diff --git a/mr/Ub2/mr2.pdf b/mr/Ub2/mr2.pdf
index d314208..8ea81ac 100644
--- a/mr/Ub2/mr2.pdf
+++ b/mr/Ub2/mr2.pdf
Binary files differ
diff --git a/mr/Ub2/mr2.tex b/mr/Ub2/mr2.tex
index a4e86dc..18e764f 100644
--- a/mr/Ub2/mr2.tex
+++ b/mr/Ub2/mr2.tex
@@ -130,6 +130,24 @@
             $\Rightarrow r(\mathbf{s})\approx -\frac{3}{2}-2\left(s_l-2l\right)+\left(s_r-l\right)$
         \end{enumerate}
      \Aufgabe{}{6}
+        \begin{enumerate}[(a)]
+            \item $\mathbf{x}_1=\begin{pmatrix}
+                    p_x\\p_y\\\Delta\theta
+                \end{pmatrix}=\begin{pmatrix}
+                    r\sin(\Delta\theta)\\
+                    -r\cos(\Delta\theta)+r\\
+                    \Delta\theta
+                \end{pmatrix}$, $\Delta\theta=\sin^{-1}\left(\frac{p_x}{r}\right)$\\
+                $p_y=-r\cos\sin^{-1}\left(\frac{p_x}{r}\right)+r=-r\sqrt{1-\frac{p_x^2}{r^2}}+r=-\sqrt{r^2-p_x^2}+r$\\
+                $(r-p_y)^2=r^2-p_x^2 \Leftrightarrow r^2-2p_yr+p_y^2=r^2-p_x^2 \Leftrightarrow r=\frac{p_x^2+p_y^2}{2p_y}$
+            \item $\Delta\theta=\sin^{-1}\left(\frac{p_x}{r}\right)=\sin^{-1}\left(p_x\cdot \frac{2p_y}{p_x^2+p_y^2}\right)=\sin^{-1}\left(\frac{2p_xp_y}{p_x^2+p_y^2}\right)$
+            \item $s_l=\left(r-\frac{l}{2}\right)\Delta\theta=\left(\frac{p_x^2+p_y^2}{2p_y}-\frac{l}{2}\right)\sin^{-1}\left(\frac{2p_xp_y}{p_x^2+p_y^2}\right)=\frac{p_x^2+p_y^2-lp_y}{2p_y}\sin^{-1}\left(\frac{2p_xp_y}{p_x^2+p_y^2}\right)$\\
+            $s_r=\left(r+\frac{l}{2}\right)\Delta\theta=\frac{p_x^2+p_y^2+lp_y}{2p_y}\sin^{-1}\left(\frac{2p_xp_y}{p_x^2+p_y^2}\right)$
+            \item $r=\frac{p_x^2+p_y^2}{2p_y}=\frac{16+4}{4}m=5m$\\
+                $\Delta\theta=\sin^{-1}\left(\frac{p_x}{r}\right)=\sin^{-1}\left(\frac{4}{5}\right)\approx 0.927$\\
+                $s_l=\left(r-\frac{l}{2}\right)\Delta\theta=\left(5m-\frac{1}{2}m\right)\sin^{-1}\left(\frac{4}{5}\right)=4.5m\cdot \sin^{-1}\left(\frac{4}{5}\right)\approx 4.173$\\
+                $s_l=\left(r+\frac{l}{2}\right)\Delta\theta=\left(5m+\frac{1}{2}m\right)\sin^{-1}\left(\frac{4}{5}\right)=5.5m\cdot \sin^{-1}\left(\frac{4}{5}\right)\approx 5.100$\\
+        \end{enumerate}
 
 \end{document}