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I want to determine the coordinates of vertices of a triangle in geometry 3D, knowing the measure of interior angles are $15^\circ$, $30^\circ$, $135^\circ$. and the length of one side is 3. Please help me. If the measure of interior angles are $30^\circ$, $30^\circ$, $120^\circ$ and the length of one side is 3, I use Maple, I find the other sides are 3 and $\sqrt{3}$. And then, I determine the coordinates of vertices of this triangle. 

    restart:
    a:=3:
    eq1:=(a^2+b^2-c^2)/(2*a*b):
    eq2:=(b^2+c^2-a^2)/(2*b*c):
    eq3:=(c^2+a^2-b^2)/(2*a*c):
    solve([eq1=cos(30*Pi/180),eq2=cos(30*Pi/180),eq3=cos(120*Pi/180)],[c,b]);
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You'll only get a triangle up to similarity. You need more than three angles. – EuYu Nov 9 '12 at 15:09
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You're really going to have to add more detail. Knowing the length of one side is not enough to recover the triangle. You want the triangle embedded in three dimensions so you're going to have to include a position and an orientation. As it stands, this question is nonsense. – EuYu Nov 9 '12 at 15:37
You will need to specify some information - for example, the location of one point (e.g. one of the vertices, the centroid); the length of one identified side (which angles are at the ends); and the direction of the normal to the plane of the triangle to give its orientation (and you need some convention as to the sign of the normal). – Mark Bennet Nov 9 '12 at 15:40
Thank you for all comments. I need to see my question carefully. – minthao_2011 Nov 9 '12 at 15:45

closed as not a real question by amWhy, EuYu, Thomas, Arkamis, Jason DeVito Nov 9 '12 at 18:08

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

1 Answer

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There are triangles of any size you want with those angles inside. You need more information.

Edit: This answer was perfectly fine before the ensuing four edits, which appended enough information to give a real answer.

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Perhaps your "answer", as it stands, would better serve as a comment? See, e.g. @EuYu 's earlier comment. – amWhy Nov 9 '12 at 15:11
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This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – Arkamis Nov 9 '12 at 16:06
@amWhy This is a perfectly fine answer for the question that existed when I posted this (the first edition). Since then it has been altered dramatically. – rschwieb Nov 9 '12 at 18:27
What you wrote was preceded by my COMMENT, which I have since I deleted... but was in plain view prior to your answer. I simply suggested where I think observations such as yours belong. It wasn't a criticism. There is still not enough info in the question, by the way, to obtain the coordinates for which the question was asked. – amWhy Nov 9 '12 at 18:30
I know there are a multitude of triangles satisfying the conditions that I asked, but I only want a triangle like that. First, I find two remaining sides. And I found $b = 3\sqrt{2}$ and $c = \frac{3\sqrt{2}(\sqrt{3}-1)}{2}$. When I have the lengh of sides $a$, $b$ and $c$, I see at math.stackexchange.com/questions/229358/… and I have triagle $O(0,0,0)$, $A(3, 0,0)$ and $B\left(\dfrac{3+3\sqrt{2}}{2};\dfrac{3-3\sqrt{2}}{2},0 \right)$. – minthao_2011 Nov 11 '12 at 2:48
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