I try to program a slope failure simulation in Matlab and now the following problem occurred:

Let's say I've got to points $A(X_1,Y_1)$ and $B(X_2,Y_2)$, which build one side $c$ of a triangle. Additionally I've got the opposite angle $\gamma$ of side $c$ and the ratio of side $a$ to $b$.

The triangle should find the center of a logarithmic spiral, that's why I know the ratio:

$$ b=a*e^{\gamma *k} $$

where $k$ is a known constant.

Now I need the third point of this triangle (or center of spiral) C. Do you have any idea how i could do this?

I thought of sine law and that the sum of the three angles should be $\pi$, but I couldn't find a solution yet.


This seems to0 simple to me, but here we go. Let's say the triangle consists of sides $a,b,c$ with opposite angles $A,B,C$, respectively. We know $c, C$, and $a/b$. We want to determine $a, b, A$ and $B$.

The law of cosines tells us that

$$ c^2=a^2+b^2-2ab\cos C\\ \frac{c^2}{ab}=\frac{a^2}{ab}+\frac{b^2}{ab}-2\cos C=\frac{a}{b}+\frac{b}{a}-2\cos C\\ $$

The only unknown is the product $ab$, hence

$$ab=\frac{c^2}{\frac{a}{b}+\frac{b}{a}-2\cos C}$$

So $ab$ now known, then

$$b=\sqrt{ab*\frac{a}{b}}\quad \&\quad a=\sqrt{ab*\frac{b}{a}}$$

So it looks like $a$ and $b$ are fully determined and we turn our attention the the law of sines to get the angles, to wit,

$$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$$

So we complete the problem with

$$ A=\sin^{-1}\left( \frac{a}{c}\sin C \right)\\ B=\sin^{-1}\left( \frac{b}{c}\sin C \right) $$


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