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I have 2 points, A + B, with vectors from the origin a and b. The vector from A to B is

c = b - a

a and b are defined in polar coordinates with a $ = (r_a,\theta_a) $, b $ = (r_b,\theta_b ) $

I want to know how to define the angle c makes with the horizontal in terms of $\theta_a $ and/or $\theta_b$.

vector angles

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If it is an isosceles triangle, $\angle A=\angle B=90-\frac{\theta_a}{2}$

Extending $c$ down to the horizontal makes a triangle with angles $\theta_b,180-(90-\frac{\theta_a}{2}),\theta_c$. With $\theta_c$ being the angle you are after. It can now be expressed in terms of $\theta_a$ and $\theta_b$

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  • $\begingroup$ Thanks for your answer, the case I'm thinking of would not have isosceles triangles although I've just realised I had a typo in the original, have corrected now. Sorry for the confusion! In general $r_a \neq r_b$ $\endgroup$ – Ciara Jan 14 '15 at 16:18
  • $\begingroup$ Ah. Then I believe the angle you are after will need to be in terms of 4 variables, the two angles and the two radii. $\endgroup$ – turkeyhundt Jan 14 '15 at 16:22
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You can find $A$ from the law of sines $\frac {\sin A}b=\frac{\sin \theta_a}c$ (but worry about whether the triangle is obtuse) or the law of cosines $b^2=a^2+c^2-2ac\cos A$ Then the angle you want is $\frac \pi2 - (\theta_a+\theta_b) - A$

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