# Angle between the sum of two vectors and the horizontal

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 $\theta_b$.

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 $\theta_b$
• 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 $r_a \neq r_b$ – Ciara Jan 14 '15 at 16:18
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 $\frac \pi2 - (\theta_a+\theta_b) - A$