Berechnen Sie die Länge der Strecke $\stackrel{‾}{}\backslash overline\{\}$$\overline{CE}\backslash overline\{CE\}$ in cm.

Empfohlene Vorgehensweise:

1.) Berechne die Strecke$\overline{DF}\backslash \; \backslash overline\{DF\}\backslash$wende den Satz des Pythagoras an.

2.) Berechne den Winkel $\alpha \backslash \; \backslash alpha\backslash$mit dem Kosinus.

3.) Berechne den Winkel$\beta \backslash \; \backslash beta\backslash$(beachte, $\overline{DF}\mathrm{\mid }\mathrm{\mid }\overline{BC})\Rightarrow \mathrm{△}ADF\text{ und }\mathrm{△}ABC\backslash overline\{DF\}||\backslash overline\{BC\})\backslash \; \backslash Rightarrow\backslash \; \backslash triangle\{ADF\}\backslash text\{\backslash \; und\}\backslash \; \backslash triangle\{ABC\}\backslash$sind ähnlich.

4.) Berechne die Strecke$\overline{CE}\backslash \; \backslash overline\{CE\}\backslash$mit dem Sinus.

$\text{zu 1.) }\backslash text\{zu\; 1.)\}\backslash$ $\overline{DF}=\sqrt{\overline{(AD{)}^2}-\overline{(AF{)}^2}}\Rightarrow \overline{DF}=\sqrt{25-16}[cm]\Rightarrow \underset{‾}{\overline{DF}=3cm}\backslash \; \backslash overline\{DF\}=\backslash sqrt\{\backslash overline\{(AD)^2\}-\backslash overline\{(AF)^2\}\}\backslash quad\backslash Rightarrow\backslash quad\backslash \; \backslash overline\{DF\}=\backslash sqrt\{25-16\}\backslash \; [cm]\backslash qquad\backslash Rightarrow\backslash quad\backslash underline\{\backslash overline\{DF\}\backslash \; =\backslash \; 3\backslash \; cm\}$

$\text{zu 2.) }\backslash text\{zu\; 2.)\}\backslash$$cos(\alpha )={\displaystyle \frac{\overline{AF}}{\overline{AD}}}\Rightarrow cos(\alpha )={\displaystyle \frac{4}{5}}\Rightarrow cos(\alpha )=\mathrm{0,8}\Rightarrow \underset{‾}{\alpha \approx {37}^{\circ }}\backslash \; cos(\backslash alpha)\backslash \; =\backslash \; \backslash dfrac\{\backslash overline\{AF\}\}\{\backslash overline\{AD\}\}\backslash quad\backslash Rightarrow\backslash quad\; cos(\backslash alpha)\backslash \; =\backslash \; \backslash dfrac\{4\}\{5\}\backslash quad\backslash Rightarrow\backslash quad\; cos(\backslash alpha)\backslash \; =\backslash \; 0\{,\}8\backslash quad\; \backslash Rightarrow\backslash quad\; \backslash underline\{\backslash alpha\backslash approx\; 37^\backslash circ\}$

$\text{zu 3.) }\backslash text\{zu\; 3.)\}\backslash$$\beta ={180}^{\circ }-{90}^{\circ }-{37}^{\circ }\Rightarrow \underset{‾}{\beta ={53}^{\circ }}\backslash beta\backslash \; =\backslash \; 180^\backslash circ-90^\backslash circ-37^\backslash circ\backslash quad\backslash Rightarrow\backslash quad\backslash underline\{\backslash beta\backslash \; =\backslash \; 53^\backslash circ\}$

$\text{zu 4.) }\backslash text\{zu\; 4.)\}\backslash$$\text{ sin}(\beta )={\displaystyle \frac{\overline{CE}}{\overline{BC}}}\backslash \; \backslash text\{sin\}(\backslash beta)\backslash \; =\backslash \; \backslash dfrac\{\backslash overline\{CE\}\}\{\backslash overline\{BC\}\}$;$3:\overline{BC}=1:3\Rightarrow \overline{BC}=9cm\backslash qquad\backslash qquad\; 3:\backslash overline\{BC\}\backslash \; =\backslash \; 1:3\backslash quad\backslash Rightarrow\backslash quad\; \backslash overline\{BC\}\backslash \; =\backslash \; 9\backslash \; cm$

$\overline{CE}=9\cdot \sin {53}^{\circ }\Rightarrow \underset{‾}{\underset{‾}{\overline{CE}\approx \mathrm{7,2}cm}}\backslash hspace\{15mm\}\; \backslash overline\{CE\}\backslash \; =\backslash \; 9\backslash cdot\; \backslash sin\; 53^\backslash circ\backslash quad\backslash Rightarrow\backslash quad\backslash underline\{\backslash underline\{\backslash overline\{CE\}\backslash \; \backslash approx7\{,\}2\backslash \; cm\}\}$

Die die Länge der Strecke $\stackrel{‾}{}\backslash overline\{\}$$\overline{CE}\backslash overline\{CE\}$ beträgt ungefähr $\mathrm{7,2}cm7\{,\}2\backslash \; cm$.$$