Lösung

$\overline{AC}=\backslash \; \backslash overline\{AC\}\backslash \; =\backslash$Körpergröße von Heike$=\mathrm{1,69}m\backslash \; =\backslash \; 1\{,\}69\backslash \; m$

$\overline{AB}=\backslash \; \backslash overline\{AB\}\backslash \; =\backslash$Länge von Heikes Schatten

Im obigen Dreieck gilt:$\text{ tan}({30}^{\circ })={\displaystyle \frac{\overline{AC}}{\overline{AB}}}\backslash \; \backslash text\{tan\}(30^\{\backslash circ\})\backslash \; =\backslash \; \backslash dfrac\{\backslash overline\{AC\}\}\{\backslash overline\{AB\}\}$

$\text{ tan}({30}^{\circ })={\displaystyle \frac{\mathrm{1,69}m}{\overline{AB}}}\Rightarrow \overline{AB}={\displaystyle \frac{\mathrm{1,69}m}{\text{tan}({30}^{\circ })}}\backslash \; \backslash text\{tan\}(30^\{\backslash circ\})\backslash \; =\backslash \; \backslash dfrac\{1\{,\}69\backslash \; m\}\{\backslash overline\{AB\}\}\backslash \; \backslash Rightarrow\backslash \; \backslash overline\{AB\}\backslash \; =\backslash \; \backslash dfrac\{1\{,\}69\backslash \; m\}\{\backslash text\{tan\}(30^\{\backslash circ\})\}$

$\overline{AB}\approx \frac{\mathrm{1,69}m}{0.5774}\backslash \; \backslash overline\{AB\}\backslash \; \backslash approx\backslash \; \backslash frac\{1\{,\}69\backslash \; m\}\{0.5774\}$

$\overline{AB}\approx \mathrm{2,93}m\backslash \; \backslash overline\{AB\}\backslash \; \backslash approx\backslash \; 2\{,\}93\backslash \; m$