diff --git a/mr/ubB/mrB.pdf b/mr/ubB/mrB.pdf
index 58c697f..8126324 100644
--- a/mr/ubB/mrB.pdf
+++ b/mr/ubB/mrB.pdf
Binary files differ
diff --git a/mr/ubB/mrB.tex b/mr/ubB/mrB.tex
index 5d77ae9..bed3353 100644
--- a/mr/ubB/mrB.tex
+++ b/mr/ubB/mrB.tex
@@ -107,11 +107,11 @@
     \Aufgabe{Reeds-Shepp-Curves}{8}
         \begin{enumerate}[(a)]
             \item As the tangent is always perpendicular to the radius there are right triangles. So for $r_1=r_2=r$ we can calculate:\\
-            $d=2\cdot \sqrt{dist(\mathbf{s'},\mathbf{g'})^2-r^2}$\\
+            $d=2\cdot \sqrt{\left(\frac{1}{2}dist(\mathbf{s'},\mathbf{g'})\right)^2-r^2}$\\
             For $r_1\not=r_2$ we can use:\\
             $d=\sqrt{dist(\mathbf{s'},\mathbf{g'})^2-(r_1+r_2)^2}$\\
             There geometrical explanation is to translate $r_2$ and append it to $r_1$. Then there is only one right triangle and we again can calculate the pythagorean.
-            \item $\delta(\alpha_r,\mathbf{s'},\mathbf{g'},\alpha_l)=r\alpha_r+d+r\alpha_l=r\alpha_r+2\cdot\sqrt{dist(\mathbf{s'},\mathbf{g'})^2-r^2}+r\alpha_l$
+            \item $\delta(\alpha_r,\mathbf{s'},\mathbf{g'},\alpha_l)=r\alpha_r+d+r\alpha_l=r\alpha_r+2\cdot\sqrt{\left(\frac{1}{2}dist(\mathbf{s'},\mathbf{g'})\right)^2-r^2}+r\alpha_l$
             \item $\delta(\alpha_r,\mathbf{s'},\mathbf{g'},\alpha_l)=r_1\alpha_r+d+r_2\alpha_l=r_1\alpha_r+\sqrt{dist(\mathbf{s'},\mathbf{g'})^2-(r_1+r_2)^2}+r_2\alpha_l$
         \end{enumerate}
 \end{document}