Geometrie-Trigonometrie-Kosinussatz

• $a^2 = b^{2} +c^{2} -2\cdot b\cdot c\cdot cos\alpha$
$a = \sqrt{b^{2} +c^{2} -2\cdot b\cdot c\cdot cos\alpha }$
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$cos\alpha = \frac{b^{2} +c^{2} -a^{2} }{2\cdot b\cdot c}$
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Beispiel Nr: 05
$\text{Gegeben:}\\\text{Länge der Seite} \qquad c \qquad [m] \\ \text{Länge der Seite} \qquad b \qquad [m] \\ \text{Länge der Seite} \qquad a \qquad [m] \\ \\ \text{Gesucht:} \\\text{Winkel} \qquad \alpha \qquad [^{\circ}] \\ \\ cos\alpha = \frac{b^{2} +c^{2} -a^{2} }{2\cdot b\cdot c}\\ \textbf{Gegeben:} \\ c=1\frac{9}{10}m \qquad b=3\frac{3}{5}m \qquad a=5\frac{1}{5}m \qquad \\ \\ \textbf{Rechnung:} \\ cos\alpha = \frac{b^{2} +c^{2} -a^{2} }{2\cdot b\cdot c} \\ c=1\frac{9}{10}m\\ b=3\frac{3}{5}m\\ a=5\frac{1}{5}m\\ cos\alpha = \frac{(3\frac{3}{5}m)^{2} +(1\frac{9}{10}m)^{2} -(5\frac{1}{5}m)^{2} }{2\cdot 3\frac{3}{5}m\cdot 1\frac{9}{10}m}\\\\\alpha=140^{\circ} \\\\\\ \small \begin{array}{|l|} \hline c=\\ \hline 1\frac{9}{10} m \\ \hline 19 dm \\ \hline 190 cm \\ \hline 1,9\cdot 10^{3} mm \\ \hline 1,9\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline b=\\ \hline 3\frac{3}{5} m \\ \hline 36 dm \\ \hline 360 cm \\ \hline 3,6\cdot 10^{3} mm \\ \hline 3,6\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline a=\\ \hline 5\frac{1}{5} m \\ \hline 52 dm \\ \hline 520 cm \\ \hline 5,2\cdot 10^{3} mm \\ \hline 5,2\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline alpha=\\ \hline 140 ° \\ \hline 8,4\cdot 10^{3} \text{'} \\ \hline 5,04\cdot 10^{5} \text{''} \\ \hline 155 gon \\ \hline 2,44 rad \\ \hline \end{array}$