65 degrees is opposite to side with length 16 54 degrees is opposite to side with length 4z-19 we know that side opposite to larger angle is larger than the side opposite to smaller angle, so 65>54 16>4z-19 25>4z 25/4>z or z<25/4
Good job! I didn't even read the question line. Grrr.
That just clicked..thank you ash2326 (:

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