Im trying to find a way to calculate the position of the one of the base points of an isocicles triangle if I know the positions of the other two points, the angle measures, and the side lengths. It must be possible since you know everything else about the triangle except that one base point. However I've yet to find success.
Is there a way to do this?
On
To do this lets break the isosceles down the middle into two right triangles. And we know that the tip of the isosceles triangle has the point $(x_1,y_1)$ and one of the base points is $(x_2,y_2)$ and you want to find the other point $(x_3,y_3)$ in terms of the other points. Well right off the bat if you know the side lengths then it can be said that
$$x_3=\left\{\begin{aligned}
&x_2+L &&: x_2 < x_3\\
&x_2-L &&: x_2 > x_3
\end{aligned}
\right.$$
$$y_3=y_2$$
Where the condition $x_2<x_3$ basically means if the point you're trying to find is to the right of the base point you know and $x_2>x_3$ essentially means the point is to the left and $L$ is the length of the base.
Now lets say you know the angle $\theta$ on the top and the two side lengths, $S$, but not the base length, then we must solve for $L$. In that case
$$L=2S\sin\left(\frac{\theta}{2}\right)$$
Hint.
It is sufficient to know the position of a base point $A=(x_A,y_A)$ , the position of the vertex $C=(x_C,y_C)$ and the leght of the basis $\overline{AB}=a$.
We have that the side of the triangle is $\overline{AC}=l= \sqrt{(x_A-x_C)^2+(y_A-y_C)^2}$ and the $B$ have to be the intersection of the two circle with center $C$ and radius $l$ and of center $A$ and radius $a$, given by the solution of the system: $$ \begin{cases} (x-x_A)^2+(y-y_A)^2=a^2\\ (x-x_C)^2+(y-y_C)^2=l^2 \end{cases} $$ Note that, in general we have two solutions.