1. Schritt

Strecke $\overline{CF}\backslash overline\{CF\}$ berechnen:

Mithilfe der Sinus Funktion wird die Strecke $\overline{CF}\backslash overline\{CF\}$ berechnet.

$\sin \epsilon ={\displaystyle \frac{\overline{CF}}{\overline{CE}}}\backslash sin\{\backslash varepsilon\}=\backslash dfrac\{\backslash overline\{CF\}\}\{\backslash overline\{CE\}\}$

$\sin {49}^{\circ }={\displaystyle \frac{\overline{CF}}{\mathrm{7,2}}}\mathrm{\mid }\cdot \mathrm{7,2}\backslash sin\{49^\backslash circ\}=\backslash dfrac\{\backslash overline\{CF\}\}\{7\{,\}2\}\backslash quad\; |\backslash text\{\backslash footnotesize\backslash ,\$\backslash cdot\; 7\{,\}2\$\}$

$\sin {49}^{\circ }\cdot \mathrm{7,2}=\overline{CF}\backslash sin\{49^\backslash circ\}\backslash cdot\; 7\{,\}2=\backslash overline\{CF\}$

$\overline{CF}=\mathrm{5,43}\backslash overline\{CF\}=5\{,\}43$

Die Länge der Strecke $\overline{CF}\backslash overline\{CF\}$ beträgt rund $\mathrm{5,43}\mathrm{c}\mathrm{m}5\{,\}43\backslash ;\backslash mathrm\{cm\}$.

2. Schritt

${\phi }_2\backslash varphi\_2$ berechnen:

Die Summe der Innenwinkel im Dreieck beträgt ${180}^{\circ }180^\backslash circ$

${\phi }_2={180}^{\circ }-{90}^{\circ }-{49}^{\circ }={41}^{\circ }\backslash varphi\_2=180^\backslash circ\; -\; 90^\backslash circ\; -\; 49^\backslash circ=41^\backslash circ$

3. Schritt

${\phi }_3\backslash varphi\_3$ berechnen:

Mithilfe der Kosinus Funktion wird der Winkel ${\phi }_3\backslash varphi\_3$ berechnet.

$\cos {\phi }_3={\displaystyle \frac{\overline{BC}}{\overline{CF}}}\backslash cos\{\backslash varphi\_3\}=\backslash dfrac\{\backslash overline\{BC\}\}\{\backslash overline\{CF\}\}$

$\cos {\phi }_3={\displaystyle \frac{\mathrm{4,6}}{\mathrm{5,43}}}\mathrm{\mid }\cos {}^{-1}\backslash cos\{\backslash varphi\_3\}=\backslash dfrac\{4\{,\}6\}\{5\{,\}43\}\backslash quad\; |\backslash text\{\backslash footnotesize\backslash ,\$\backslash cos\{\}^\{-1\}\$\}$

${\phi }_3={\mathrm{32,1}}^{\circ }\backslash varphi\_3=32\{,\}1^\backslash circ$

4. Schritt

${\phi }_1\backslash varphi\_1$ berechnen

Der Winkel $\mathrm{\angle }DCB\backslash angle\; DCB$ ist ein rechter Winkel und hat demnach ${90}^{\circ }90^\backslash circ$.

${\phi }_1={90}^{\circ }-{41}^{\circ }-{\mathrm{32,1}}^{\circ }={\mathrm{16,9}}^{\circ }\backslash varphi\_1=90^\backslash circ\; -\; 41^\backslash circ\; -32\{,\}1^\backslash circ=16\{,\}9^\backslash circ$

5. Schritt

Strecke $\overline{CD}\backslash overline\{CD\}$ berechnen:

Mithilfe der Kosinus Funktion wird die Länge der Strecke $\overline{CD}\backslash overline\{CD\}$ berechnet.

Strecke $\overline{CD}\backslash overline\{CD\}$ berechnen:

$\cos {\phi }_1={\displaystyle \frac{\overline{CD}}{\overline{CE}}}\backslash cos\{\backslash varphi\_1\}=\backslash dfrac\{\backslash overline\{CD\}\}\{\backslash overline\{CE\}\}$

$\cos {\mathrm{16,9}}^{\circ }={\displaystyle \frac{\overline{CD}}{\mathrm{7,2}}}\mathrm{\mid }\cdot \mathrm{7,2}\backslash cos\{16\{,\}9^\backslash circ\}=\backslash dfrac\{\backslash overline\{CD\}\}\{7\{,\}2\}\backslash quad\; |\; \backslash text\{\backslash footnotesize\backslash ,\$\backslash cdot\; 7\{,\}2\$\}$

$\cos {\mathrm{16,9}}^{\circ }\cdot \mathrm{7,2}=\overline{CD}\backslash cos\{16\{,\}9^\backslash circ\}\backslash cdot\; 7\{,\}2=\backslash overline\{CD\}$

$\overline{CD}=\mathrm{6,89}\backslash overline\{CD\}=6\{,\}89$

Die Länge der Strecke $\overline{CD}\backslash overline\{CD\}$ beträgt rund $\mathrm{6,89}\mathrm{c}\mathrm{m}6\{,\}89\backslash ;\backslash mathrm\{cm\}$.

6. Schritt

Umfang des Vierecks $ABCDABCD$ berechnen:

$U=2\cdot \overline{BC}+2\cdot \overline{CD}U=2\backslash cdot\backslash overline\{BC\}+2\backslash cdot\backslash overline\{CD\}$

$U=2\cdot \mathrm{4,6}+2\cdot \mathrm{6,89}=\mathrm{22,98}\approx 23U=2\backslash cdot\; 4\{,\}6+2\backslash cdot\; 6\{,\}89=22\{,\}98\backslash approx23$

Der Umfang des Vierecks $ABCDABCD$ beträgt rund $23\mathrm{c}\mathrm{m}23\backslash ;\backslash mathrm\{cm\}$.