Zeige rechnerisch, dass für die Länge der Strecken $\overline{D{P}_n}\backslash overline\{DP\_n\}$ in Abhängigkeit von $\phi \backslash varphi$ gilt: $\mathrm{\mid }\overline{D{P}_n}\mathrm{\mid }(\phi )={\displaystyle \frac{11\cdot \sin \phi }{\sin ({\mathrm{35,54}}^{\circ }+\phi )}}\mathrm{c}\mathrm{m}|\backslash overline\{DP\_n\}|(\backslash varphi)=\backslash dfrac\{11\backslash cdot\; \backslash sin\backslash varphi\}\{\; \backslash sin(35\{,\}54^\backslash circ\; +\backslash varphi)\}\backslash ;\backslash mathrm\{cm\}$

In beliebigen Dreiecken gilt der Sinussatz:

${\displaystyle \frac{a}{\sin \left(\alpha \right)}}={\displaystyle \frac{b}{\sin \left(\beta \right)}}={\displaystyle \frac{c}{\sin \left(\gamma \right)}}\backslash dfrac\{a\}\{\backslash sin\backslash left(\backslash alpha\backslash right)\}=\backslash dfrac\{b\}\{\backslash sin\backslash left(\backslash beta\backslash right)\}=\backslash dfrac\{c\}\{\backslash sin\backslash left(\backslash gamma\backslash right)\}$

Angewandt auf das Dreieck $CD{P}_{n}CDP\_n$ folgt:

${\displaystyle \frac{\mathrm{\mid }\overline{D{P}_n}\mathrm{\mid }}{\sin \phi }}={\displaystyle \frac{\mathrm{\mid }\overline{CD}\mathrm{\mid }}{\sin \mathrm{\sphericalangle }C{P}_{n}D}}\backslash dfrac\{|\backslash overline\{DP\_n\}|\}\{\backslash sin\; \backslash varphi\}=\backslash dfrac\{|\backslash overline\{CD\}|\}\{\backslash sin\backslash sphericalangle\; CP\_nD\}$

Weiterhin gilt: $\mathrm{\sphericalangle }C{P}_{n}D={180}^{\circ }-(\mathrm{\sphericalangle }BDC+\phi )\backslash sphericalangle\; CP\_nD=180^\backslash circ\; -(\backslash sphericalangle\; BDC+\backslash varphi)$

Für den Winkel $\mathrm{\sphericalangle }BDC\backslash sphericalangle\; BDC$ gilt: $\mathrm{\sphericalangle }BDC={90}^{\circ }-\mathrm{\sphericalangle }ADB\backslash sphericalangle\; BDC=90^\backslash circ\; -\backslash sphericalangle\; ADB$

Der Winkel $\mathrm{\sphericalangle }ADB\backslash sphericalangle\; ADB$ befindet sich im rechtwinkligen Dreieck $ABDABD$, sodass er berechnet werden kann.

$\tan \mathrm{\sphericalangle }ADB={\displaystyle \frac{\mathrm{\mid }\overline{AB}\mathrm{\mid }}{\mathrm{\mid }\overline{AD}\mathrm{\mid }}}={\displaystyle \frac{7}{5}}=\mathrm{1,4}\Rightarrow \mathrm{\sphericalangle }ADB={\tan }^{-1}(\mathrm{1,4})\approx {\mathrm{54,46}}^{\circ }\backslash tan\; \backslash sphericalangle\; ADB=\backslash dfrac\{|\backslash overline\{AB\}|\}\{|\backslash overline\{AD\}|\}=\backslash dfrac\{7\}\{5\}=1\{,\}4\backslash ;\backslash Rightarrow\backslash ;\backslash sphericalangle\; ADB=\backslash tan^\{-1\}(1\{,\}4)\backslash approx54\{,\}46^\backslash circ$

Dann folgt für den Winkel $\mathrm{\sphericalangle }BDC={90}^{\circ }-\mathrm{\sphericalangle }ADB={90}^{\circ }-{\mathrm{54,46}}^{\circ }={\mathrm{35,54}}^{\circ }\backslash sphericalangle\; BDC=90^\backslash circ\; -\backslash sphericalangle\; ADB=90^\backslash circ-54\{,\}46^\backslash circ=35\{,\}54^\backslash circ$

Nun kann auch der Winkel $\mathrm{\sphericalangle }C{P}_{n}D={180}^{\circ }-(\mathrm{\sphericalangle }BDC+\phi )\backslash sphericalangle\; CP\_nD=180^\backslash circ\; -(\backslash sphericalangle\; BDC+\backslash varphi)$ angegeben werden:

$\mathrm{\sphericalangle }C{P}_{n}D={180}^{\circ }-({\mathrm{35,54}}^{\circ }+\phi )\backslash sphericalangle\; CP\_nD=180^\backslash circ\; -(35\{,\}54^\backslash circ+\backslash varphi)$

Für den Ansatz mit dem Sinussatz erhält man nun:

${\displaystyle \frac{\mathrm{\mid }\overline{D{P}_n}\mathrm{\mid }}{\sin \phi }}={\displaystyle \frac{\mathrm{\mid }\overline{CD}\mathrm{\mid }}{\sin \mathrm{\sphericalangle }C{P}_{n}D}}={\displaystyle \frac{\mathrm{\mid }\overline{CD}\mathrm{\mid }}{\sin ({180}^{\circ }-({\mathrm{35,54}}^{\circ }+\phi ))}}\backslash dfrac\{|\backslash overline\{DP\_n\}|\}\{\backslash sin\; \backslash varphi\}=\backslash dfrac\{|\backslash overline\{CD\}|\}\{\backslash sin\backslash sphericalangle\; CP\_nD\}=\backslash dfrac\{|\backslash overline\{CD\}|\}\{\backslash sin(180^\backslash circ\; -(35\{,\}54^\backslash circ+\backslash varphi))\}$

Wegen $\sin ({180}^{\circ }-\alpha )=\sin \alpha \backslash sin(180^\backslash circ\; -\backslash alpha)=\backslash sin\; \backslash alpha$ kann die rechte Seite der Gleichung noch umgeformt werden:

${\displaystyle \frac{\mathrm{\mid }\overline{D{P}_n}\mathrm{\mid }}{\sin \phi }}={\displaystyle \frac{\mathrm{\mid }\overline{CD}\mathrm{\mid }}{\sin ({180}^{\circ }-({\mathrm{35,54}}^{\circ }+\phi ))}}={\displaystyle \frac{\mathrm{\mid }\overline{CD}\mathrm{\mid }}{\sin ({\mathrm{35,54}}^{\circ }+\phi ))}}\backslash dfrac\{|\backslash overline\{DP\_n\}|\}\{\backslash sin\; \backslash varphi\}=\backslash dfrac\{|\backslash overline\{CD\}|\}\{\backslash sin(180^\backslash circ\; -(35\{,\}54^\backslash circ+\backslash varphi))\}=\backslash dfrac\{|\backslash overline\{CD\}|\}\{\backslash sin(35\{,\}54^\backslash circ+\backslash varphi))\}$

Aufgelöst nach $\mathrm{\mid }\overline{D{P}_n}\mathrm{\mid }|\backslash overline\{DP\_n\}|$ erhält man:

$\mathrm{\mid }\overline{D{P}_n}\mathrm{\mid }={\displaystyle \frac{\mathrm{\mid }\overline{CD}\mathrm{\mid }\cdot \sin \phi }{\sin ({\mathrm{35,54}}^{\circ }+\phi ))}}|\backslash overline\{DP\_n\}|=\backslash dfrac\{|\backslash overline\{CD\}|\backslash cdot\backslash sin\; \backslash varphi\}\{\backslash sin(35\{,\}54^\backslash circ+\backslash varphi))\}$

Setzt man nun noch $\mathrm{\mid }\overline{CD}\mathrm{\mid }=11\mathrm{c}\mathrm{m}|\backslash overline\{CD\}|=\; 11\backslash ;\backslash mathrm\{cm\}$ ein, dann folgt:

$\mathrm{\mid }\overline{D{P}_n}\mathrm{\mid }={\displaystyle \frac{11\mathrm{c}\mathrm{m}\cdot \sin \phi }{\sin ({\mathrm{35,54}}^{\circ }+\phi ))}}\Rightarrow \mathrm{\mid }\overline{D{P}_n}\mathrm{\mid }(\phi )={\displaystyle \frac{11\cdot \sin \phi }{\sin ({\mathrm{35,54}}^{\circ }+\phi )}}\mathrm{c}\mathrm{m}|\backslash overline\{DP\_n\}|=\backslash dfrac\{\; 11\backslash ;\backslash mathrm\{cm\}\backslash cdot\backslash sin\; \backslash varphi\}\{\backslash sin(35\{,\}54^\backslash circ+\backslash varphi))\}\backslash ;\backslash Rightarrow\backslash ;|\backslash overline\{DP\_n\}|(\backslash varphi)=\backslash dfrac\{11\backslash cdot\; \backslash sin\backslash varphi\}\{\; \backslash sin(35\{,\}54^\backslash circ\; +\backslash varphi)\}\backslash ;\backslash mathrm\{cm\}$

Berechne die Länge der Strecke $\overline{D{P}_1}\backslash overline\{DP\_1\}$

Der Winkel $\phi \backslash varphi$ ist hier ${20}^{\circ }20^\backslash circ$.

Eingesetzt in $\mathrm{\mid }\overline{D{P}_n}\mathrm{\mid }(\phi )\Rightarrow \mathrm{\mid }\overline{D{P}_1}\mathrm{\mid }({20}^{\circ })={\displaystyle \frac{11\cdot \sin {20}^{\circ }}{\sin ({\mathrm{35,54}}^{\circ }+{20}^{\circ })}}\approx \mathrm{4,56}|\backslash overline\{DP\_n\}|(\backslash varphi)\backslash ;\backslash Rightarrow\backslash ;|\backslash overline\{DP\_1\}|(20^\backslash circ)=\backslash dfrac\{11\backslash cdot\; \backslash sin20^\backslash circ\}\{\; \backslash sin(35\{,\}54^\backslash circ\; +20^\backslash circ)\}\backslash approx4\{,\}56$

Die Länge der Strecke $\overline{D{P}_1}\backslash overline\{DP\_1\}$ beträgt rund $\mathrm{4,56}\mathrm{c}\mathrm{m}4\{,\}56\backslash ;\backslash mathrm\{cm\}$.

Ergänze das Dreieck $CD{P}_{2}CDP\_2$ in der Zeichnung zur Aufgabenstellung

Das Dreieck $CD{P}_{2}CDP\_2$ ist gleichschenklig mit der Basis $\overline{CD}\backslash overline\{CD\}$.

Errichte die Mittelsenkrechte über der Strecke $\overline{CD}\backslash overline\{CD\}$.

Die Mittelsenkrechte schneidet die Strecke $\overline{BD}\backslash overline\{BD\}$ in Punkt $P_{2}P\_2$.

Siehe Zeichnung: