Heft 2 - B1

Berechnen des Winkels $\beta \backslash beta$ mit Hilfe des Tangens.

$\mathrm{t}\mathrm{a}\mathrm{n}(\beta )={\displaystyle \frac{\mathrm{G}\mathrm{e}\mathrm{g}\mathrm{e}\mathrm{n}\mathrm{k}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{t}\mathrm{e}}{\mathrm{A}\mathrm{n}\mathrm{k}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{t}\mathrm{e}}}\backslash mathrm\{tan(\backslash beta)=\backslash dfrac\{Gegenkathete\}\{Ankathete\}\}$

Im Dreieck $\text{ABC }\backslash text\{ABC\}\backslash$gilt: $\mathrm{G}\mathrm{e}\mathrm{g}\mathrm{e}\mathrm{n}\mathrm{k}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{t}\mathrm{e}=\mathrm{b}=5\text{ cm}\backslash \; \backslash mathrm\{Gegenkathete=b=5\backslash cm\}$

$\mathrm{A}\mathrm{n}\mathrm{k}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{t}\mathrm{e}=\mathrm{a}=\mathrm{7,5}\text{ cm}\backslash hspace\{35mm\}\backslash mathrm\{Ankathete=a=7\{,\}5\backslash cm\}$

Setze die Werte in die Formel ein und berechne den Winkel $\beta \backslash beta$.

$\mathrm{t}\mathrm{a}\mathrm{n}(\beta )={\displaystyle \frac{5}{\mathrm{7,5}}}\Rightarrow \beta \approx {\mathrm{33,7}}^{\circ }\backslash mathrm\{tan(\backslash beta)=\backslash dfrac\{5\}\{7\{,\}5\}\}\backslash \; \backslash Rightarrow\backslash beta\backslash approx33\{,\}7^\backslash circ$

Der Winkel $\beta \backslash beta$ beträgt ungefähr $\boxed{{\displaystyle {\mathrm{33,7}}^{\circ }}}\backslash \; \backslash boxed\{33\{,\}7^\backslash circ\}$

Berechnen der Länge der Strecke $\text{w}\backslash text\{w\}$ mit Hilfe des Kosinus.

$\mathrm{c}\mathrm{o}\mathrm{s}\left({\displaystyle \frac{\beta }{2}}\right)={\displaystyle \frac{\mathrm{A}\mathrm{n}\mathrm{k}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{t}\mathrm{e}}{\mathrm{H}\mathrm{y}\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{u}\mathrm{s}\mathrm{e}}}\backslash mathrm\{cos\backslash left(\backslash dfrac\{\backslash beta\}\{2\}\backslash right)=\backslash dfrac\{Ankathete\}\{Hypotenuse\}\}$

Im Dreieck$\text{ DBC }\backslash \; \backslash text\{DBC\}\backslash$gilt: $\mathrm{A}\mathrm{n}\mathrm{k}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{t}\mathrm{e}=\mathrm{a}=\mathrm{7,5}\text{ cm}\backslash \; \backslash mathrm\{Ankathete=a=7\{,\}5\backslash cm\}$

$\mathrm{H}\mathrm{y}\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{u}\mathrm{s}\mathrm{e}=\mathrm{w}\backslash hspace\{29mm\}\backslash mathrm\{Hypotenuse=w\}$

$\left({\displaystyle \frac{\beta }{2}}\right)={\mathrm{16,85}}^{\circ }\backslash hspace\{38mm\}\backslash left(\backslash dfrac\{\backslash beta\}\{2\}\backslash right)=16\{,\}85^\backslash circ$

Löse die Formel nach der Hypotenuse $(\text{w})(\backslash text\{w\})$ auf und berechne die Länge der Seite $\text{w}\backslash text\{w\}$.

$\mathrm{w}={\displaystyle \frac{\mathrm{A}\mathrm{n}\mathrm{k}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{t}\mathrm{e}}{(\cos {\displaystyle \frac{\beta }{2}})}}\Rightarrow \mathrm{w}={\displaystyle \frac{\mathrm{7,5}}{\cos {\mathrm{16,85}}^{\circ }}}[\mathrm{c}\mathrm{m}]\backslash mathrm\{w=\backslash dfrac\{Ankathete\}\{\backslash left(\backslash cos\backslash dfrac\{\backslash beta\}\{2\}\backslash right)\}\backslash Rightarrow\backslash \; w=\backslash dfrac\{7\{,\}5\}\{\backslash cos16\{,\}85^\backslash circ\}\backslash \; [cm]\}$

$\mathrm{w}\approx \mathrm{7,84}\text{ cm}\backslash mathrm\{w\backslash approx7\{,\}84\backslash cm\}$

Die Länge der Seite $\text{w}\backslash text\{w\}$ beträgt ungefähr $\boxed{{\displaystyle \mathrm{7,84}\text{ cm}}}\backslash \; \backslash boxed\{7\{,\}84\backslash cm\}$

Berechnen der Seite$\overline{\mathrm{D}\mathrm{C}}\backslash \; \backslash mathrm\{\backslash overline\{DC\}\}$ mit Hilfe des Tangens.

$\mathrm{t}\mathrm{a}\mathrm{n}(\beta )={\displaystyle \frac{\mathrm{G}\mathrm{e}\mathrm{g}\mathrm{e}\mathrm{n}\mathrm{k}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{t}\mathrm{e}}{\mathrm{A}\mathrm{n}\mathrm{k}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{t}\mathrm{e}}}\backslash mathrm\{tan(\backslash beta)=\backslash dfrac\{Gegenkathete\}\{Ankathete\}\}$

Im Dreieck $\text{DBC }\backslash text\{DBC\}\backslash$gilt: $\mathrm{G}\mathrm{e}\mathrm{g}\mathrm{e}\mathrm{n}\mathrm{k}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{t}\mathrm{e}=\overline{\mathrm{D}\mathrm{C}}\backslash \; \backslash mathrm\{Gegenkathete=\backslash overline\{DC\}\}$

$\mathrm{A}\mathrm{n}\mathrm{k}\mathrm{a}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{t}\mathrm{e}=\mathrm{a}=\mathrm{7,5}\text{ cm}\backslash hspace\{36mm\}\backslash mathrm\{Ankathete=a=7\{,\}5\backslash cm\}$

Löse die Formel nach der Gegenkathete $(\overline{\mathrm{D}\mathrm{C}})(\backslash mathrm\{\backslash overline\{DC\}\})$ auf und berechne die Länge der Seite $\overline{\mathrm{D}\mathrm{C}}\backslash mathrm\{\backslash overline\{DC\}\}$.

$\overline{\mathrm{D}\mathrm{C}}=\mathrm{a}\cdot \tan \left({\displaystyle \frac{\beta }{2}}\right)\Rightarrow \overline{\mathrm{D}\mathrm{C}}=\mathrm{7,5}\cdot \mathrm{0,3029}[\mathrm{c}\mathrm{m}]\backslash mathrm\{\backslash overline\{DC\}=a\backslash cdot\; \backslash tan\backslash left(\{\backslash dfrac\{\backslash beta\}\{2\}\}\backslash right)\backslash Rightarrow\backslash overline\{DC\}=7\{,\}5\backslash cdot0\{,\}3029\backslash \; \backslash \; [cm]\}$

$\overline{\mathrm{D}\mathrm{C}}\approx \mathrm{2,27}\text{ cm}\backslash mathrm\{\backslash overline\{DC\}\backslash approx2\{,\}27\}\backslash cm$

${\displaystyle \frac{\overline{\mathrm{A}\mathrm{C}}}{2}}=\mathrm{2,5}\text{ cm}\backslash mathrm\{\backslash dfrac\{\backslash overline\{AC\}\}\{2\}\}=2\{,\}5\backslash cm$

Die Strecke$\overline{\mathrm{D}\mathrm{C}}\backslash \; \backslash mathrm\{\backslash overline\{DC\}\}\backslash$ist nicht halb so lang wie die Strecke $\overline{\mathrm{A}\mathrm{C}}\backslash mathrm\{\backslash overline\{AC\}\}$.

Berechnen der Höhe $\text{h}\backslash text\{h\}$

$\cos (\alpha )=\frac{\text{Ankathete}}{\text{Hypotenuse}}\backslash cos(\backslash alpha)=\backslash frac\{\backslash text\{Ankathete\}\}\{\backslash text\{Hypotenuse\}\}$

$\mathrm{c}\mathrm{o}\mathrm{s}({50}^{\circ })={\displaystyle \frac{\left({\displaystyle \frac{\mathrm{a}}{2}}\right)}{\mathrm{h}}}\Rightarrow \mathrm{h}={\displaystyle \frac{115}{\mathrm{c}\mathrm{o}\mathrm{s}({50}^{\circ })}}\backslash mathrm\{cos(50^\backslash circ)=\backslash dfrac\{\backslash left(\backslash dfrac\{a\}\{2\}\backslash right)\}\{h\}\backslash \; \backslash Rightarrow\; h=\backslash dfrac\{115\}\{cos(50^\backslash circ)\}\}$

$\text{ h}={\displaystyle \frac{115}{\mathrm{0,6428}}}\backslash h=\backslash dfrac\{115\}\{0\{,\}6428\}$

$\text{ h}\approx 179\text{ m}\backslash h\backslash approx179\backslash m$

Berechnen der Mantelfläche der Pyramide.

Berechnen der Fläche eines Dreiecks.

${\mathrm{A}}_{\mathrm{\Delta }}={\displaystyle \frac{\mathrm{a}\cdot \mathrm{h}}{2}}\Rightarrow {\mathrm{A}}_{\mathrm{\Delta }}={\displaystyle \frac{230\cdot 179}{2}}[{\mathrm{m}}^2]\backslash mathrm\{A\_\{\backslash Delta\}=\backslash dfrac\{a\backslash cdot\; h\}\{2\}\backslash Rightarrow\; A\_\{\backslash Delta\}=\backslash dfrac\{230\backslash cdot\; 179\}\{2\}\backslash \; [m^2]\}$

${\mathrm{A}}_{\mathrm{\Delta }}=20585{\text{ m}}^2\backslash mathrm\{A\_\{\backslash Delta\}=20585\backslash m^2\}$

Die Mantelfläche $\text{M}\backslash text\{M\}$ der Pyramide beträgt dann:

$\mathrm{M}=4\cdot {\mathrm{A}}_{\mathrm{\Delta }}\Rightarrow \mathrm{M}=4\cdot 20585{\mathrm{m}}^2\backslash mathrm\{M=4\backslash cdot\; A\_\{\backslash Delta\}\backslash \; \backslash Rightarrow\backslash \; M=4\backslash cdot\; 20585\backslash \; m^2\}$

$\boxed{{\displaystyle \mathrm{M}=82340{\text{ m}}^2}}\backslash boxed\{\backslash mathrm\{M=82340\backslash m^2\}\}$

Berechnen der Höhe$\text{ k }\backslash \; \backslash text\{k\}\backslash$für den Winkel${50}^{\circ }\backslash \; 50^\backslash circ\backslash$

$\mathrm{t}\mathrm{a}\mathrm{n}({50}^{\circ })={\displaystyle \frac{{\mathrm{k}}_{50}}{\left({\displaystyle \frac{\mathrm{a}}{2}}\right)}}\backslash mathrm\{tan(50^\backslash circ)=\backslash dfrac\{k\_\{50\}\}\{\backslash left(\backslash dfrac\{a\}\{2\}\backslash right)\}\}$

$\mathrm{t}\mathrm{a}\mathrm{n}({50}^{\circ })={\displaystyle \frac{{\mathrm{k}}_{50}}{115}}\Rightarrow {\mathrm{k}}_{50}=\mathrm{t}\mathrm{a}\mathrm{n}({50}^{\circ })\cdot 115\mathrm{m}\backslash mathrm\{tan(50^\backslash circ)=\backslash dfrac\{k\_\{50\}\}\{115\}\backslash \; \backslash Rightarrow\backslash \; k\_\{50\}=tan(50^\backslash circ)\backslash cdot\; 115\backslash \; m\}$

${\mathrm{k}}_{50}\approx \mathrm{137,1}\text{ m}\backslash mathrm\{k\_\{50\}\backslash approx137\{,\}1\backslash m\}$

Berechnen der Höhe$\text{ k }\backslash \; \backslash text\{k\}\backslash$für den Winkel${10}^{\circ }\backslash \; 10^\backslash circ\backslash$

$\mathrm{t}\mathrm{a}\mathrm{n}({10}^{\circ })={\displaystyle \frac{{\mathrm{k}}_{10}}{\left({\displaystyle \frac{\mathrm{a}}{2}}\right)}}\backslash mathrm\{tan(10^\backslash circ)=\backslash dfrac\{k\_\{10\}\}\{\backslash left(\backslash dfrac\{a\}\{2\}\backslash right)\}\}$

$\mathrm{t}\mathrm{a}\mathrm{n}({10}^{\circ })={\displaystyle \frac{{\mathrm{k}}_{10}}{115}}\Rightarrow {\mathrm{k}}_{10}=\mathrm{t}\mathrm{a}\mathrm{n}({10}^{\circ })\cdot 115\mathrm{m}\backslash mathrm\{tan(10^\backslash circ)=\backslash dfrac\{k\_\{10\}\}\{115\}\backslash \; \backslash Rightarrow\backslash \; k\_\{10\}=tan(10^\backslash circ)\backslash cdot\; 115\backslash \; m\}$

${\mathrm{k}}_{10}\approx \mathrm{20,3}\text{ m}\backslash mathrm\{k\_\{10\}\backslash approx20\{,\}3\backslash m\}$

Die Aussage stimmt nicht, da:

$\mathrm{137,1}>\mathrm{101,5}(5\cdot \mathrm{20,3})[\mathrm{m}]\backslash mathrm\{137\{,\}1>101\{,\}5\backslash \; (5\backslash cdot\; 20\{,\}3)\backslash \; [m]\}$