I read this paper: http://people.cs.uchicago.edu/~niyogi/papersps/NiySmaWeiHom.pdf .
I have a problem with the proof of the Lemma 4.1. I don't know how to get the following formula $A=b\sin(\theta)+\sqrt{\epsilon^{2}-b^{2}\cos^{2}(\theta)}$.
I suppose that I should use some formulas with trigonometry and Pythagorean theorem, but I don't see it.
Please, help me with my problem.
On
Ok. Consider this.
$A=(A-q)+q$ where q is the vertical distance above the x-axis.
Create a line perpendicular to A that passes through q. This will divide the triangle into two other triangles
Part 1: The bottom trialngle Consider the bottom triangle. It's a simple right triangle. From here we see that $q=b \sin(\theta)$.
Part 2. The top triangle is also a simple right triangle. Note that from the bottom triangle, we also have that the distance of the line dividing the large triangle into two smaller triangles is $b \cos(\theta)$. If we want to get distance $A-q$, we need only apply the Pythagorean Theorem again. This gives $A-q=\sqrt{\epsilon^2-{b \cos(\theta)}^2}$.
To get distance A, then just add $A-q$ to $q$.
Use the orthogonal projection of $q$ onto $T_p^{\perp}$ to split $A$ into two line segments. The bottom one has length $b\sin \theta$, while for the top one you need to apply Pythagoras' Theorem.
Credits: Fig. 1 from the linked paper.