Berechnen des Flächeninhalts des Dreiecks $\text{ACD}\backslash text\{ACD\}$.

Zur Berechnung des Flächeninhalts des Dreiecks $\text{ABD}\backslash text\{ABD\}$ verwenden wir die Formel:

$\boxed{{\displaystyle {\mathrm{A}}_{\mathrm{A}\mathrm{C}\mathrm{D}}=\frac{1}{2}\cdot \overline{\mathrm{A}\mathrm{D}}\cdot \overline{\mathrm{A}\mathrm{C}}\cdot \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{\sphericalangle }\mathrm{D}\mathrm{A}\mathrm{C}}}\backslash boxed\{\backslash mathrm\{A\_\{ACD\}=\backslash dfrac\{1\}\{2\}\backslash cdot\; \backslash overline\{AD\}\backslash cdot\; \backslash overline\{AC\}\backslash cdot\; sin\backslash sphericalangle\{DAC\}\}\}$

Wir berechnen zuerst $\overline{\mathrm{A}\mathrm{D}}\backslash mathrm\{\backslash overline\{AD\}\}$ mit dem Sinussatz.

Der Sinussatz, angewendet auf unser Dreieck, lautet:

${\displaystyle \frac{\overline{\mathrm{A}\mathrm{D}}}{\sin {118}^{\circ }}}={\displaystyle \frac{53}{\sin \mathrm{\sphericalangle }\mathrm{C}\mathrm{D}\mathrm{A}}}\backslash mathrm\{\backslash dfrac\{\backslash overline\{AD\}\}\{\backslash sin118^\backslash circ\}=\backslash dfrac\{53\}\{\backslash sin\backslash sphericalangle\{CDA\}\}\}$

Berechnen von $\mathrm{\sphericalangle }\mathrm{C}\mathrm{D}\mathrm{A}\backslash mathrm\{\backslash sphericalangle\{CDA\}\}$ .

$\mathrm{\sphericalangle }\mathrm{C}\mathrm{D}\mathrm{A}={180}^{\circ }-{118}^{\circ }-{24}^{\circ }\backslash mathrm\{\backslash sphericalangle\{CDA\}=180^\backslash circ-118^\backslash circ-24^\backslash circ\}$

$\mathrm{\sphericalangle }\mathrm{C}\mathrm{D}\mathrm{A}={38}^{\circ }\backslash mathrm\{\backslash sphericalangle\{CDA\}=38^\backslash circ\}$

Einsetzen der Werte in die Formel:

${\displaystyle \frac{\overline{\mathrm{A}\mathrm{D}}}{\sin {118}^{\circ }}}={\displaystyle \frac{53}{\sin {38}^{\circ }}}\Rightarrow \overline{\mathrm{A}\mathrm{D}}={\displaystyle \frac{53\cdot \sin {118}^{\circ }}{\sin {38}^{\circ }}}\backslash mathrm\{\backslash dfrac\{\backslash overline\{AD\}\}\{\backslash sin118^\backslash circ\}=\backslash dfrac\{53\}\{\backslash sin38^\backslash circ\}\backslash Rightarrow\backslash \; \backslash overline\{AD\}=\backslash dfrac\{53\backslash cdot\; \backslash sin118^\backslash circ\}\{\backslash sin38^\backslash circ\}\}$

$\overline{\mathrm{A}\mathrm{D}}=\mathrm{76,00}\text{ m}\backslash mathrm\{\backslash overline\{AD\}=76\{,\}00\backslash m\}$

Berechnen des Flächeninhalts des Dreiecks $\text{ACD}\backslash text\{ACD\}$ .

${\mathrm{A}}_{\mathrm{A}\mathrm{C}\mathrm{D}}={\displaystyle \frac{1}{2}}\cdot 76\cdot 53\cdot \mathrm{s}\mathrm{i}\mathrm{n}{24}^{\circ }\backslash mathrm\{A\_\{ACD\}=\backslash dfrac\{1\}\{2\}\backslash cdot\; 76\backslash cdot\; 53\backslash cdot\; sin24^\backslash circ\}$

${\mathrm{A}}_{\mathrm{A}\mathrm{C}\mathrm{D}}=2014\cdot \mathrm{0,4067}[{\mathrm{m}}^2]\backslash mathrm\{A\_\{ACD\}=2014\backslash cdot\; 0\{,\}4067\backslash \; \backslash \; [m^2]\}$

${\mathrm{A}}_{\mathrm{A}\mathrm{C}\mathrm{D}}\approx \mathrm{819,1}{\text{ m}}^2\backslash mathrm\{A\_\{ACD\}\backslash approx819\{,\}1\backslash m^2\}$

Der Flächeninhalt des Dreiecks $\text{ACD}\backslash text\{ACD\}$ beträgt ungefähr: $\boxed{{\displaystyle 819{\text{ m}}^2}}\backslash boxed\{819\backslash m^2\}$

Einzeichnen der Strecke $\overline{\mathrm{B}\mathrm{E}}\backslash mathrm\{\backslash overline\{BE\}\}$