1. Für den Sinus im Dreieck $\mathrm{D}\mathrm{C}\mathrm{A}\backslash mathrm\{DCA\}$ gilt :$\sin (\mathrm{\measuredangle }\mathrm{D}\mathrm{C}\mathrm{A})={\displaystyle \frac{\text{Gegenkathete}}{\text{Hypotenuse}}}=\boxed{{\displaystyle \frac{\mathrm{7,1}}{9}}}\backslash \; \backslash mathrm\{\backslash sin\; (\backslash measuredangle\; D\; C\; A)\}=\backslash dfrac\{\backslash text\{Gegenkathete\}\}\{\backslash text\{Hypotenuse\}\}=\backslash boxed\{\backslash dfrac\{7\{,\}1\}\{9\}\}$ $\mathrm{\measuredangle }DCA\approx {\mathrm{52,1}}^{\circ }\backslash measuredangle\; DCA\backslash approx52\{,\}1^\backslash circ$

2. ${120}^{\circ }=\mathrm{\measuredangle }\mathrm{D}\mathrm{C}\mathrm{A}+\mathrm{\measuredangle }\mathrm{A}\mathrm{C}\mathrm{B}\Rightarrow \mathrm{\measuredangle }\boxed{{\displaystyle \text{ACB}}}={120}^{\circ }-\mathrm{\measuredangle }\mathrm{D}\mathrm{C}\mathrm{A}\approx {\mathrm{67,9}}^{\circ }\backslash mathrm\{120^\backslash circ=\backslash measuredangle\{DCA+\backslash measuredangle\{ACB\}\}\backslash quad\backslash Rightarrow\; \backslash measuredangle\{\backslash boxed\{\backslash text\{ACB\}\}\}=120^\backslash circ-\backslash measuredangle\{DCA\}\}\backslash approx\; 67\{,\}9^\backslash circ$

3. Für den Kosinussatz im Dreieck $\mathrm{D}\mathrm{C}\mathrm{A}\backslash mathrm\{DCA\}$ gilt: $\mathrm{\mid }\overline{\mathrm{A}\mathrm{B}}{\mathrm{\mid }}^2=\boxed{{\displaystyle 9^2+6^2-2\cdot 9\cdot 6\cdot \mathrm{c}\mathrm{o}\mathrm{s}(\mathrm{\measuredangle }\mathrm{A}\mathrm{C}\mathrm{B})}}\Rightarrow \mathrm{\mid }\overline{\mathrm{A}\mathrm{B}}\mathrm{\mid }\approx \mathrm{8,74}\mathrm{c}\mathrm{m}\backslash mathrm\{|\backslash overline\{A\; B\}|^\{2\}=\backslash boxed\{9^2+6^2-2\backslash cdot\; 9\backslash cdot6\backslash cdot\; \backslash mathrm\{cos\{(\backslash measuredangle\{ACB\})\}\}\}\}\; \backslash quad\backslash Rightarrow\backslash mathrm\{|\backslash overline\{AB\}|\backslash approx8\{,\}74\backslash \; cm\}$

Berechnen des Umfangs des Vierecks $\text{ABCD}\backslash text\{ABCD\}$

${\mathrm{U}}_{\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}}=\overline{\mathrm{A}\mathrm{B}}+\overline{\mathrm{B}\mathrm{C}}+\overline{\mathrm{C}\mathrm{D}}+\overline{\mathrm{D}\mathrm{A}}\backslash mathrm\{U\_\{ABCD\}=\backslash overline\{AB\}+\backslash overline\{BC\}+\backslash overline\{CD\}+\backslash overline\{DA\}\}$ setze die Zahlenwerte in die Formel ein und $\backslash hspace\{44mm\}$berechne den Umfang.

${\mathrm{U}}_{\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}}=\mathrm{8,74}+6+\mathrm{5,53}+\mathrm{7,1}[\mathrm{c}\mathrm{m}]\approx \mathrm{27,37}\mathrm{c}\mathrm{m}\backslash mathrm\{U\_\{ABCD\}=8\{,\}74+6+5\{,\}53+7\{,\}1\backslash \; [cm]\backslash approx27\{,\}37\backslash \; cm\}$

Der Umfang des Vierecks $\text{ABCD}\backslash text\{ABCD\}$ beträgt ca. $\boxed{{\displaystyle \mathrm{27,37}\mathrm{c}\mathrm{m}}}\mathrm{.}\backslash boxed\{\backslash mathrm\{27\{,\}37\backslash \; cm\}\}.$

Schreibe den Wert in die Lücke.

Andere Möglichkeit die Länge der Strecke $\overline{\mathrm{A}\mathrm{B}}\backslash mathrm\{\backslash overline\{AB\}\}$ rechnerisch zu bestimmen:

• Berechne den Winkel $\text{DCA}\backslash text\{DCA\}$:

$\sin (\mathrm{\measuredangle }DCA)=\left({\displaystyle \frac{\mathrm{7,1}}{9}}\right)\backslash sin\; (\backslash measuredangle\; D\; C\; A)=\backslash left(\backslash dfrac\{7\{,\}1\}\{9\}\backslash right)$

• Berechne den Winkel $\text{ACB}:\backslash text\{ACB\}:$

$\mathrm{\measuredangle }\mathrm{A}\mathrm{C}\mathrm{B}=120\mathrm{°}-\mathrm{\measuredangle }\mathrm{D}\mathrm{C}\mathrm{A}\backslash quad\backslash mathrm\{\backslash measuredangle\{ACB\}=120°-\backslash measuredangle\{DCA\}\}$

• Berechne die Strecke $\overline{\mathrm{A}\mathrm{B}}\backslash mathrm\{\backslash overline\{AB\}\}$ mit dem Sinussatz:

${\displaystyle \frac{\overline{\mathrm{A}\mathrm{B}}}{\mathrm{s}\mathrm{i}\mathrm{n}(\mathrm{\measuredangle }\mathrm{A}\mathrm{C}\mathrm{B})}}={\displaystyle \frac{6}{\mathrm{s}\mathrm{i}\mathrm{n}(\mathrm{40,41}\mathrm{°})}}\backslash quad\backslash mathrm\{\backslash dfrac\{\backslash overline\{AB\}\}\{sin(\backslash measuredangle\{ACB\})\}=\backslash dfrac\{6\}\{sin(40\{,\}41°)\}\}$