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:

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