I’ve been looking at triangulation calculations, and I’ve become a bit stumped as to how the authors of the attached document have come to the following calculations.
In the page it explains how one can get the slope length for value C-B. I don’t know why they are using the 02 value as this is over 90 degrees and would not be used in trigonometry as it wouldn’t make up one of the internal angles.
Even if they subtracted 90 degrees from this angle to get the angle from north left the opposite and adjacent lengths are reversed. I hope this makes slight sense!
You can see that $$\tan(\theta_2-90^\circ)=\frac{y_2-y_3}{x_3-x_2}=-\frac{y_3-y_2}{x_3-x_2}$$ And using that $\tan(\theta_2-90^\circ)=-\tan \theta_2$ we get that $\tan \theta_2=\frac{y_3-y_2}{x_3-x_2}$