MHB Determining the Degrees of an Angle Given Three X and Y Coordinates

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To determine the degrees of an angle formed by three given X and Y coordinates, define two vectors from the points. The first vector, a, is from point 1 to point 2, and the second vector, b, is from point 2 to point 3. The angle θ can be calculated using the formula θ = (180/π) * arccos((a·b) / (|a||b|)), where a·b is the dot product of the vectors and |a| and |b| are their magnitudes. This method provides a straightforward way to compute the angle using basic vector operations. The discussion emphasizes the need for a clear formula to implement in code.
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How do I determine the degrees of an angle if I three X and Y coordinates? I honestly just need a formula to plug into some code. Thank you in advance.
 
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xyle said:
How do I determine the degrees of an angle if I three X and Y coordinates? I honestly just need a formula to plug into some code. Thank you in advance.

Suppose the 3 points are given by:

$$\left(x_i,y_i\right)$$ where $$i\in\{1,2,3\}$$

Now further suppose we wish to make one line segment from point 1 to point 2, and another from point 2 to point 3, and then find the angle, in degrees, subtended by the two segments. I would begin by defining the vectors:

$$a=\left\langle x_2-x_1,y_2-y_1 \right\rangle$$

$$b=\left\langle x_3-x_2,y_3-y_2 \right\rangle$$

And then, from the dot product of the two vectors, we may write:

$$\theta=\frac{180^{\circ}}{\pi}\arccos\left(\frac{a\cdot b}{|a||b|}\right)$$

where:

$$a\cdot b=\left(x_2-x_1\right)\left(x_3-x_2\right)+\left(y_2-y_1\right)\left(y_3-y_2\right)$$

$$|a|=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$$

$$|b|=\sqrt{\left(x_3-x_2\right)^2+\left(y_3-y_2\right)^2}$$

Does that make sense?
 
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