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I need to calculate the angle between two sides, I have the length of A & B sides, but don't know how to find the angle…
Both sides are the same length.
I can get the start and end vectors of each side, I can also get the center values of each side.
Here's an image better describing my question:
Consider the following triangle $\triangle ABC$, the $ \color{maroon} {\text{poly 1}}$ below, with sides $\color{maroon}{\overline{AB}=c}$ and $\color{maroon}{\overline{AC}=b}$ known. Further the angle between them, $\color{green}\alpha$ is known.
$\hskip{2 in}$
Then, the law of cosines tell you that $$\color{maroon}{a^2=b^2+c^2-2bc\;\cos }\color{blue}{\alpha}$$
Draw in the radii from the center to the endpoints of $a$ to form two congruent right triangles, sharing a leg of length $r=1$ and each having another leg of length $\sqrt{3}$. From right triangle trigonometry, the angles in these right triangles at the center of the circle has measure $\arctan(\sqrt{3})=\frac{\pi}{3},$ so the measure of the entire central angle (angle at the center of the circle) that subtends the same arc as the chord $a$ is $2\arctan(\sqrt{3})=\frac{2\pi}{3},$ and the measure of the inscribed angle $x$ is half of that, $$\arctan(\sqrt{3})=\frac{\pi}{3}.$$
Best Answer
Assuming you have points: $$A=(A_x, A_y),\ B=(B_x,B_y)$$ And two equal sides with length $l$ originating from a shared point $O=(0,0)$, then the angle between $AO$ and $BO$ will be: $$\cos\theta = \frac{AO\cdot BO}{l^2}$$ $$\theta = \cos^{-1} \left(\frac{A_xB_x+A_yB_y}{l^2}\right)$$ This is just a particular case of the dot product.