Q: Splitting vector into x and y components.

Question

I have struck a problem within the aspects of simple middle school geometry. I'll just cut to the case.

Visual exmaple of problem

We have a triangle of any size or form. We know angles A, B and C. We have a vector, v along one of the triangles sides. We know the magnitude of vector v. We have a smaller triangle representing the x and y components. We do not know the angles of the smaller triangle, other than it is a right angled triangle. Keep in mind that the original triangle can be of any size or form, so in the case of the picture, extending the vector triangle to get one of the angles is not an option. Is it possible to acquire the x and y components? If so, is there any general formula to achieve this?

Sorry for the horrible picture. Any help is highly appreciated.

Edit: I forgot to inform that we also know the x and y coordinates of the original triangles corners.

Answer 1

OK, your edit makes all the difference in the world. Let $B(x_B,y_B)$ be the start of the vector and $A(x_A,y_A)$ is a vertex on the same side of the triangle where the vector is. Then you can easily find tangent of the vector $m=\frac{Y_B-Y_A}{X_B-X_A}$ (beware of division by zero). Then $|x|=\frac{1}{\sqrt{1+m^2}}v$, $|y|=\frac{|m|}{\sqrt{1+m^2}}v$ where $v$ is the magnitude of vector $\vec{v}$. You can determine whether $x$ and $y$ are positive or negative by finding components of vector $\vec{BA}(x_A-x_B,y_A-y_B)$. They will have the same signs.

Answer 2

Not from the information provided. In particular, any rotation of the triangle shown would give different $x$ and $y$ values, but the same $A$, $B$, and $C$, so there is no possible function giving $x$ and $y$ in terms of $A$, $B$, and $C$ alone.

Answer 3

Since you know the coordinates of the vertices, you can ignore the angles. The coordinates of $\vec v$ can be found as a proportion of the coordinates of the appropriate edge. For the illustrated case, $$\vec v = {\|\vec v\|\over\|A-C\|}(A-C).$$