Teil A

$y=a\cdot {\left(\begin{array}{c}1+{\displaystyle \frac{p}{100}}\end{array}\right)}^{X}y=a\backslash cdot\backslash begin\{pmatrix\}\; 1+\backslash dfrac\{p\}\{100\}\backslash end\{pmatrix\}^X\backslash$setzte, $y=\mathrm{2,50};a=\mathrm{2,20}\text{ und }x=2\backslash \; y=2\{,\}50;\backslash \; a=2\{,\}20\backslash \; \backslash text\{und\}\backslash \; x=2\backslash$in die Gleichung ein

$\backslash hspace\{30mm\}$und berechne $p\backslash \; \backslash text\{\}p$.

Preissteigerung bestimmen

$p_1=\mathrm{6,60}\mathbb{L}=\{\mathrm{6,60}\}\backslash hspace\{5mm\}p\_\{1\}\backslash \; =\backslash \; 6\{,\}60\backslash hspace\{15mm\}\backslash mathbb\{L\}=\backslash \{6\{,\}60\backslash \}$

$(p_2=-\mathrm{206,60})\backslash hspace\{5mm\}(p\_\{2\}\backslash \; =\backslash \; -206\{,\}60)$ :$p_2\mathrm{\notin }{\mathbb{R}}^{+}\backslash \; \{p\_\{2\}\}\backslash notin\{\backslash mathbb\{R^+\}\}$

Preis der Rose

Berechne jetzt, wie viel eine Rose im Juni 2027 voraussichtlich kostet.

$f:y=\mathrm{2,20}\cdot {\left(\begin{array}{c}1+{\displaystyle \frac{\mathrm{6,60}}{100}}\end{array}\right)}^X\mathbb{G}={\mathbb{R}}^{+}\times {\mathbb{R}}^{+}f:y=2\{,\}20\backslash cdot\backslash begin\{pmatrix\}\; 1+\backslash dfrac\{6\{,\}60\}\{100\}\backslash end\{pmatrix\}^X\backslash \; \backslash hspace\{15mm\}\backslash mathbb\{G\}=\backslash mathbb\{R^+\}\backslash times\backslash mathbb\{R^+\}$

$f(7)=\mathrm{2,20}\cdot {\left(\begin{array}{c}1+{\displaystyle \frac{\mathrm{6,60}}{100}}\end{array}\right)}^{7}f(7)=2\{,\}20\backslash cdot\backslash begin\{pmatrix\}\; 1+\backslash dfrac\{6\{,\}60\}\{100\}\backslash end\{pmatrix\}^7\backslash$

$f(7)=\mathrm{2,20}\cdot {\mathrm{1,066}}^7\Rightarrow f(7)=\mathrm{2,200}\cdot \mathrm{1,564}f(7)=2\{,\}20\backslash cdot1\{,\}066^7\backslash \; \backslash Rightarrow\backslash \; f(7)=2\{,\}200\backslash cdot1\{,\}564$

$f(7)=\mathrm{3,44}f(7)=3\{,\}44$

Eine Rose würde voraussichtlich$\mathrm{3,44}\text{ € }\backslash \; 3\{,\}44\backslash \; €\backslash$kosten.

Berechnung der Anzahl der Jahre

$2\cdot \mathrm{2,20}=\mathrm{2,20}\cdot {(1+{\displaystyle \frac{\mathrm{6,60}}{100}})}^{X}x\in {\mathbb{R}}^{+}\backslash \; 2\backslash cdot2\{,\}20=2\{,\}20\backslash cdot\backslash left(1+\backslash dfrac\{6\{,\}60\}\{100\}\backslash right)^X\backslash qquad\; x\backslash in\backslash mathbb\{R\}^+$

$\mathrm{4,40}=\mathrm{2,20}\cdot {(1+{\displaystyle \frac{\mathrm{6,60}}{100}})}^X\backslash qquad4\{,\}40=2\{,\}20\backslash cdot\backslash left(1+\backslash dfrac\{6\{,\}60\}\{100\}\backslash right)^X$

Mia müsste im Jahr 2031 erstmals mehr als doppelt so viel für eine Rose bezahlen, wie am 1. Juni 2020.

Berechnung des Winkels

Für den Winkel ist die Strecke $[AM][AM]$ die Gegenkathete und die Strecke $[MS][MS]$ die Ankathete.

geg.: $MS=7cmMS=7\; cm$

ges.: Rechnerischer Nachweis für $\overline{M{P}_n}(\phi )=\frac{\mathrm{5,53}}{sin(\phi +\mathrm{52,13}\mathrm{°})}cm\backslash overline\; \{MP\_n\}(\phi )=\backslash frac\{5\{,\}53\}\{sin(\phi +52\{,\}13°)\}cm$

Nachweis von $\overline{M{P}_n}(\phi )\backslash overline\; \{MP\_n\}(\phi )$

Im Dreieck $M{P}_{n}SMP\_nS$ ist der Winkel bei $SS$ und der bei $MM$ bekannt. Daher ist der Winkel bei $P_{n}P\_n$ gerade ${180}^{\circ }-({\mathrm{52,13}}^{\circ }+\phi )180^\backslash circ-(52\{,\}13^\backslash circ+\backslash varphi)$.

Verwende nun die Beziehung $\sin ({180}^{\circ }-\alpha )=\sin \alpha \backslash sin(180^\backslash circ-\backslash alpha)=\backslash sin\; \backslash alpha$.

1. Nachweis für $[M{P}_n(\phi )][MP\_n(\backslash varphi\; )]$ durch Bildung von Verhältnis der Strecken- und Sinus-Werte

${\displaystyle \phantom{\Rightarrow }\frac{\overline{M{P}_n}}{\overline{MS}}=\frac{\sin ({\mathrm{52,53}}^{\circ })}{\sin (\phi +{\mathrm{52,13}}^{\circ })}}\backslash displaystyle\backslash hphantom\{\backslash Rightarrow\}\; \backslash frac\{\backslash overline\; \{MP\_n\}\}\{\backslash overline\; \{MS\}\}=\backslash frac\{\backslash sin(52\{,\}53^\backslash circ)\}\{\backslash sin(\phi +52\{,\}13^\backslash circ)\}$

${\displaystyle \Rightarrow \frac{\overline{M{P}_n}}{\overline{7cm}}=\frac{\sin ({\mathrm{52,53}}^{\circ })}{\sin (\phi +{\mathrm{52,13}}^{\circ })}\mathrm{\mid }}\backslash displaystyle\backslash Rightarrow\; \backslash frac\{\backslash overline\; \{MP\_n\}\}\{\backslash overline\; \{7cm\}\}=\backslash frac\{\backslash sin(52\{,\}53^\backslash circ)\}\{\backslash sin(\phi +52\{,\}13^\backslash circ)\}\backslash qquad|$auflösen nach $\overline{M{P}_n}(\phi )\backslash color\{blue\}\{\backslash overline\; \{MP\_n\}(\phi )\}$

${\displaystyle \Rightarrow \overline{\mathbf{M}{\mathbf{P}}_{\mathbf{n}}}(\mathbf{\phi })=\frac{7cm\cdot \mathrm{0,78941}}{\sin (\phi +{\mathrm{52,13}}^{\circ })}=\frac{\mathbf{5,53}}{\sin (\mathbf{\phi }+{\mathbf{52,13}}^{\circ })}cm}\backslash displaystyle\backslash Rightarrow\; \backslash mathbf\{\backslash overline\; \{MP\_n\}(\phi )\}=\backslash frac\{7cm\backslash cdot0\{,\}78941\}\{\backslash sin(\phi +52\{,\}13^\backslash circ)\}=\backslash mathbf\{\backslash frac\{5\{,\}53\}\{\backslash sin(\phi +52\{,\}13^\backslash circ)\}\}cm\; \backslash qquad$$\phi \in ]0^{\circ };{90}^{\circ }]\backslash varphi\backslash in]0^\backslash circ;90^\backslash circ]\backslash qquad$

Zugehöriger Wert für $\phi \backslash varphi$, damit $\overline{M{P}_0}(\phi )\backslash overline\{MP\_0\}(\phi )$ minimal ist

Der Bruch wird am kleinsten, wenn der Nenner am größten ist. Bestimme $\phi \backslash varphi$ so, dass der Sinus von ${90}^{\circ }90^\backslash circ$ dort steht:

$\mathbf{\phi }={90}^{\circ }-{\mathrm{52,13}}^{\circ }={\mathbf{37,87}}^{\circ }\backslash mathbf\{\backslash varphi\}=90^\backslash circ-52\{,\}13^\backslash circ=\backslash mathbf\{37\{,\}87^\backslash circ\}$

geg.: ${\displaystyle \overline{\mathbf{M}{\mathbf{P}}_{\mathbf{n}}}(\mathbf{\phi })=\frac{\mathbf{5,53}}{\mathbf{s}\mathbf{i}\mathbf{n}(\mathbf{\phi }+{\mathbf{52,13}}^{\circ })}cm}\backslash displaystyle\backslash mathbf\{\backslash overline\; \{MP\_n\}(\phi )\}=\backslash mathbf\{\backslash frac\{5\{,\}53\}\{sin(\phi +52\{,\}13^\backslash circ)\}\}cm$, Identität $cos(90\mathrm{°}-\phi )cos(90°-\backslash varphi)$ ≙ $sin(\phi )sin(\backslash varphi\; )$

ges.:

1. Länge der Strecke $\overline{M{Q}_n}(\phi )\backslash overline\{MQ\_n\}(\phi )$

2. Volumen der Pyramide $V_{\text{Pyr}}(\phi ={60}^{\circ })V\_\{\backslash text\{Pyr\}\}(\backslash varphi=60^\backslash circ)$

Berechnung Länge der Strecke $\overline{M{Q}_n}(\phi )\backslash overline\{MQ\_n\}(\phi )$

Es gilt: ${\displaystyle \mathbf{c}\mathbf{o}\mathbf{s}({\mathbf{90}}^{\circ }-\mathbf{\phi })=\left[\frac{AK}{HY}\right]=\frac{\overline{\mathbf{M}{\mathbf{Q}}_{\mathbf{n}}}}{\overline{\mathbf{M}{\mathbf{P}}_{\mathbf{n}}}}}\backslash displaystyle\backslash mathbf\{cos(90^\backslash circ-\backslash varphi)\}=\backslash biggl[\backslash frac\; \{AK\}\{HY\}\backslash biggl\; ]=\backslash mathbf\{\backslash frac\{\backslash overline\; \{MQ\_n\}\}\{\backslash overline\; \{MP\_n\}\}\}\backslash qquad$ für $\phi \in ]0^{\circ };{90}^{\circ }]\backslash varphi\; \backslash in]0^\backslash circ;90^\backslash circ]$

Mit gegebenem $\overline{M{P}_n}(\phi )\backslash overline\{MP\_n\}(\phi )$ und Identität: $cos({90}^{\circ }-\alpha )=sin(\alpha )\backslash color\{blue\}cos(90^\backslash circ\; -\; \alpha )\; =\; sin(\alpha )$

$\Rightarrow \backslash Rightarrow$ ${\displaystyle \frac{\overline{M{Q}_n}}{\overline{M{P}_n}}=\sin (\phi )}\backslash displaystyle\backslash frac\{\backslash overline\; \{MQ\_n\}\}\{\backslash overline\; \{MP\_n\}\}=\backslash sin(\backslash varphi)\; \backslash qquad$

$\Rightarrow \overline{M{Q}_n}(\phi )=\overline{M{P}_n}\cdot \sin (\phi )\backslash Rightarrow\; \{\backslash overline\; \{MQ\_n\}(\phi )\}=\{\backslash overline\; \{MP\_n\}\}\backslash cdot\; \backslash sin(\backslash varphi)$

${\displaystyle \Rightarrow \overline{M{Q}_n}(\phi )=\frac{\mathrm{5,53}}{\sin (\phi +{\mathrm{52,13}}^{\circ })}\cdot \sin (\phi )}\backslash displaystyle\backslash Rightarrow\; \backslash overline\; \{MQ\_n\}(\phi )=\{\backslash frac\{5\{,\}53\}\{\backslash sin(\phi +52\{,\}13^\backslash circ)\}\}\backslash cdot\; \backslash sin(\phi )\backslash quad$

${\displaystyle \Rightarrow \overline{\mathbf{M}{\mathbf{Q}}_{\mathbf{n}}}(\mathbf{\phi })=\frac{\mathbf{5,53}\cdot \sin (\mathbf{\phi })}{\mathbf{s}\mathbf{i}\mathbf{n}(\mathbf{\phi }+{\mathbf{52,13}}^{\circ })}}\backslash displaystyle\backslash mathbf\{\backslash Rightarrow\; \backslash overline\; \{MQ\_n\}(\phi )=\{\backslash frac\{5\{,\}53\backslash cdot\; \backslash sin(\backslash varphi)\}\{sin(\phi +52\{,\}13^\backslash circ)\}\}\}\backslash qquad$q.e.d.

Volumen der Pyramide $V_{Pyr}(\phi =60\mathrm{°})V\_\{Pyr\}(\backslash varphi=60°)$ berechnen:

${\mathbf{V}}_{\mathbf{B}\mathbf{C}{\mathbf{Q}}_{\mathbf{1}}\mathbf{S}}=\frac{\mathbf{1}}{\mathbf{3}}\cdot \mathbf{G}\cdot \mathbf{h}\backslash mathbf\{V\_\{BCQ\_1S\}=\backslash frac\{1\}\{3\}\backslash cdot\; G\backslash cdot\; h\}$

Für $GG$ gilt: $G=\frac{1}{2}\cdot \overline{M{Q}_1}\cdot \overline{BC}[c{m}^2];\phi =60\mathrm{°}G=\backslash frac\{1\}\{2\}\backslash cdot\; \backslash overline\{MQ\_1\}\backslash cdot\; \backslash overline\; \{BC\}\backslash :\; [cm^2];\; \backslash quad\backslash varphi=60°$

Mit ${\displaystyle \overline{M{Q}_1}(60\mathrm{°})=\frac{\mathrm{5,53}\cdot \sin ({60}^{\circ })}{\sin ({60}^{\circ }+{\mathrm{52,13}}^{\circ })}=\frac{\mathrm{5,53}\cdot \mathrm{0,866}}{\mathrm{0,926}}=\frac{\mathrm{4,788}}{\mathrm{0,926}}=\mathrm{5,17}cm}\backslash displaystyle\; \backslash overline\; \{MQ\_1\}(60°)=\backslash frac\{5\{,\}53\backslash cdot\; \backslash sin\; (60^\backslash circ)\}\{\backslash sin(60^\backslash circ+52\{,\}13^\backslash circ)\}=\backslash frac\{5\{,\}53\backslash cdot\; 0\{,\}866\}\{0\{,\}926\}=\backslash frac\{4\{,\}788\}\{0\{,\}926\}=5\{,\}17\backslash :\; cm$,

mit $BC=10cmBC=10\backslash :cm$ und mit $MS\mathrm{\stackrel{\frown}{=}}h=7cmMS\stackrel{\wedge}{=}h=7\backslash :\; cm$

$\Rightarrow {\mathbf{V}}_{\mathbf{B}\mathbf{C}{\mathbf{Q}}_{\mathbf{1}}\mathbf{S}}=\frac{1}{3}\cdot \frac{1}{2}\cdot 10cm\cdot 7cm\cdot \mathrm{5,17}cm=\mathbf{60,32}\mathbf{c}{\mathbf{m}}^{\mathbf{3}}\backslash mathbf\{\backslash Rightarrow\; V\_\{BCQ\_1S\}\}=\backslash frac\{1\}\{3\}\backslash cdot\backslash frac\{1\}\{2\}\backslash cdot10\backslash :cm\backslash cdot\; 7\backslash :cm\backslash cdot\; 5\{,\}17\backslash :\; cm=\backslash mathbf\{60\{,\}32\backslash :\; cm^3\}$

geg.: $\overrightarrow{O{P}_n}(\phi )=\left(\begin{array}{c}5\cdot \sin \phi \\ 5\cdot \cos \phi \end{array}\right)\backslash overrightarrow\; \{OP\_n\}(\backslash varphi)=\; \backslash begin\{pmatrix\}\; 5\backslash cdot\; \backslash sin\backslash varphi\; \backslash \backslash \; 5\backslash cdot\; \backslash cos\backslash varphi\; \backslash end\{pmatrix\}$; $\overrightarrow{O{Q}_n}(\phi )=\left(\begin{array}{c}-2\cdot {\sin }^{\mathbf{2}}\mathbf{\phi }\\ {\displaystyle \frac{4}{\sin \phi }}\end{array}\right)\backslash overrightarrow\; \{OQ\_n\}(\backslash varphi)=\; \backslash begin\{pmatrix\}\; -2\backslash cdot\; \backslash mathbf\{\backslash sin^2\backslash varphi\}\; \backslash \backslash \; \backslash dfrac\{4\}\{\backslash sin\backslash varphi\}\; \backslash end\{pmatrix\}$; $\phi =80\mathrm{°}\backslash varphi=80°$

ges.: Koordinaten $\overrightarrow{O{P}_1}\backslash overrightarrow\; \{OP\_1\}$ und $\overrightarrow{O{Q}_1}\backslash overrightarrow\; \{OQ\_1\}$ für $\phi =80\mathrm{°}\backslash varphi=80°$; Zeichnung Dreieck $O{P}_1{Q}_{1}OP\_1Q\_1$

Berechnung der Koordinaten für $\overrightarrow{O{P}_1}(\phi =80\mathrm{°})\backslash overrightarrow\; \{OP\_1\}(\backslash varphi=80°)$:

$\overrightarrow{O{P}_1}(\phi =80\mathrm{°})=\left(\begin{array}{c}5\cdot \sin 80\mathrm{°}\\ 5\cdot \cos 80\mathrm{°}\end{array}\right)=\left(\begin{array}{c}5\cdot \mathrm{0,98481}\\ 5\cdot \mathrm{0,17365}\end{array}\right)=\left(\begin{array}{c}\mathbf{4,92}\\ \mathbf{0,87}\end{array}\right)\backslash overrightarrow\; \{OP\_1\}(\backslash varphi=80°)=\; \backslash begin\{pmatrix\}\; 5\backslash cdot\; \{\backslash sin80°\}\; \backslash \backslash \; 5\backslash cdot\; \{\backslash cos\; 80°\}\; \backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\; 5\backslash cdot\; \{0\{,\}98481\}\; \backslash \backslash \; 5\backslash cdot\; \{0\{,\}17365\}\; \backslash end\{pmatrix\}=\backslash mathbf\{\backslash begin\{pmatrix\}\; 4\{,\}92\; \backslash \backslash \; 0\{,\}87\; \backslash end\{pmatrix\}\}$

Berechnung der Koordinaten für $\overrightarrow{O{Q}_1}(\phi =80\mathrm{°})\backslash overrightarrow\; \{OQ\_1\}(\backslash varphi=80°)$

$\Rightarrow \overrightarrow{O{Q}_1}(\phi =80\mathrm{°})=\left(\begin{array}{c}-2\cdot {\sin }^2(80\mathrm{°})\\ {\displaystyle \frac{4}{\sin 80\mathrm{°}}}\end{array}\right)\backslash Rightarrow\; \backslash overrightarrow\; \{OQ\_1\}(\backslash varphi=80°)=\; \backslash begin\{pmatrix\}\; -2\backslash cdot\; \backslash sin^2(80°)\; \backslash \backslash \; \backslash dfrac\{4\}\{\backslash sin\; 80°\}\; \backslash end\{pmatrix\}$

$\phantom{\Rightarrow }=\left(\begin{array}{c}-2\cdot {\mathrm{0,98481}}^2\\ {\displaystyle \frac{4}{\mathrm{0,98481}}}\end{array}\right)=\left(\begin{array}{c}-\mathbf{1,94}\\ \mathbf{4,06}\end{array}\right)\backslash hphantom\{\backslash Rightarrow\}=\backslash begin\{pmatrix\}\; \{-2\backslash cdot\; 0\{,\}98481^2\}\; \backslash \backslash \; \backslash dfrac\{4\}\{0\{,\}98481\}\; \backslash end\{pmatrix\}=\backslash mathbf\; \{\backslash begin\{pmatrix\}\; \{-1\{,\}94\}\; \backslash \backslash \; 4\{,\}06\; \backslash end\{pmatrix\}\}$

Einzeichnen des Dreiecks $O{P}_1{Q}_{1}OP\_1Q\_1$ in das KOSY

geg.: Behauptung : Strecke $[O{P}_n]=const\mathrm{.}[OP\_n]=const.$

ges.: Begründung durch rechnerischen Nachweis

Berechnung der Strecken aus den Koordinaten

Mit dem Satz des Pythagoras gilt:

$\overrightarrow{O{P}_n}(\phi )=\left(\begin{array}{c}5\cdot \sin \phi \\ 5\cdot \cos \phi \end{array}\right)\backslash overrightarrow\; \{OP\_n\}(\backslash varphi)=\; \backslash begin\{pmatrix\}\; 5\backslash cdot\; \backslash sin\backslash varphi\; \backslash \backslash \; 5\backslash cdot\; \backslash cos\backslash varphi\; \backslash end\{pmatrix\}\backslash \backslash$$\overline{\mathbf{O}{\mathbf{P}}_{\mathbf{n}}}(\mathbf{\phi }{)}^{\mathbf{2}}=(\mathbf{5}\cdot \mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\phi }{)}^{\mathbf{2}}+(\mathbf{5}\cdot \mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\phi }{)}^{\mathbf{2}}\backslash mathbf\{\backslash overline\{OP\_n\}(\phi )^2=(5\backslash cdot\; sin\backslash varphi)^2+(5\backslash cdot\; cos\backslash varphi)^2\}$

${\overline{O{P}_n}}^2(\phi )=25\cdot (\sin \phi {)}^2+25\cdot (\cos \phi {)}^2\{\backslash overline\{OP\_n\}^2(\phi )=\{25\backslash cdot\; (\backslash sin\backslash varphi)^2+25\backslash cdot\; (\backslash cos\backslash varphi)^2\}\}$

$\phantom{{\overline{O{P}_n}}^2(\phi )}=25\cdot ({\sin }^2\phi +{\cos }^2\phi )\backslash hphantom\{\{\backslash overline\{OP\_n\}^2(\phi )\}\}=25\; \backslash cdot(\backslash sin^2\; \backslash varphi+\backslash cos^2\; \backslash varphi)$

$\phantom{{\overline{O{P}_n}}^2(\phi )}=25\backslash hphantom\{\{\backslash overline\{OP\_n\}^2(\phi )\}\}=25$ (trigonometrischer Pythagoras)

Daher ist:

$\overline{\mathbf{O}{\mathbf{P}}_{\mathbf{n}}}(\mathbf{\phi })=\sqrt{\mathbf{25}}\mathbf{L}\mathbf{E}=\mathbf{5}\mathbf{L}\mathbf{E}\backslash mathbf\{\; \backslash overline\{OP\_n\}(\phi )=\backslash sqrt\{25\}\backslash \; LE\; =5\backslash \; LE\}$ für alle $\phi \in ]0\mathrm{°};90\mathrm{°}]\backslash varphi\; \backslash in]0°;90°]\backslash qquad$q.e.d.

geg.: Vektoren $\overline{O{P}_1}(80\mathrm{°})=\left(\begin{array}{c}\mathrm{4,92}\\ \mathrm{0,87}\end{array}\right)\backslash overline\{OP\_1\}(80°)=\backslash begin\{pmatrix\}\; 4\{,\}92\; \backslash \backslash \; 0\{,\}87\; \backslash end\{pmatrix\}$ und $\overline{O{Q}_1}(80\mathrm{°})=\left(\begin{array}{c}-\mathrm{1,94}\\ \mathrm{4,06}\end{array}\right)\backslash overline\{OQ\_1\}(80°)=\backslash begin\{pmatrix\}\; -1\{,\}94\; \backslash \backslash \; 4\{,\}06\; \backslash end\{pmatrix\}$ gem. Aufgabe 3a

ges.: $\mathrm{\measuredangle }{P}_{1}O{Q}_1\measuredangle \; P\_1OQ\_1$

Berechnung mit Kosinussatz aufgelöst nach Kosinus $\gamma \gamma$

$c^2=a^2+b^2-2ab\cdot \cos (\gamma )c^2=a^2+b^2-2ab\backslash cdot\; \backslash cos(\gamma )\backslash qquad\backslash qquad$|$-c^2-c^2$

$0=c^2+a^2+b^2-2ab\cdot \cos (\gamma )0=c^2+a^2+b^2-2ab\backslash cdot\; \backslash cos(\gamma )\backslash qquad\backslash :\backslash :$|$+2ab\cdot \cos (\gamma )+\; 2ab\backslash cdot\; \backslash cos(\gamma )$

$2ab\cdot \cos (\gamma )=-c^2+a^2+b^{2}2ab\backslash cdot\; \backslash cos(\gamma )=-c^2+a^2+b^2\backslash qquad\backslash quad$ | $:2ab:\; 2ab$

${\displaystyle \cos (\gamma )=\frac{-c^2+a^2+b^2}{2ab}}\backslash displaystyle\backslash cos(\gamma )=\backslash frac\{-c^2+a^2+b^2\}\{2ab\}$

Berechnung von $a,b,ca,\; b,\; c$ und $cos(\gamma )cos(\gamma )$

$\Rightarrow a:\mathrm{\stackrel{\frown}{=}}\mathrm{\mid }\overline{O{P}_1}\mathrm{\mid }=5\backslash Rightarrow\; a:\; \stackrel{\wedge}{=}\; |\backslash overline\; \{OP\_1\}|=5\backslash :\backslash qquad$für alle $\phi \backslash varphi$ nach Teil b)

$\Rightarrow b:\mathrm{\stackrel{\frown}{=}}\mathrm{\mid }\overline{O{Q}_1}\mathrm{\mid }=\sqrt{-{\mathrm{1,94}}^2+{\mathrm{4,06}}^2}=\sqrt{\mathrm{3,76}+\mathrm{16,48}}=\sqrt{\mathrm{20,24}}=\mathrm{4,50}\backslash Rightarrow\; b:\; \stackrel{\wedge}{=}\; |\backslash overline\; \{OQ\_1\}|=\backslash sqrt\{-1\{,\}94^2+4\{,\}06^2\}=\backslash sqrt\{3\{,\}76+16\{,\}48\}=\backslash sqrt\{20\{,\}24\}=4\{,\}50$

Berechnung Vektor $\overrightarrow{P_1{Q}_1}\backslash overrightarrow\; \{P\_1Q\_1\}$ mit Vektoraddition $\overrightarrow{O{P}_1}\oplus \overrightarrow{P_1{Q}_1}=\overrightarrow{O{Q}_1}\backslash overrightarrow\{OP\_1\}\backslash oplus\backslash overrightarrow\{P\_1Q\_1\}=\backslash overrightarrow\; \{OQ\_1\}$

umgestellt nach $\overrightarrow{P_1{Q}_1}=\overrightarrow{O{Q}_1}-\overrightarrow{O{P}_1}=\left(\begin{array}{c}-\mathrm{1,94}\\ \mathrm{4,06}\end{array}\right)-\left(\begin{array}{c}\mathrm{4,92}\\ \mathrm{0,87}\end{array}\right)=\left(\begin{array}{c}-\mathrm{6,86}\\ \mathrm{3,19}\end{array}\right)\backslash overrightarrow\{P\_1Q\_1\}=\backslash overrightarrow\; \{OQ\_1\}-\backslash overrightarrow\{OP\_1\}=\backslash begin\{pmatrix\}\; -1\{,\}94\; \backslash \backslash \; 4\{,\}06\; \backslash end\{pmatrix\}-\backslash begin\{pmatrix\}\; 4\{,\}92\; \backslash \backslash \; 0\{,\}87\; \backslash end\{pmatrix\}=\backslash begin\{pmatrix\}\; -6\{,\}86\; \backslash \backslash \; 3\{,\}19\; \backslash end\{pmatrix\}$

$c^2=\overrightarrow{\mathbf{\mid }{\mathbf{P}}_{\mathbf{1}}{\mathbf{Q}}_{\mathbf{1}}{\mathbf{\mid }}^{\mathbf{2}}}=(-\mathrm{6,86}{)}^2+{\mathrm{3,22}}^2=\mathrm{47,06}+\mathrm{10,18}=\mathrm{57,24}c^2=\; \backslash mathbf\{\backslash overrightarrow\{|P\_1Q\_1|^2\}\}=\{(-6\{,\}86)^2+3\{,\}22^2\}=\{47\{,\}06+10\{,\}18\}=\{57\{,\}24\}$

Die Werte von $a^{2}a^2$ und $b^{2}b^2$ sind schon oben berechnet worden.

${\displaystyle \Rightarrow \cos (\mathbf{\measuredangle }{\mathbf{P}}_{\mathbf{1}}\mathbf{O}{\mathbf{Q}}_{\mathbf{1}})=\frac{-c^2+a^2+b^2}{2ab}}\backslash displaystyle\backslash Rightarrow\; \backslash mathbf\{\backslash cos(\measuredangle P\_1\; OQ\_1\; )\}=\backslash frac\{-c^2+a^2+b^2\}\{2ab\}$

${\displaystyle =\frac{-\mathrm{57,46}+25+\mathrm{20,24}}{2\cdot 5\cdot \mathrm{4,50}}=\frac{-\mathrm{12,22}}{45}=-\mathbf{0,27}}\backslash displaystyle=\backslash frac\{-57\{,\}46+25+20\{,\}24\}\{2\backslash cdot5\backslash cdot4\{,\}50\}\; =\backslash frac\{-12\{,\}22\}\{45\}=\backslash mathbf\{-0\{,\}27\}$

Falls $\cos (\mathrm{\measuredangle }{P}_{1}O{Q}_1)\backslash cos(\measuredangle P\_1OQ\_1\; )$ negativ, so ist $\mathrm{\measuredangle }{P}_{1}O{Q}_1\measuredangle P\_1OQ\_1$ zwischen $90\mathrm{°}90°$ und $180\mathrm{°}180°$ groß.

Kosinuswert $\mathbf{\measuredangle }{\mathbf{P}}_{\mathbf{1}}\mathbf{O}{\mathbf{Q}}_{\mathbf{1}}=-\mathbf{0,27}\Rightarrow \mathbf{\measuredangle }{\mathbf{P}}_{\mathbf{1}}\mathbf{O}{\mathbf{Q}}_{\mathbf{1}}={\mathbf{105,76}}^{\circ }\backslash mathbf\{\measuredangle P\_1OQ\_1=-0\{,\}27\; \backslash qquad\; \backslash Rightarrow\; \measuredangle P\_1\; OQ\_1=105\{,\}76^\backslash circ\}$