Berechnen der Strecke $\overline{\mathrm{A}\mathrm{B}}\backslash mathrm\{\backslash overline\{AB\}\}$

Das Dreieck $\text{ABD}\backslash text\{ABD\}$ ist ein rechtwinkliges Dreieck.

In diesem Dreieck gilt: ${\displaystyle \frac{\overline{\mathrm{A}\mathrm{B}}}{\overline{\mathrm{B}\mathrm{D}}}}=\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{\sphericalangle }\mathrm{A}\mathrm{D}\mathrm{B}\Rightarrow \overline{\mathrm{A}\mathrm{B}}=\sin \mathrm{\sphericalangle }\mathrm{A}\mathrm{D}\mathrm{B}\cdot \overline{\mathrm{B}\mathrm{D}}\backslash \; \backslash mathrm\{\backslash dfrac\{\backslash overline\{AB\}\}\{\backslash overline\{BD\}\}=sin\; \backslash sphericalangle\{ADB\}\backslash quad\backslash Rightarrow\backslash overline\{AB\}=\backslash sin\backslash sphericalangle\{ADB\}\backslash cdot\; \backslash overline\{BD\}\}$

$\overline{\mathrm{A}\mathrm{B}}=\sin {55}^{\circ }\cdot 1500\mathrm{m}\backslash hspace\{34mm\}\backslash mathrm\{\backslash overline\{AB\}=\backslash sin55^\backslash circ\backslash cdot\; 1500\backslash \; m\}$

$\overline{\mathrm{A}\mathrm{B}}=\mathrm{1228,73}\mathrm{m}\backslash hspace\{34mm\}\backslash mathrm\{\backslash overline\{AB\}=1228\{,\}73\backslash \; m\}$

Die Strecke $\overline{\mathrm{A}\mathrm{B}}\backslash mathrm\{\backslash overline\{AB\}\}$ beträgt ca. $\boxed{{\displaystyle 1229\mathrm{m}}}\mathrm{.}\backslash boxed\{\backslash mathrm\{1229\backslash \; m\}\}.$

Berechnen der Strecke $\overline{\mathrm{A}\mathrm{D}}\backslash mathrm\{\backslash overline\{AD\}\}$ mit dem Kosinus.

Das Dreieck $\text{ABD}\backslash text\{ABD\}$ ist ein rechtwinkliges Dreieck.

In diesem Dreieck gilt:${\displaystyle \frac{\overline{\mathrm{A}\mathrm{D}}}{\overline{\mathrm{B}\mathrm{D}}}}=\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{\sphericalangle }\mathrm{A}\mathrm{D}\mathrm{B}\Rightarrow \overline{\mathrm{A}\mathrm{D}}=\cos \mathrm{\sphericalangle }\mathrm{A}\mathrm{D}\mathrm{B}\cdot \overline{\mathrm{B}\mathrm{D}}\backslash \; \backslash mathrm\{\backslash dfrac\{\backslash overline\{AD\}\}\{\backslash overline\{BD\}\}=cos\; \backslash sphericalangle\{ADB\}\backslash quad\backslash Rightarrow\backslash overline\{AD\}=\backslash cos\backslash sphericalangle\{ADB\}\backslash cdot\; \backslash overline\{BD\}\}$

$\overline{\mathrm{A}\mathrm{D}}=\cos {55}^{\circ }\cdot 1500\mathrm{m}\backslash hspace\{34mm\}\backslash mathrm\{\backslash overline\{AD\}=\backslash cos55^\backslash circ\backslash cdot\; 1500\backslash \; m\}$

$\overline{\mathrm{A}\mathrm{D}}=\mathrm{860,36}\mathrm{m}\backslash hspace\{34mm\}\backslash mathrm\{\backslash overline\{AD\}=860\{,\}36\backslash \; m\}$

Die Strecke $\overline{\mathrm{A}\mathrm{D}}\backslash mathrm\{\backslash overline\{AD\}\}$ beträgt ca. $\boxed{{\displaystyle 860\mathrm{m}}}\mathrm{.}\backslash boxed\{\backslash mathrm\{860\backslash \; m\}\}.$

Berechnen der Strecke $\overline{\mathrm{A}\mathrm{D}}\backslash mathrm\{\backslash overline\{AD\}\}$ mit dem Satz des Pythagoras.

Satz des Pythagoras angewandt auf das Dreieck $\text{ABD}:\backslash text\{ABD\}:$

${\overline{\mathrm{B}\mathrm{D}}}^2={\overline{\mathrm{A}\mathrm{B}}}^2+{\overline{\mathrm{A}\mathrm{D}}}^2\Rightarrow \overline{\mathrm{A}\mathrm{D}}=\sqrt{{\overline{\mathrm{B}\mathrm{D}}}^2-{\overline{\mathrm{A}\mathrm{B}}}^2}\backslash mathrm\{\backslash overline\{BD\}^2=\backslash overline\{AB\}^2+\backslash overline\{AD\}^2\backslash quad\backslash Rightarrow\backslash overline\{AD\}=\backslash sqrt\{\backslash overline\{BD\}^2-\backslash overline\{AB\}^2\}\}$

$\overline{\mathrm{A}\mathrm{D}}=\sqrt{{1500}^2-{\mathrm{1228,73}}^2}[\mathrm{m}]\backslash hspace\{36mm\}\backslash mathrm\{\backslash overline\{AD\}=\backslash sqrt\{1500^2-1228\{,\}73^2\}\backslash \; \backslash \; [m]\}$

$\overline{\mathrm{A}\mathrm{D}}=\sqrt{\mathrm{740222,6}}[\mathrm{m}]\backslash hspace\{36mm\}\backslash mathrm\{\backslash overline\{AD\}=\backslash sqrt\{740222\{,\}6\}\backslash \; \backslash \; [m]\}$

$\overline{\mathrm{A}\mathrm{D}}=\mathrm{860,36}\mathrm{m}\backslash hspace\{36mm\}\backslash mathrm\{\backslash overline\{AD\}=860\{,\}36\backslash \; m\}$

Berechnen des Winkels $\mathrm{\sphericalangle }\text{DBC}\backslash sphericalangle\{\backslash text\{DBC\}\}$

Berechne zuerst den Winkel $\mathrm{\sphericalangle }\text{ABD}\backslash sphericalangle\{\backslash text\{ABD\}\}$

In einem Dreieck ist die Winkelsumme ${180}^{\circ }180^\backslash circ$

$\mathrm{\sphericalangle }\text{ABD}={180}^{\circ }-{90}^{\circ }-{55}^{\circ }\backslash sphericalangle\{\backslash text\{ABD\}\}=180^\backslash circ-90^\backslash circ-55^\backslash circ$

$\mathrm{\sphericalangle }\text{ABD}={35}^{\circ }\backslash sphericalangle\{\backslash text\{ABD\}\}=35^\backslash circ$

$\mathrm{\sphericalangle }\text{DBC}=\mathrm{\sphericalangle }\text{ABC}-\mathrm{\sphericalangle }\text{ABD}\backslash sphericalangle\{\backslash text\{DBC\}\}=\backslash sphericalangle\{\backslash text\{ABC\}\}-\backslash sphericalangle\{\backslash text\{ABD\}\}$

$\mathrm{\sphericalangle }\text{DBC}={109}^{\circ }-{35}^{\circ }\backslash sphericalangle\{\backslash text\{DBC\}\}=109^\backslash circ-35^\backslash circ$

$\mathrm{\sphericalangle }\text{DBC}={74}^{\circ }\backslash sphericalangle\{\backslash text\{DBC\}\}=74^\backslash circ$

Da die Basiswinkel des Dreiecks $\text{DBC}\backslash text\{DBC\}$ unterschiedlich groß sind, ist die Aussage nicht richtig.

Kreuze entsprechend an.