Gesuchte : Länge der Strecke $[AC][AC]$ in cm.

Folgende Grössen sind bekannt:

$\overline{AB}=46cm;\alpha =28\mathrm{°};\beta =140\mathrm{°}\backslash overline\{AB\}=46\backslash \; cm;\backslash \; \alpha =28°;\backslash \; \beta =140°$

Berechne zuerst die Strecke $\overline{BD}\backslash overline\{BD\}\backslash$

${\displaystyle \frac{\overline{BD}}{\overline{AB}}}=\sin \alpha \Rightarrow \overline{BD}=46cm\cdot \sin 28\mathrm{°}\backslash dfrac\{\backslash overline\{BD\}\}\{\backslash overline\{AB\}\}\backslash \; =\backslash \; \backslash sin\{\backslash alpha\}\backslash quad\backslash Rightarrow\backslash quad\; \backslash overline\{BD\}\backslash \; =\backslash \; 46\backslash \; cm\backslash cdot\backslash sin28°$

$\overline{BD}\approx \mathrm{21,6}cm\backslash overline\{BD\}\backslash \; \backslash approx\backslash \; 21\{,\}6\backslash \; cm$

Berechne die Strecke $\overline{AD}\backslash overline\{AD\}\backslash$

${\displaystyle \frac{\overline{AD}}{\overline{AB}}}=\cos \alpha \Rightarrow \overline{AD}=46cm\cdot \cos 28\mathrm{°}\backslash dfrac\{\backslash overline\{AD\}\}\{\backslash overline\{AB\}\}\backslash \; =\backslash \; \backslash cos\{\backslash alpha\}\backslash quad\backslash Rightarrow\backslash quad\; \backslash overline\{AD\}\backslash \; =\backslash \; 46\backslash \; cm\backslash cdot\backslash cos28°$

$\overline{AD}\approx \mathrm{40,6}cm\backslash overline\{AD\}\backslash \; \backslash approx\backslash \; 40\{,\}6\backslash \; cm$

Berechne die Strecke $\overline{CD}\backslash overline\{CD\}\backslash$

${\displaystyle \frac{\overline{BD}}{\overline{CD}}}=\tan \mathrm{\sphericalangle }BCD\mathrm{\sphericalangle }BCD=180\mathrm{°}-28\mathrm{°}-140\mathrm{°}=12\mathrm{°}\backslash dfrac\{\backslash overline\{BD\}\}\{\backslash overline\{CD\}\}\backslash \; =\backslash \; \backslash tan\backslash sphericalangle\; BCD\backslash hspace\{3cm\}\backslash sphericalangle\; BCD\backslash \; =\backslash \; 180°-28°-140°\backslash \; =\backslash \; 12°$

$\overline{CD}={\displaystyle \frac{\overline{BD}}{\mathrm{\sphericalangle }BCD}}\Rightarrow \overline{CD}={\displaystyle \frac{\mathrm{21,6}cm}{\tan 12\mathrm{°}}}\backslash overline\{CD\}\backslash \; =\backslash \; \backslash dfrac\{\backslash overline\{BD\}\}\{\backslash sphericalangle\; BCD\}\backslash qquad\backslash Rightarrow\backslash qquad\backslash overline\{CD\}\backslash \; =\; \backslash \; \backslash dfrac\{21\{,\}6\backslash \; cm\}\{\backslash tan12°\}$

$\overline{CD}\approx 101.6cm\backslash overline\{CD\}\backslash \; \backslash approx\backslash \; 101.6\backslash \; cm$

Länge der Strecke $[AC]=\overline{AD}+\overline{CD}\Rightarrow \overline{AC}\approx \mathrm{40,6}cm+\mathrm{101,6}cm[AC]\backslash \; =\backslash \; \backslash overline\{AD\}\backslash \; +\backslash \; \backslash overline\{CD\}\backslash quad\backslash Rightarrow\backslash quad\backslash overline\{AC\}\backslash approx40\{,\}6\backslash \; cm\backslash \; +\backslash \; 101\{,\}6\backslash \; cm$

$\overline{AC}\approx \mathrm{142,2}cm\backslash overline\{AC\}\backslash approx142\{,\}2\backslash \; cm$

Die Strecke $[AC][AC]\backslash$ist ungefähr $\mathrm{142,2}cm142\{,\}2\backslash \; cm$ lang.

$\text{Hinweis: Andere L}\ddot{\text{o}}\text{sungswege sind m}\ddot{\text{o}}\text{glich.}\backslash text\{Hinweis:\; Andere\; Lösungswege\; sind\; möglich.\}$