Der Kosinussatz angewandt auf die Vorlage lautet:

${\left(\overline{BC}\right)}^2={\left(\overline{AB}\right)}^2+{\left(\overline{AC}\right)}^2-2\cdot {\left(\overline{AB}\right)}^2\cdot {\left(\overline{AC}\right)}^2\cdot \cos \mathrm{\sphericalangle }\text{BAC}\backslash \; \backslash left(\backslash overline\{BC\}\backslash right)^2=\backslash left(\backslash overline\{AB\}\backslash right)^2+\backslash left(\backslash overline\{AC\}\backslash right)^2-2\backslash cdot\backslash left(\backslash overline\{AB\}\backslash right)^2\backslash cdot\backslash left(\backslash overline\{AC\}\backslash right)^2\backslash cdot\backslash cos\backslash sphericalangle\; \backslash text\{BAC\}$

$\cos \mathrm{\sphericalangle }\text{BAC}={\displaystyle \frac{{\left(\overline{AB}\right)}^2+{\left(\overline{AC}\right)}^2-{\left(\overline{BC}\right)}^2}{2\cdot \overline{AB}\cdot \overline{AC}}}\backslash \; \backslash cos\backslash sphericalangle\; \backslash text\{BAC\}=\backslash dfrac\{\backslash left(\backslash overline\{AB\}\backslash right)^2+\backslash left(\backslash overline\{AC\}\backslash right)^2-\backslash left(\backslash overline\{BC\}\backslash right)^2\}\{2\backslash cdot\backslash overline\{AB\}\backslash cdot\backslash overline\{AC\}\}$

$\cos \mathrm{\sphericalangle }\text{BAC}={\displaystyle \frac{(16+144-100){\text{cm}}^2}{(2\cdot 4\cdot 12){\text{cm}}^2}}\backslash \; \backslash cos\backslash sphericalangle\; \backslash text\{BAC\}=\backslash dfrac\{(16+144-100)\backslash ;\backslash text\{cm\}^2\; \backslash \; \}\{(2\backslash cdot4\backslash cdot12)\backslash ;\backslash text\{cm\}^2\}$

$\cos \mathrm{\sphericalangle }\text{BAC}=\mathrm{0,625}\Rightarrow \mathrm{\sphericalangle }\text{BAC}={\mathrm{51,32}}^{\circ }\backslash \; \backslash cos\backslash sphericalangle\; \backslash text\{BAC\}=0\{,\}625\backslash \; \backslash Rightarrow\backslash \; \backslash sphericalangle\; \backslash text\{BAC\}=51\{,\}32^\{\backslash circ\}$

Der Winkel$\mathrm{\sphericalangle }\text{BAC }\backslash \; \backslash sphericalangle\{\backslash text\{BAC\}\}\backslash$beträgt:$\boxed{{\displaystyle {\mathrm{51,32}}^{\circ }}}\backslash \; \backslash boxed\{51\{,\}32^\{\backslash circ\}\}$

Der Punkt$\text{ D }\backslash \; \backslash text\{D\}\backslash$liegt auf dem Thaleskreis über$\overline{\text{AC}}\backslash \; \backslash overline\{\backslash text\{AC\}\}$, deshalb ist$\mathrm{\sphericalangle }\text{ADC}={90}^{\circ }\mathrm{.}\backslash \; \backslash sphericalangle\{\backslash text\{ADC\}\}=90^\backslash circ.$

Da im Dreieck $\text{ AMD }\backslash \; \backslash text\{AMD\}\backslash$ gilt:$\overline{\text{AM}}=\overline{\text{DM}}=\overline{\text{AD}}=6\text{ cm},\backslash \; \backslash overline\{\backslash text\{AM\}\}=\backslash overline\{\backslash text\{DM\}\}=\backslash overline\{\backslash text\{AD\}\}=6\backslash \; \backslash text\{cm\},\backslash$ist das Dreieck gleichseitig.

Somit ist $\mathrm{\sphericalangle }\text{DMA}={60}^{\circ }\mathrm{.}\backslash sphericalangle\{\backslash text\{DMA\}\}=60^\backslash circ.$

$\boxed{{\displaystyle A_{\text{Figur}}=A_{\text{Sektor}}+A_{\mathrm{△}\text{ABC}}+A_{\mathrm{△}\text{CDM}}}}\backslash \; \backslash boxed\{A\_\{\backslash text\{Figur\}\}=A\_\{\backslash text\{Sektor\}\}+A\_\{\backslash triangle\backslash text\{ABC\}\}+A\_\{\backslash triangle\backslash text\{CDM\}\}\}$

$A_{\text{Sektor}}={\displaystyle \frac{\mathrm{\sphericalangle }\text{AMD}}{{360}^{\circ }}}\cdot {\overline{\text{AM}}}^2\cdot \pi A\_\{\backslash text\{Sektor\}\}=\backslash dfrac\{\backslash sphericalangle\{\backslash text\{AMD\}\}\}\{360^\backslash circ\}\backslash cdot\backslash overline\{\backslash text\{AM\}\}^2\backslash cdot\backslash pi$

$A_{\text{Sektor}}={\displaystyle \frac{{60}^{\circ }}{{360}^{\circ }}}\cdot 36\cdot \pi \cdot {\text{cm}}^2\Rightarrow A_{\text{Sektor}}=\underset{‾}{\mathrm{18,85}{\text{cm}}^2}A\_\{\backslash text\{Sektor\}\}=\backslash dfrac\{60^\backslash circ\}\{360^\backslash circ\}\backslash cdot\; 36\backslash cdot\backslash pi\backslash cdot\backslash \; \backslash text\{cm\}^2\backslash Rightarrow\; A\_\{\backslash text\{Sektor\}\}=\backslash underline\{18\{,\}85\backslash \; \backslash text\{cm\}^2\}$

$\backslash rule\{10cm\}\{o.5mm\}$

$A_{\mathrm{△}\text{ABC}}={\displaystyle \frac{1}{2}}\cdot \overline{\text{AB}}\cdot \overline{\text{AC}}\cdot \sin \mathrm{\sphericalangle }\text{BAC}\backslash \; A\_\{\backslash triangle\backslash text\{ABC\}\}=\backslash dfrac\{1\}\{2\}\backslash cdot\backslash overline\{\backslash text\{AB\}\}\backslash cdot\backslash overline\{\backslash text\{AC\}\}\backslash cdot\backslash sin\backslash sphericalangle\{\backslash text\{BAC\}\}$

$A_{\mathrm{△}\text{ABC}}={\displaystyle \frac{1}{2}}\cdot 4\cdot 12\cdot \sin {\mathrm{51,32}}^{\circ }{\text{cm}}^2\Rightarrow A_{\mathrm{△}\text{ABC}}=\underset{‾}{\mathrm{18,74}{\text{cm}}^2}\backslash \; A\_\{\backslash triangle\backslash text\{ABC\}\}=\backslash dfrac\{1\}\{2\}\backslash cdot4\backslash cdot12\backslash cdot\backslash sin51\{,\}32^\backslash circ\backslash \; \backslash text\{cm\}^2\backslash \; \backslash Rightarrow\backslash \; A\_\{\backslash triangle\backslash text\{ABC\}\}=\backslash underline\{18\{,\}74\backslash \; \backslash text\{cm\}^2\}$

$\backslash rule\{10cm\}\{o.5mm\}$

$A_{\mathrm{△}\text{CDM}}={\displaystyle \frac{1}{2}}\cdot \overline{\text{DM}}\cdot \overline{\text{MC}}\cdot \sin \mathrm{\sphericalangle }\text{DMC}\backslash \; A\_\{\backslash triangle\backslash text\{CDM\}\}=\backslash dfrac\{1\}\{2\}\backslash cdot\backslash overline\{\backslash text\{DM\}\}\backslash cdot\backslash overline\{\backslash text\{MC\}\}\backslash cdot\backslash sin\backslash sphericalangle\{\backslash text\{DMC\}\}$

$A_{\mathrm{△}\text{CDM}}={\displaystyle \frac{1}{2}}\cdot 6\cdot 6\cdot \sin ({180}^{\circ }-{60}^{\circ }){\text{cm}}^2\Rightarrow A_{\mathrm{△}\text{CDM}}=\underset{‾}{\mathrm{15,59}{\text{cm}}^2}\backslash \; A\_\{\backslash triangle\backslash text\{CDM\}\}=\backslash dfrac\{1\}\{2\}\backslash cdot6\backslash cdot6\backslash cdot\backslash sin(180^\backslash circ-60^\backslash circ)\backslash \; \backslash text\{cm\}^2\backslash \; \backslash Rightarrow\backslash \; A\_\{\backslash triangle\backslash text\{CDM\}\}=\backslash underline\{15\{,\}59\backslash \; \backslash text\{cm\}^2\}$

$\backslash rule\{10cm\}\{o.5mm\}$

$A_{\text{Figur}}=(\mathrm{18,85}+\mathrm{18,74}+\mathrm{15,59}){\text{cm}}^2\backslash \; A\_\{\backslash text\{Figur\}\}=(18\{,\}85+18\{,\}74+15\{,\}59)\backslash \; \backslash text\{cm\}^2$

$A_{\text{Figur}}=\mathrm{53,18}{\text{cm}}^2\backslash \; A\_\{\backslash text\{Figur\}\}=53\{,\}18\backslash text\{cm\}^2$

Der Flächeninhalt der Figur beträgt:$\boxed{{\displaystyle \mathrm{53,18}{\text{cm}}^2}}\backslash \; \backslash boxed\{53\{,\}18\backslash text\{cm\}^2\}\backslash$