$\alpha \backslash alpha$ berechnen

$\beta \backslash beta$ berechnen

$\beta \backslash beta$ kannst du ausrechnen, indem du alle gegebenen Winkel von der Gesamtsumme aller Winkel in einem Dreieck $\left({180}^{\circ }\right)\backslash left(180^\backslash circ\backslash right)$ abziehst.

$\beta ={180}^{\circ }-{90}^{\circ }-{\mathrm{30,7}}^{\circ }\backslash beta=180^\backslash circ-90^\backslash circ-30\{,\}7^\backslash circ$

$\beta ={\mathrm{59,3}}^{\circ }\backslash beta=59\{,\}3^\backslash circ$

$bb$ berechnen

$bb$ mithilfe des Satzes des Pythagoras ausrechnen.

${\left(\mathrm{24,9}\mathrm{c}\mathrm{m}\right)}^2={\left(\mathrm{12,7}\mathrm{c}\mathrm{m}\right)}^2+b^2\backslash left(24\{,\}9\backslash ,\backslash mathrm\{cm\}\backslash right)^2=\backslash left(12\{,\}7\backslash ,\backslash mathrm\{cm\}\backslash right)^2+b^2$

$\mathrm{\mid }-{\left(\mathrm{12,7}\mathrm{c}\mathrm{m}\right)}^2|\{\}-\backslash left(12\{,\}7\backslash ,\backslash mathrm\{cm\}\backslash right)^2$

$b^2={\left(\mathrm{24,9}\mathrm{c}\mathrm{m}\right)}^2-{\left(\mathrm{12,7}\mathrm{c}\mathrm{m}\right)}^{2}b^2=\backslash left(24\{,\}9\backslash ,\backslash mathrm\{cm\}\backslash right)^2-\backslash left(12\{,\}7\backslash ,\backslash mathrm\{cm\}\backslash right)^2$

$b^2=\mathrm{458,72}{\mathrm{c}\mathrm{m}}^{2}b^2=458\{,\}72\backslash ,\backslash mathrm\{cm\}^2$

Wurzel ziehen.

$b\approx \mathrm{21,4}\mathrm{c}\mathrm{m}b\backslash approx21\{,\}4\backslash ,\backslash mathrm\{cm\}$

$\beta \backslash beta$ berechnen

$\beta ={\mathrm{40,6}}^{\circ }\backslash beta=40\{,\}6^\backslash circ$

$\gamma \backslash gamma$ berechnen

$\gamma \backslash gamma$ kannst du ausrechnen, indem du alle gegebenen Winkel von der Gesamtsumme aller Winkel in einem Dreieck $\left({180}^{\circ }\right)\backslash left(180^\backslash circ\backslash right)$ abziehst.

$\gamma ={180}^{\circ }-{90}^{\circ }-{\mathrm{40,6}}^{\circ }={\mathrm{49,4}}^{\circ }\backslash gamma=180^\backslash circ-90^\backslash circ-40\{,\}6^\backslash circ=49\{,\}4^\backslash circ$

$cc$ berechnen

$cc$ mithilfe des Satzes des Pythagoras ausrechnen.

Beachte hier insbesondere, dass $cc$ nicht die Hypotenuse des Dreiecks ist, sondern $aa$ (siehe Bild oben), sodass die Form des Pythagoras zwar ähnlich bleibt, aber nun $aa$, $bb$ und $cc$ nicht die bekannten Rollen, also $aa$,$bb$ die Kathethen und $cc$ die Hypotenuse sind, einnehmen.

$(645\mathrm{m}{)}^2=(420\mathrm{m}{)}^2+c^2(645\backslash ,\backslash mathrm\; m)^2=(420\backslash ,\backslash mathrm\; m)^2+c^2$

$c^2=(645\mathrm{m}{)}^2-(420\mathrm{m}{)}^{2}c^2=\; (645\backslash ,\backslash mathrm\; m)^2\; -\; (420\backslash ,\backslash mathrm\; m)^2$

$c^2=239625{\mathrm{m}}^{2}c^2=239\backslash ,625\backslash ,\backslash mathrm\; m^2$

Wurzel ziehen.

$c\approx 490\mathrm{m}c\backslash approx490\backslash ,\backslash mathrm\; m$

$bb$ berechnen

$bb$ mithilfe des Satzes des Pythagoras ausrechnen.

$b^2={\left(\mathrm{30,7}\mathrm{c}\mathrm{m}\right)}^2+{\left(\mathrm{15,8}\mathrm{c}\mathrm{m}\right)}^{2}b^2=\backslash left(30\{,\}7\backslash ,\backslash mathrm\{cm\}\backslash right)^2+\backslash left(15\{,\}8\backslash ,\backslash mathrm\{cm\}\backslash right)^2$

$b^2=\mathrm{1192,13}{\mathrm{c}\mathrm{m}}^2\mathrm{\mid }\sqrt{}b^2=1192\{,\}13\backslash ,\backslash mathrm\{cm\}^2\; \backslash ;\backslash ;\backslash ;\backslash ;|\backslash sqrt\{\}$

$b\approx \mathrm{34,5}\mathrm{c}\mathrm{m}b\backslash approx34\{,\}5\backslash ,\backslash mathrm\{cm\}$

$\alpha \backslash alpha$ berechnen

$\tan \left(\alpha \right)=\frac{a}{c}\backslash tan\backslash left(\backslash alpha\backslash right)=\backslash frac\{a\}\{c\}$

$\tan \left(\alpha \right)={\displaystyle \frac{\mathrm{30,7}\mathrm{c}\mathrm{m}}{\mathrm{15,8}\mathrm{c}\mathrm{m}}}\backslash tan\backslash left(\backslash alpha\backslash right)=\backslash dfrac\{30\{,\}7\backslash ,\backslash mathrm\{cm\}\}\{15\{,\}8\backslash ,\backslash mathrm\{cm\}\}$

$\alpha \approx {\mathrm{62,7}}^{\circ }\backslash alpha\backslash approx62\{,\}7^\backslash circ$

$\gamma \backslash gamma$ berechnen

$\gamma \backslash gamma$ kannst du ausrechnen, indem du alle gegebenen Winkel von der Gesamtsumme aller Winkel in einem Dreieck $\left({180}^{\circ }\right)\backslash left(180^\backslash circ\backslash right)$ abziehst.

$\gamma \approx {180}^{\circ }-{90}^{\circ }-{\mathrm{62,7}}^{\circ }\backslash gamma\backslash approx\; 180^\backslash circ-90^\backslash circ-62\{,\}7^\backslash circ$

$\gamma \approx {\mathrm{27,3}}^{\circ }\backslash gamma\; \backslash approx\; 27\{,\}3^\backslash circ$

$\beta \backslash beta$ berechnen

$\beta \backslash beta$ kannst du ausrechnen, indem du alle gegebenen Winkel von der Gesamtsumme aller Winkel in einem Dreieck $\left({180}^{\circ }\right)\backslash left(180^\backslash circ\backslash right)$ abziehst.

$\beta ={180}^{\circ }-{90}^{\circ }-{35}^{\circ }\backslash beta=180^\backslash circ-90^\backslash circ-35^\backslash circ$

$\beta ={55}^{\circ }\backslash beta=55^\backslash circ$

$aa$ berechnen

$\sin \left({35}^{\circ }\right)=\frac{a}{\mathrm{12,5}\mathrm{c}\mathrm{m}}\mathrm{\mid }\cdot \mathrm{12,5}\mathrm{c}\mathrm{m}\backslash sin\backslash left(35^\backslash circ\backslash right)=\backslash frac\{a\}\{12\{,\}5\backslash ,\backslash mathrm\{cm\}\}\; \backslash :\backslash :\backslash ;\backslash ;\backslash ;\backslash ;|\{\}\backslash cdot12\{,\}5\backslash ,\backslash mathrm\{cm\}$

$a=\sin \left({35}^{\circ }\right)\cdot \mathrm{12,5}\mathrm{c}\mathrm{m}a=\backslash sin\backslash left(35^\backslash circ\backslash right)\backslash cdot12\{,\}5\backslash ,\backslash mathrm\{cm\}$

$a=\mathrm{7,2}\mathrm{c}\mathrm{m}a=7\{,\}2\backslash ,\backslash mathrm\{cm\}$

$bb$ berechnen

$bb$ mithilfe des Satzes des Pythagoras ausrechnen.

$c^2=a^2+b^{2}c^2=a^2+b^2$

${\left(\mathrm{12,5}\mathrm{c}\mathrm{m}\right)}^2={\left(\mathrm{7,2}\mathrm{c}\mathrm{m}\right)}^2+b^2\mathrm{\mid }-{\left(\mathrm{7,2}\mathrm{c}\mathrm{m}\right)}^2\backslash left(12\{,\}5\backslash ,\backslash mathrm\{cm\}\backslash right)^2=\backslash left(7\{,\}2\backslash ,\backslash mathrm\{cm\}\backslash right)^2+b^2\; \backslash :\backslash :\backslash ;\backslash ;\backslash ;\backslash ;|\{\}-\backslash left(7\{,\}2\backslash ,\backslash mathrm\{cm\}\backslash right)^2$

$b^2={\left(\mathrm{12,5}\mathrm{c}\mathrm{m}\right)}^2-{\left(\mathrm{7,2}\mathrm{c}\mathrm{m}\right)}^{2}b^2=\backslash left(12\{,\}5\backslash ,\backslash mathrm\{cm\}\backslash right)^2-\backslash left(7\{,\}2\backslash ,\backslash mathrm\{cm\}\backslash right)^2$

$b^2\approx \mathrm{104,4}{\mathrm{c}\mathrm{m}}^{2}b^2\backslash approx104\{,\}4\backslash ,\backslash mathrm\{cm\}^2$

Wurzel ziehen.

$b\approx \mathrm{10,2}\mathrm{c}\mathrm{m}b\backslash approx10\{,\}2\backslash ,\backslash mathrm\{cm\}$

$bb$ berechnen (alternative Lösung mit dem Kosinus)

$\cos \left(\alpha \right)=\frac{b}{c}\backslash cos\backslash left(\backslash alpha\backslash right)=\backslash frac\; bc$

Nach $bb$ umstellen.

$b=\cos \left(\alpha \right)\cdot c\approx \mathrm{10,2}\mathrm{c}\mathrm{m}b\; =\; \backslash cos\backslash left(\; \backslash alpha\; \backslash right)\; \backslash cdot\; c\; \backslash approx\; 10\{,\}2\backslash ,\backslash mathrm\{cm\}$

$\beta \backslash beta$ berechnen

$\beta \backslash beta$ kannst du ausrechnen, indem du alle gegebenen Winkel von der Gesamtsumme aller Winkel in einem Dreieck $\left({180}^{\circ }\right)\backslash left(180^\backslash circ\backslash right)$ abziehst.

$\beta ={180}^{\circ }-{90}^{\circ }-{\mathrm{40,3}}^{\circ }\backslash beta=180^\backslash circ-90^\backslash circ-40\{,\}3^\backslash circ$

$\beta ={\mathrm{49,7}}^{\circ }\backslash beta=49\{,\}7^\backslash circ$

$bb$ berechnen

$\sin \left({\mathrm{49,7}}^{\circ }\right)=\frac{b}{\mathrm{10,5}\mathrm{c}\mathrm{m}}\mathrm{\mid }\cdot \mathrm{10,5}\mathrm{c}\mathrm{m}\backslash sin\backslash left(49\{,\}7^\backslash circ\backslash right)=\backslash frac\{b\}\{10\{,\}5\backslash ,\backslash mathrm\{cm\}\}\backslash :\backslash :\backslash ;\backslash ;\backslash ;\backslash ;\; |\{\}\backslash cdot10\{,\}5\backslash ,\backslash mathrm\{cm\}$

$b=\sin \left({\mathrm{49,7}}^{\circ }\right)\cdot \mathrm{10,5}\mathrm{c}\mathrm{m}b=\backslash sin\backslash left(49\{,\}7^\backslash circ\backslash right)\backslash cdot10\{,\}5\backslash ,\backslash mathrm\{cm\}$

$b\approx 8\mathrm{c}\mathrm{m}b\backslash approx8\backslash ,\backslash mathrm\{cm\}$

$cc$ berechnen

$cc$ mithilfe des Satzes des Pythagoras ausrechnen.

${\left(\mathrm{10,5}\mathrm{c}\mathrm{m}\right)}^2=c^2+{\left(8\mathrm{c}\mathrm{m}\right)}^2\backslash left(10\{,\}5\backslash ,\backslash mathrm\{cm\}\backslash right)^2=c^2+\backslash left(8\backslash ,\backslash mathrm\{cm\}\backslash right)^2$

$c^2=\mathrm{46,25}{\mathrm{c}\mathrm{m}}^{2}c^2=46\{,\}25\backslash ,\backslash mathrm\{cm\}^2$

Wurzel ziehen.

$c\approx \mathrm{6,8}\mathrm{c}\mathrm{m}c\backslash approx6\{,\}8\backslash ,\backslash mathrm\{cm\}$