Kosinussatz: $c^2=a^2+b^2-2ab\cdot \cos \left(\gamma \right)c^2=a^2+b^2-2ab\backslash cdot\backslash cos\backslash left(\backslash gamma\backslash right)$

$a=b=5\text{ cm}a=b=5\backslash cm$

$\gamma =90\mathrm{°}-\alpha =90\mathrm{°}-75\mathrm{°}=15\mathrm{°}\backslash gamma=90°-\backslash alpha=90°-75°=15°$

Einsetzen:

${\overline{BC}}^2=5^2+5^2-2\cdot 5\cdot 5\cdot \cos (15\mathrm{°})\backslash overline\{BC\}^2=5^2+5^2-2\backslash cdot\; 5\backslash cdot5\backslash cdot\; \backslash cos(15°)$

${\overline{BC}}^2=50-50\cdot \cos (15\mathrm{°})=50-50\cdot \mathrm{0,9659}=\mathrm{1,705}\backslash overline\{BC\}^2=50-50\backslash cdot\; \backslash cos(15°)=50-50\backslash cdot\; 0\{,\}9659=1\{,\}705$

$\overline{BC}\approx \mathrm{1,3}\backslash overline\{BC\}\backslash approx\; 1\{,\}3$

Kosinussatz: $c^2=a^2+b^2-2ab\cdot \cos \left(\alpha \right)c^2=a^2+b^2-2ab\backslash cdot\backslash cos\backslash left(\backslash alpha\backslash right)$

$a=b=5\text{ cm}a=b=5\backslash cm$

$\alpha =90\mathrm{°}\backslash alpha=90°$

Einsetzen:

$\mathrm{\mid }\overline{AC}{\mathrm{\mid }}^2=5^2+5^2-2\cdot 5\cdot 5\cdot \cos (90\mathrm{°})|\backslash overline\{AC\}|^2=5^2+5^2-2\backslash cdot\; 5\backslash cdot5\backslash cdot\; \backslash cos(90°)$

$\mathrm{\mid }\overline{AC}{\mathrm{\mid }}^2=50-50\cdot \cos (90\mathrm{°})=50|\backslash overline\{AC\}|^2=50-50\backslash cdot\; \backslash cos(90°)=50$

$\mathrm{\mid }\overline{AC}\mathrm{\mid }=\sqrt{50}=\sqrt{2\cdot 25}=\sqrt{2}\cdot 5|\backslash overline\{AC\}|=\backslash sqrt\{50\}=\backslash sqrt\{2\backslash cdot\; 25\}=\backslash sqrt\{2\}\backslash cdot\; 5$

Der Abstand $aa$ beträgt $\sqrt{2}\cdot 5\text{ cm}\backslash sqrt\{2\}\backslash cdot\; 5\backslash cm$.

Alternative:

Das Dreieck $ACMACM$ist bei $MM$ rechtwinklig.

Daher gilt mit dem Satz des Pythagoras

$\mathrm{\mid }AC{\mathrm{\mid }}^2=\mathrm{\mid }AM{\mathrm{\mid }}^2+\mathrm{\mid }CM{\mathrm{\mid }}^2=5^2+5^2=50|AC|^2=|AM|^2+|CM|^2=5^2+5^2=50$

$\mathrm{\mid }AC\mathrm{\mid }=\sqrt{50}=2\sqrt{5}|AC|=\backslash sqrt\{50\}=2\backslash sqrt\{5\}$

Kosinussatz: $c^2=a^2+b^2-2ab\cdot \cos \left(\alpha \right)c^2=a^2+b^2-2ab\backslash cdot\backslash cos\backslash left(\backslash alpha\backslash right)$

$\mathrm{\mid }\overline{AB}\mathrm{\mid }=c=5\text{ cm}|\backslash overline\{AB\}|=c=5\backslash cm$

$a=b=5\text{ cm}a=b=5\backslash cm$

Einsetzen:

$5^2=50-50\cdot \cos (\alpha )5^2=50-50\backslash cdot\; \backslash cos(\backslash alpha)$

$\mathrm{0,5}=\cos (\alpha )0\{,\}5=\backslash cos(\backslash alpha)$

$\alpha =60\mathrm{°}\backslash alpha=60°$

Die Größe des Winkels $\alpha \backslash alpha$ beträgt 60°.

Alternative:

Das Dreieck $ABMABM$ ist gleichseitig, da alle Seitenlängen $5\text{ cm}5\backslash cm$ betragen. Daher sind alle Innenwinkel ${60}^{\circ }60^\backslash circ$.