Benutzer:Antonsusi/Heptagon

Gleichschenkliges Dreieck AMN

1.0$\triangle {AMN}$

Gegeben: ${\overline {AM}}=r$ ; ${\overline {AN}}=r$ ; ${\overline {MN}}={\frac {r}{3}}$

 

1.1   $\angle {MAN}$ Mit dem Kosinussatz ergibt sich:

 

   $cos(\angle {MAN})=1-{\frac {({\frac {r}{3}})^{2}}{2r^{2}}}=1-{\frac {1}{18}}={\frac {17}{18}}$

 

   $\alpha =\angle {MAN}=\arccos \left({\frac {17}{18}}\right)\approx 19{,}1881364537209^{\circ }$

1.2 Höhe zur Seite ${\overline {AM}}$

   ${\overline {VN}}=r\,sin{\alpha }=r{\sqrt {1-\left({\frac {17}{18}}\right)^{2}}}={\frac {\sqrt {35}}{18}}r\approx 0{,}3286710990610898$

 

1.3    ${\overline {MV}}={\sqrt {{\overline {NM}}^{2}-{\overline {NV}}^{2}}}={\frac {1}{18}}r$

Rechtwinkeliges Dreieck LJN

2.0 $\triangle {LJN}$

Gegeben:

• ${\overline {MV}}={\frac {1}{18}}r$ (aus 1.3)
• ${\overline {LM}}={\frac {1}{3}}r$
• ${\overline {JL}}={\frac {2}{3}}r$
• ${\overline {NV}}={\frac {\sqrt {35}}{18}}r$ (aus 1.2)

2.1 ${\overline {LV}}={\overline {LM}}+{\overline {MV}}\Rightarrow {\overline {LV}}={\frac {7}{18}}r$

2.2 Hypotenuse ${\overline {LN}}={\sqrt {{\overline {LV}}^{2}+{\overline {NV}}^{2}}}\Rightarrow {\overline {LN}}={\sqrt {\frac {7}{27}}}\,r$

Winkel $\beta =\angle {JLN}=\angle {ALP}$ ergibt sich aus

2.3 $\beta =\arcsin {\frac {\overline {NV}}{\overline {LN}}}=\arcsin \left({\frac {\frac {\sqrt {35}}{18}}{\sqrt {\frac {7}{27}}}}\right)=\arcsin \left({\frac {\sqrt {15}}{6}}\right)$

$\beta \approx 40{,}202965886569769^{\circ }$

2.4 ${\overline {JN}}={\sqrt {{\overline {LJ}}^{2}-{\overline {LN}}^{2}}}={\sqrt {\frac {5}{27}}}\,r$

Rechtwinkeliges Dreieck LAP

3.0 $\triangle {LAP}$

Gegeben:

• $\triangle {LAP}$ ähnlich zu $\triangle {LJN}$ (aus 2.0)
• ${\overline {AL}}={\frac {4}{3}}r=2{\overline {LJ}}$
• ${\overline {NL}}={\sqrt {\frac {7}{27}}}\,r$ (aus 2.2)
• ${\overline {JN}}={\sqrt {\frac {5}{27}}}\,r$ (aus 2.4)

3.1 ${\overline {PL}}=2{\overline {NL}}\Rightarrow {\overline {PL}}=2{\sqrt {\frac {7}{27}}}\,r$

3.2 ${\overline {AP}}=2{\overline {JN}}\Rightarrow {\overline {AP}}=2{\sqrt {\frac {5}{27}}}\,r$

3.3 $\gamma =\angle {PAL}=90^{\circ }-\beta$

oder:

$\gamma =\arccos {\sqrt {\frac {5}{12}}}=\arcsin {\sqrt {\frac {7}{12}}}$

 

3.3 $\gamma \approx 49{,}797034113430231^{\circ }$

Rechtwinkeliges Dreieck JAQ

4.0 $\triangle {JAQ}$

Gegeben:

• ${\overline {JA}}={\frac {2}{3}}r$
• $\angle {QAJ}=\angle {PAL}=\gamma$ (aus 3.3)

4.1

${\overline {QJ}}={\overline {JA}}\tan(\gamma )={\overline {JA}}{\frac {\sin \gamma }{\cos \gamma }}$

${\overline {QJ}}={\overline {JA}}{\frac {\sqrt {\frac {7}{12}}}{\sqrt {\frac {5}{12}}}}$

${\overline {QJ}}={\overline {JA}}{\sqrt {\frac {7}{5}}}={\frac {2{\sqrt {7}}}{3{\sqrt {5}}}}r$

${\overline {QJ}}\approx 0{,}788810637746615r$

4.2.

${\overline {AQ}}={\overline {JA}}\,{\frac {\overline {JA}}{\cos \gamma }}={\overline {JA}}\,{\sqrt {\frac {12}{5}}}$

${\overline {AQ}}={\frac {2{\sqrt {12}}}{3{\sqrt {5}}}}r={\sqrt {\frac {16}{15}}}r\approx 1,032795558988645r$

Gleichseitiges Dreieck HMR

5.0 $\triangle {HMR}$

Gegeben:

• ${\overline {HM}}=r$
• ${\overline {SM}}={\frac {1}{2}}r$
• $\angle {RMS}=\delta =60^{\circ }$

5.2 Höhe:

${\overline {RS}}={\overline {HM}}*{\frac {\sqrt {3}}{2}}={\frac {\sqrt {3}}{2}}r\approx 0,866025403784439\cdot r$

Rechtwinkeliges Dreieck HAO

6.0 $\triangle {HAO}$

Gegeben:

• $\triangle {HAO}$ ähnlich $\triangle {LJN}$ (aus 2.0) also ${\frac {\overline {HA}}{\overline {LJ}}}={\frac {\overline {AO}}{\overline {JN}}}$
• ${\overline {AQ}}={\sqrt {\frac {16}{15}}}r$ (aus 4.2)
• ${\overline {HA}}=2r$ ; ${\overline {LJ}}={\frac {2}{3}}\cdot r$ ; ${\overline {JN}}={\sqrt {\frac {5}{27}}}\cdot r$ (aus 2.4)

6.1 ${\overline {AO}}={\frac {{\overline {JN}}\cdot {\overline {HA}}}{\overline {LJ}}}={\frac {{\sqrt {\frac {5}{27}}}\cdot 2}{\frac {2}{3}}}\,r=3\cdot {\sqrt {\frac {5}{27}}}\,r={\sqrt {\frac {5}{3}}}\cdot r\approx 1{,}290994448735805\cdot r$

6.2 ${\overline {QO}}={\overline {AO}}-{\overline {AQ}}=\left({\sqrt {\frac {5}{3}}}-{\sqrt {\frac {16}{15}}}\right)r=\left({\sqrt {\frac {25}{15}}}-{\sqrt {\frac {16}{15}}}\right)r={\frac {{\sqrt {25}}-{\sqrt {16}}}{\sqrt {15}}}r={\frac {1}{\sqrt {15}}}r\approx 0{,}258198889747161r$

Stumpfwinkeliges Dreieck RST

7.0 $\triangle {RST}$

Gegeben:

• ${\overline {TR}}={\overline {QO}}={\frac {1}{\sqrt {15}}}r$ (aus 6.2)
• ${\overline {ST}}=r$
• ${\overline {RS}}={\frac {\sqrt {3}}{2}}r$ (aus 5.2)

7.1 Nach dem Kosinussatz gilt:

$\cos(\angle {SRT})=\cos(\epsilon )={\frac {{\overline {TR}}^{2}+{\overline {RS}}^{2}-{\overline {ST}}^{2}}{2{\overline {TR}}\cdot {\overline {RS}}}}={\frac {{\frac {1}{15}}+{\frac {3}{4}}-1}{2\cdot {\frac {1}{\sqrt {15}}}\cdot {\frac {\sqrt {3}}{2}}}}={\frac {\frac {4+45-60}{60}}{\frac {1}{\sqrt {5}}}}={\frac {-11{\sqrt {5}}}{60}}$

$\epsilon =\arccos \left({\frac {-11{\sqrt {5}}}{60}}\right)\approx 114{,}201429828962958^{\circ }$

7.2 Berechnung $\angle {WTR}$ und $\cos \angle {WTR}$

(Berechnung indirekt, um den arithm. Ausdruck zu erhalten)

$\angle {WTR}=\zeta =\epsilon -90^{\circ }$

Mit dem Additionstheorem:

$\cos(x-y)=\cos x\;\cos y+\sin x\;\sin y$

ergibt sich:

$\cos \zeta =\sin \epsilon ={\sqrt {1-\cos ^{2}(\epsilon )}}={\sqrt {1-\cos ^{2}(\epsilon )}}={\frac {\sqrt {2995}}{60}}$

$\zeta \approx 24{,}201429828962958$

7.3 Höhe ${\overline {WT}}$ von $\triangle {RST}$ zur Seite ${\overline {RS}}$

${\overline {WT}}={\overline {TR}}\cdot \cos \zeta ={\frac {1}{\sqrt {15}}}\cdot {\frac {\sqrt {2995}}{60}}\cdot r={\frac {\sqrt {599}}{60{\sqrt {3}}}}\cdot r$

${\overline {WT}}\approx 0{,}235505759935852\cdot r$

Rechtwinkeliges Dreieck TWR

8.0 $\triangle {TWR}$

Gegeben:

• ${\overline {RT}}={\overline {QO}}={\frac {1}{\sqrt {15}}}\cdot r$ (aus 6.2)
• ${\overline {WT}}={\frac {\sqrt {599}}{60{\sqrt {3}}}}\cdot r$ (aus 7.3)

8.1

${\overline {WR}}={\sqrt {{\overline {RT}}^{2}-{\overline {WT}}^{2}}}={\sqrt {{\frac {1}{15}}-{\frac {599}{60^{2}\cdot 3}}}}\cdot r={\sqrt {\frac {720-599}{60^{2}\cdot 3}}}\cdot r={\sqrt {\frac {121}{60^{2}\cdot 3}}}\cdot r={\frac {11}{60{\sqrt {3}}}}\cdot r$

${\overline {WR}}\approx 0{,}105847549351431\cdot r$

Rechtwinkeliges Dreieck QXT

9.0 $\triangle {QXT}$

Gegeben:

• ${\overline {WT}}={\frac {\sqrt {599}}{60{\sqrt {3}}}}\cdot r$ (aus 7.3)
• ${\overline {SJ}}={\frac {5}{6}}\cdot r$ (aus Zeichnung)

9.1

${\overline {XQ}}={\overline {SJ}}-{\overline {WT}}=\left({\frac {5}{6}}-{\frac {\sqrt {599}}{60{\sqrt {3}}}}\right)\cdot r={\frac {50{\sqrt {3}}-{\sqrt {599}}}{60{\sqrt {3}}}}\cdot r$

${\overline {XQ}}\approx 0{,}597827573397482\cdot r$

9.2

${\overline {XT}}={\overline {RS}}+{\overline {WR}}-{\overline {QJ}}$

mit

• ${\overline {RS}}={\frac {\sqrt {3}}{2}}r$ (aus 5.2)
• ${\overline {WR}}={\frac {11}{60{\sqrt {3}}}}\cdot r$ (aus 8.1)
• ${\overline {QJ}}={\frac {2{\sqrt {7}}}{3{\sqrt {5}}}}\cdot r$ (aus 4.1)

ergibt sich:

${\overline {XT}}=\left({\frac {\sqrt {3}}{2}}+{\frac {11}{60{\sqrt {3}}}}-{\frac {2{\sqrt {7}}}{3{\sqrt {5}}}}\right)\cdot r=\left({\frac {90+11}{60{\sqrt {3}}}}-{\frac {2{\sqrt {7}}}{3{\sqrt {5}}}}\right)\cdot r=\left({\frac {101-8{\sqrt {105}}}{60{\sqrt {3}}}}\right)\cdot r$

${\overline {XT}}\approx 0{,}183062315389255\cdot r$

9.3 Berechnung $\angle {XTQ}=\eta$

$\tan(\eta )={\frac {\overline {XQ}}{\overline {XT}}}={\frac {50{\sqrt {3}}-{\sqrt {599}}}{101-8{\sqrt {105}}}}$

$\eta =\arctan \left({\frac {\overline {XQ}}{\overline {XT}}}={\frac {50{\sqrt {3}}-{\sqrt {599}}}{101-8{\sqrt {105}}}}\right)$

$\eta \approx 72{,}974753574847806^{\circ }$

Rechtwinkeliges Dreieck QPU

10.0 $\triangle {QPU}$

Gegeben:

• ${\overline {AQ}}={\sqrt {\frac {16}{15}}}\,r$ (aus 4.2)
• ${\overline {AP}}=2{\sqrt {\frac {5}{27}}}\,r$ (aus 3.2)
• ${\overline {PL}}=2{\sqrt {\frac {7}{27}}}\,r$ (aus 3.1)
• $\tan \eta ={\frac {50{\sqrt {3}}-{\sqrt {599}}}{101-8{\sqrt {105}}}}$ (aus 9.3)

10.1

${\overline {QP}}={\overline {AQ}}-{\overline {AP}}=\left({\frac {\sqrt {16}}{\sqrt {15}}}-{\frac {2\cdot {\sqrt {5}}}{\sqrt {27}}}\right)r=\left({\frac {4\cdot 3-2\cdot 5}{\sqrt {27\cdot 5}}}\right)r={\frac {2}{\sqrt {135}}}\cdot r$

${\overline {QP}}\approx 0{,}172132593164774\cdot r$

10.2 Berechnung $\tan \angle {PQU}=\theta$

$\tan \beta ={\frac {\overline {AP}}{\overline {LP}}}={\frac {2{\sqrt {\frac {5}{27}}}}{2{\sqrt {\frac {7}{27}}}}}={\frac {\sqrt {5}}{\sqrt {7}}}$

$\theta =\eta -\beta$

Mit dem Additionstheorem

$\tan(\eta -\beta )={\frac {\tan \eta -\tan \beta }{1+\tan \eta \;\tan \beta }}$

ergibt sich:

$\tan \theta ={\frac {{\frac {50{\sqrt {3}}-{\sqrt {599}}}{101-8{\sqrt {105}}}}-{\frac {\sqrt {5}}{\sqrt {7}}}}{1+{\frac {50{\sqrt {3}}-{\sqrt {599}}}{101-8{\sqrt {105}}}}\cdot {\frac {\sqrt {5}}{\sqrt {7}}}}}$

$\tan \theta ={\frac {{\sqrt {7}}\cdot (50{\sqrt {3}}-{\sqrt {599}})-{\sqrt {5}}\cdot (101-8{\sqrt {105}})}{{\sqrt {5}}\cdot (50{\sqrt {3}}-{\sqrt {599}})+{\sqrt {7}}\cdot (101-8{\sqrt {105}})}}$

$\tan \theta \approx 0{,}643759341824444$

$\theta \approx 32{,}771787688277979^{\circ }$

10.3

${\overline {PU}}={\overline {QP}}\cdot \tan \theta ={\frac {2}{\sqrt {135}}}\cdot {\frac {{\sqrt {7}}\cdot (50{\sqrt {3}}-{\sqrt {599}})-{\sqrt {5}}\cdot (101-8{\sqrt {105}})}{{\sqrt {5}}\cdot (50{\sqrt {3}}-{\sqrt {599}})+{\sqrt {7}}\cdot (101-8{\sqrt {105}})}}\cdot r$

${\overline {PU}}\approx 0{,}110811964882290\cdot r$

Rechtwinkeliges Dreieck APU

11.0 $\triangle {APU}$

Gegeben:

• ${\overline {AP}}=2{\sqrt {\frac {5}{27}}}\,r$ (aus 3.2)
• ${\overline {PU}}={\frac {2}{\sqrt {135}}}\cdot {\frac {{\sqrt {7}}\cdot (50{\sqrt {3}}-{\sqrt {599}})-{\sqrt {5}}\cdot (101-8{\sqrt {105}})}{{\sqrt {5}}\cdot (50{\sqrt {3}}-{\sqrt {599}})+{\sqrt {7}}\cdot (101-8{\sqrt {105}})}}\cdot r$ (aus 10.3)

11.1

Die Länge der Siebeneckseite entspricht AU und beträgt:

${\overline {AU}}={\sqrt {{\overline {AP}}^{2}+{\overline {PU}}^{2}}}={\sqrt {{\frac {20}{27}}+{\frac {4}{135}}\cdot {\frac {7(8099-100{\sqrt {1797}})+5(16921-1616{\sqrt {105}})-2{\sqrt {35}}(50{\sqrt {3}}-{\sqrt {599}})(101-8{\sqrt {105}})}{5(8099-100{\sqrt {1797}})+7(16921-1616{\sqrt {105}})+2{\sqrt {35}}(50{\sqrt {3}}-{\sqrt {599}})(101-8{\sqrt {105}})}}}}\cdot r$

$s={\overline {AU}}\approx 0{,}86776726851259754370992295342964\cdot r$

Mit $p=50{\sqrt {3}}-{\sqrt {599}}$ und $q=101-8{\sqrt {105}}$ gilt:

${\overline {AU}}={\sqrt {{\frac {20}{27}}+{\frac {4}{135}}\cdot {\frac {7p^{2}+5q^{2}-2{\sqrt {35}}\cdot p\cdot q}{5p^{2}+7q^{2}+2{\sqrt {35}}\cdot p\cdot q}}}}\cdot r$

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<math>\triangle{}
: <math>\overline{} = </math>
:\overline{}
\frac{}{}
 \sqrt{}
: 
^\circ
\left( \right)
<math>\angle{}</math>
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