# Coordinates of vertices of a triangle with known interior angles & length of one sides [closed]

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
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 DeVitoNov 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.

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