Bestimme die gesuchten Größen mit dem Sinussatz

Berechnung von$\mathrm{\sphericalangle }\mathbf{\text{DBA}}\backslash \; \backslash sphericalangle\; \backslash textbf\{DBA\}$ :

${\displaystyle \frac{\overline{\text{AD}}}{\text{sin}\mathrm{\sphericalangle }\text{DBA}}}={\displaystyle \frac{\overline{\text{AB}}}{\text{sin}\mathrm{\sphericalangle }\text{ADB}}}\Rightarrow \text{sin}\mathrm{\sphericalangle }\text{DBA}={\displaystyle \frac{\overline{\text{AD}}\cdot \text{sin}\mathrm{\sphericalangle }\text{ADB}}{\overline{\text{AB}}}}\backslash dfrac\{\backslash overline\{\backslash text\{AD\}\}\}\{\backslash text\{sin\}\backslash sphericalangle\; \backslash text\{DBA\}\}=\backslash dfrac\{\backslash overline\{\backslash text\{AB\}\}\}\{\backslash text\{sin\}\backslash sphericalangle\; \backslash text\{ADB\}\}\backslash Rightarrow\{\backslash text\{sin\}\backslash sphericalangle\; \backslash text\{DBA\}\}=\backslash dfrac\{\{\backslash overline\{\backslash text\{AD\}\}\backslash cdot\backslash text\{sin\}\backslash sphericalangle\{\backslash text\{ADB\}\}\}\}\{\backslash overline\{\backslash text\{AB\}\}\}$

$\text{sin}\mathrm{\sphericalangle }\text{DBA}={\displaystyle \frac{3\cdot \text{sin }{80}^{\circ }}{\text{8}}}=\mathrm{0,3693}\Rightarrow {\text{sin}}^{-1}(\mathrm{0,3693})={\mathrm{21,67}}^{\circ }\backslash \; \{\backslash text\{sin\}\backslash sphericalangle\; \backslash text\{DBA\}\}=\backslash dfrac\{\{3\}\backslash cdot\backslash text\{sin\}\backslash \; 80^\backslash circ\}\{\{\backslash text\{8\}\}\}=0\{,\}3693\backslash Rightarrow\backslash text\{sin\}^\{-1\}(0\{,\}3693)=21\{,\}67^\backslash circ$

$\underset{‾}{{\text{Das Maß des Winkels DBA betr}\ddot{\text{a}}\text{gt: 21,67}}^{\circ }}\backslash underline\{\backslash text\{Das\backslash \; Maß\backslash \; des\backslash \; Winkels\backslash \; DBA\backslash \; beträgt:\backslash \; 21\{,\}67\}^\backslash circ\}$

Berechnung von$\overline{\mathbf{\text{BD}}}\backslash \; \backslash overline\{\backslash textbf\{BD\}\}\backslash$:

${\displaystyle \frac{\overline{\text{BD}}}{\text{sin}\mathrm{\sphericalangle }\text{BAD}}}={\displaystyle \frac{\overline{\text{AB}}}{\text{sin}\mathrm{\sphericalangle }\text{ADB}}}\mathrm{\sphericalangle }\text{BAD}={180}^{\circ }-{80}^{\circ }-{\mathrm{21,67}}^{\circ }={\mathrm{78,33}}^{\circ }\backslash \; \backslash dfrac\{\backslash overline\{\backslash text\{BD\}\}\}\{\backslash text\{sin\}\backslash sphericalangle\; \backslash text\{BAD\}\}=\backslash dfrac\{\backslash overline\{\backslash text\{AB\}\}\}\{\backslash text\{sin\}\backslash sphericalangle\; \backslash text\{ADB\}\}\backslash qquad\backslash sphericalangle\; \{\backslash text\{BAD\}\}=180^\backslash circ-80^\backslash circ-21\{,\}67^\backslash circ=78\{,\}33^\backslash circ$

$\mathrm{\sphericalangle }\text{BAD}={\mathrm{78,33}}^{\circ }\backslash \; \backslash sphericalangle\{\backslash text\{BAD\}\}=78\{,\}33^\backslash circ$

$\overline{\text{BD}}={\displaystyle \frac{\overline{\text{AB}}\cdot \text{sin}\mathrm{\sphericalangle }\text{BAD}}{\text{sin}\mathrm{\sphericalangle }\text{ADB}}}\Rightarrow \overline{\text{BD}}={\displaystyle \frac{8\cdot \text{sin }{\mathrm{78,33}}^{\circ }}{\text{sin }{80}^{\circ }}}\backslash overline\{\backslash text\{BD\}\}=\backslash dfrac\{\backslash overline\{\backslash text\{AB\}\}\backslash cdot\backslash text\{sin\}\backslash sphericalangle\{\backslash text\{BAD\}\}\}\{\backslash text\{sin\}\backslash sphericalangle\{\backslash text\{ADB\}\}\}\backslash Rightarrow\backslash overline\{\backslash text\{BD\}\}=\backslash dfrac\{\{8\}\backslash cdot\backslash text\{sin\}\backslash \; \{78\{,\}33^\backslash circ\}\}\{\backslash text\{sin\}\backslash \; \{\{80^\backslash circ\}\}\}$

$\overline{\text{BD}}=\mathrm{7,96}\text{ cm}\backslash overline\{\backslash text\{BD\}\}=7\{,\}96\backslash \; \backslash text\{cm\}$

$\underset{‾}{\text{Die L}\ddot{\text{a}}\text{nge der Strecke [BD] betr}\ddot{\text{a}}\text{gt: 7,96}\text{cm}}\backslash underline\{\backslash text\{Die\backslash \; Länge\backslash \; der\backslash \; Strecke\backslash \; [BD]\backslash \; beträgt:\backslash \; 7\{,\}96\backslash ;\backslash text\{cm\}\}\}$

Zerlegung in zwei Dreiecke

Das Viereck$\text{ ABCD }\backslash \; \backslash text\{ABCD\}\backslash$ lässt sich in beiden Dreiecke$\text{ ABD }\backslash \; \backslash text\{ABD\}\backslash$und$\text{ DBC }\backslash \; \backslash text\{DBC\}\backslash$zerlegen.

Flächeninhalt Dreieck $ABDABD$

Berechne den Flächeninhalt mithilfe des Sinussatzes.

${\text{A}}_{\text{ABD}}={\displaystyle \frac{1}{2}}\cdot \overline{\text{AB}}\cdot \overline{\text{BD}}\cdot \sin (\mathrm{\sphericalangle }\text{ABD})\backslash \; \backslash text\{A\}\_\{\backslash text\{ABD\}\}=\backslash dfrac\{1\}\{2\}\backslash cdot\backslash overline\{\backslash text\{AB\}\}\backslash cdot\backslash overline\{\backslash text\{BD\}\}\backslash cdot\backslash sin(\backslash sphericalangle\{\backslash text\{ABD\}\})$

${\text{A}}_{\text{ABD}}={\displaystyle \frac{1}{2}}\cdot 8\cdot \mathrm{7,96}\cdot \sin ({\mathrm{21,67}}^{\circ }){\text{cm}}^2\backslash \; \backslash text\{A\}\_\{\backslash text\{ABD\}\}=\backslash dfrac\{1\}\{2\}\backslash cdot8\backslash cdot7\{,\}96\backslash cdot\; \backslash sin(21\{,\}67^\backslash circ)\backslash ;\backslash text\{cm\}^2$

Flächeninhalt Dreieck $DBCDBC$

Berechne den Flächeninhalt mithilfe des Sinussatzes.

${\text{A}}_{\text{BCD}}={\displaystyle \frac{1}{2}}\cdot \overline{\text{BD}}\cdot \overline{\text{BC}}\cdot \sin (\mathrm{\sphericalangle }\text{DBC})\mathrm{\sphericalangle }\text{DBC}={110}^{\circ }-{\mathrm{21,67}}^{\circ }={\mathrm{88,33}}^{\circ }\backslash \; \backslash text\{A\}\_\{\backslash text\{BCD\}\}=\backslash dfrac\{1\}\{2\}\backslash cdot\backslash overline\{\backslash text\{BD\}\}\backslash cdot\backslash overline\{\backslash text\{BC\}\}\backslash cdot\backslash sin(\backslash sphericalangle\{\backslash text\{DBC\}\})\backslash qquad\backslash sphericalangle\{\backslash text\{DBC\}\}=110^\backslash circ-21\{,\}67^\backslash circ=88\{,\}33^\backslash circ$

${\text{A}}_{\text{BCD}}={\displaystyle \frac{1}{2}}\cdot \mathrm{7,96}\cdot 8\cdot \sin ({\mathrm{88,33}}^{\circ }){\text{cm}}^2\backslash \; \backslash text\{A\}\_\{\backslash text\{BCD\}\}=\backslash dfrac\{1\}\{2\}\backslash cdot7\{,\}96\backslash cdot8\backslash cdot\; \backslash sin(88\{,\}33^\backslash circ)\backslash ;\backslash text\{cm\}^2$

Flächeninhalt des Vierecks

$A_{\text{ABCD}}=({\displaystyle \frac{1}{2}}\cdot 8\cdot \mathrm{7,96}\cdot \sin ({\mathrm{21,67}}^{\circ })+{\displaystyle \frac{1}{2}}\cdot \mathrm{7,96}\cdot 8\cdot \sin ({\mathrm{88,33}}^{\circ })){\text{cm}}^2\backslash \; A\_\{\backslash text\{ABCD\}\}=\backslash left(\backslash dfrac\{1\}\{2\}\backslash cdot8\backslash cdot7\{,\}96\backslash cdot\; \backslash sin(21\{,\}67^\backslash circ)+\backslash dfrac\{1\}\{2\}\backslash cdot7\{,\}96\backslash cdot8\backslash cdot\; \backslash sin(\; 88\{,\}33^\backslash circ)\backslash right)\backslash \; \backslash text\{cm\}^2$

$A_{\text{ABCD}}={\displaystyle \frac{1}{2}}\cdot 8\cdot \mathrm{7,96}\cdot (\sin ({\mathrm{21,67}}^{\circ })+\sin ({\mathrm{88,33}}^{\circ })){\text{cm}}^2\backslash \; A\_\{\backslash text\{ABCD\}\}=\backslash dfrac\{1\}\{2\}\backslash cdot8\backslash cdot7\{,\}96\backslash cdot\; \backslash left(\backslash sin(21\{,\}67^\backslash circ)+\backslash sin(88\{,\}33^\backslash circ)\backslash right)\backslash \; \backslash text\{cm\}^2$

$\underset{‾}{A_{\text{ABCD}}\approx \mathrm{43,58}{\text{cm}}^2}\backslash \; \backslash underline\{A\_\{\backslash text\{ABCD\}\}\backslash approx\; 43\{,\}58\backslash \; \backslash text\{cm\}^2\}$

Der Flächeninhalt des Vierecks $\text{ABCD }\backslash text\{ABCD\}\backslash$beträgt:$\boxed{{\displaystyle \mathrm{43,58}{\text{cm}}^2}}\backslash \; \backslash boxed\{43\{,\}58\backslash \; \backslash text\{cm\}^2\}$

Der Winkel$\text{ CME }\backslash \; \backslash text\{CME\}\backslash$ist ein Stufenwinkel zum Winkel$\text{ CBA }\backslash \; \backslash text\{CBA\}\backslash$an den Parallelen$\text{ AB }\backslash \; \backslash text\{AB\}\; \backslash$und$\text{ EM }\backslash \; \backslash text\{EM\}\backslash$.

Stufenwinkel an parallelen Geraden sind gleich groß.

Somit gilt: ⁣$\mathrm{\sphericalangle }\text{CBA}=\mathrm{\sphericalangle }\text{CME}={110}^{\circ }\Rightarrow \mathrm{\sphericalangle }\text{EMB}={180}^{\circ }-{110}^{\circ }={70}^{\circ }\backslash \; \backslash sphericalangle\{\backslash text\{CBA\}\}=\backslash sphericalangle\{\backslash text\{CME\}\}=110^\backslash circ\backslash Rightarrow\backslash sphericalangle\{\backslash text\{EMB\}\}=180^\backslash circ-110^\backslash circ=70^\backslash circ$.

$\underset{‾}{\text{Berechnung der Bogenl}\ddot{\text{a}}\text{nge des Kreisbogens}\text{ mit dem Mittelpunkt M.}}\backslash underline\{\backslash text\{Berechnung\; der\; Bogenlänge\; des\; Kreisbogens\}\backslash \; \backslash overset\backslash frown\{\backslash text\{EB\}\}\backslash \; \backslash text\{mit\; dem\; Mittelpunkt\; M.\}\}$

$=\overline{\text{BC}}\cdot \pi \cdot {\displaystyle \frac{\mathrm{\sphericalangle }\text{EMB}}{{360}^{\circ }}}\text{ cm}\Rightarrow =8\cdot \pi \cdot {\displaystyle \frac{{70}^{\circ }}{{360}^{\circ }}}\text{cm}\backslash \; \backslash overset\backslash frown\{\backslash text\{EB\}\}=\backslash overline\{\backslash text\{BC\}\}\backslash cdot\backslash pi\backslash cdot\backslash dfrac\{\backslash sphericalangle\{\backslash text\{EMB\}\}\}\{360^\backslash circ\}\backslash \; \backslash text\{cm\}\backslash Rightarrow\backslash overset\backslash frown\{\backslash text\{EB\}\}=8\backslash cdot\backslash pi\backslash cdot\backslash dfrac\{70^\backslash circ\}\{360^\backslash circ\}\backslash ;\backslash text\{cm\}$

$\underset{‾}{=\mathrm{4,89}\text{ cm}}\backslash underline\{\backslash \; \backslash overset\backslash frown\{\backslash text\{EB\}\}=4\{,\}89\backslash \; \backslash text\{cm\}\}$

Die Bogenlänge des Kreisbogens$\backslash \; \backslash overset\backslash frown\{\backslash text\{EB\}\}\backslash$beträgt:$\mathrm{4,89}\text{ cm}\mathrm{.}\backslash \; 4\{,\}89\backslash \; \backslash text\{cm\}.$

Die Figur bildet einen Kreissektor.

$\underset{‾}{\text{Berechnung des Fl}\ddot{\text{a}}\text{cheninhaltes des Kreissektors der durch den Kreisbogen}}\underset{‾}{\text{und den Strecken [EM] und [BM] begrenzt wird.}}\backslash underline\{\backslash \; \backslash text\{Berechnung\; des\; Flächeninhaltes\; des\; Kreissektors\; der\; durch\; den\; Kreisbogen\}\backslash \; \backslash overset\backslash frown\{\backslash text\{EB\}\}\}\backslash \; \backslash \backslash \backslash \; \backslash underline\{\backslash text\{und\; den\; Strecken\; [EM]\; und\; [BM]\; begrenzt\; wird.\}\}$

$A_{\text{Sektor}}={\left({\displaystyle \frac{\overline{\text{BC}}}{2}}\right)}^2\cdot \pi \cdot {\displaystyle \frac{\mathrm{\sphericalangle }\text{EMB}}{360}}{\text{cm}}^2\Rightarrow A_{\text{Sektor}}=16\cdot \pi \cdot {\displaystyle \frac{{70}^{\circ }}{{360}^{\circ }}}{\text{cm}}^2\backslash \; A\_\{\backslash text\{Sektor\}\}=\backslash left(\backslash dfrac\{\backslash overline\{\backslash text\{BC\}\}\}\{2\}\backslash right)^2\backslash cdot\backslash pi\backslash cdot\backslash dfrac\{\backslash sphericalangle\{\backslash text\{EMB\}\}\}\{360\}\backslash ;\backslash text\{cm\}^2\backslash Rightarrow\; A\_\{\backslash text\{Sektor\}\}=16\backslash cdot\backslash pi\backslash cdot\backslash dfrac\{70^\backslash circ\}\{360^\backslash circ\}\backslash ;\backslash text\{cm\}^2$

$\underset{‾}{A_{\text{Sektor}}=\mathrm{9,77}{\text{cm}}^2}\backslash underline\{\backslash \; A\_\{\backslash text\{Sektor\}\}=9\{,\}77\backslash \; \backslash text\{cm\}^2\}$

Der Flächeninhalt der Figur, die durch den Kreisbogen$\backslash \; \backslash overset\backslash frown\{\backslash text\{EB\}\}\backslash$und den Strecken$[\text{EM}]\backslash \; [\backslash text\{EM\}]\backslash$

und$[\text{BM}]\backslash \; [\backslash text\{BM\}]\backslash$begrenzt wird beträgt:$\boxed{{\displaystyle \mathrm{9,77}{\text{cm}}^2}}\backslash \; \backslash boxed\{9\{,\}77\backslash \; \backslash text\{cm\}^2\}$

$\underset{‾}{\text{Bestimmung des prozentualen Anteils dieses Fl}\ddot{\text{a}}\text{cheninhalts}}\underset{‾}{\text{ am Fl}\ddot{\text{a}}\text{cheninhalt des Vierecks ABCD.}}\backslash underline\{\backslash \; \backslash text\{Bestimmung\; des\; prozentualen\; Anteils\; dieses\; Flächeninhalts\}\}\backslash \backslash \backslash underline\{\backslash text\{\; am\; Flächeninhalt\; des\; Vierecks\; ABCD.\}\}$

$\text{ p}={\displaystyle \frac{\text{W}}{\text{G}}}\text{W}=\mathrm{9,77}{\text{cm}}^2;\text{G}=\mathrm{43,58}{\text{cm}}^2\backslash \; \backslash text\{p\}=\backslash dfrac\{\backslash text\{W\}\}\{\backslash text\{G\}\}\backslash qquad\backslash qquad\; \backslash text\{W\}=9\{,\}77\backslash \; \backslash text\{cm\}^2;\backslash qquad\; \backslash text\{G\}=43\{,\}58\backslash \; \backslash text\{cm\}^2$

$\text{ p}={\displaystyle \frac{\mathrm{9,77}{\text{cm}}^2}{\mathrm{43,58}{\text{cm}}^2}}\Rightarrow \underset{‾}{\text{p}=\mathrm{0,2242}}\backslash \; \backslash text\{p\}=\backslash dfrac\{9\{,\}77\backslash \; \backslash text\{cm\}^2\}\{43\{,\}58\backslash \; \backslash text\{cm\}^2\}\backslash Rightarrow\backslash underline\{\backslash text\{p\}=0\{,\}2242\}$

Der prozentuale Anteil dieses Flächeninhalts am Flächeninhalt des Vierecks$ABCDABCD$

ist dann: $\mathrm{0,2242}\cdot 100\mathrm{\%}=\boxed{{\displaystyle \mathrm{22,42}\mathrm{\%}}}\backslash \; 0\{,\}2242\backslash cdot100\backslash ;\backslash \%=\backslash boxed\{22\{,\}42\backslash ;\backslash \%\}$