\(\cos(45°)=\dfrac{1}{\sqrt{2}}\)
Somit ergibt sich \(\dfrac{1}{\sqrt{2}}=\dfrac{24+2z}{\sqrt{36}\sqrt{36+z^2}}\Leftrightarrow \dfrac{1}{\sqrt{2}}=\dfrac{12+z}{3\sqrt{36+z^2}}\)
Dann die Reziproke bilden:
\(\sqrt{2}=\dfrac{3\sqrt{36+z^2}}{12+z}\) |* 12+z
\(\Leftrightarrow \sqrt{2}\cdot (12+z)=3\sqrt{36+z^2}\) | ^2
\(\Leftrightarrow 2 z^2 + 48 z + 288=324+9z^2 \)
\(\Leftrightarrow -7 z^2 + 48 z - 36 = 0\) | pq/abc Formel
\(\longrightarrow z_1=6,\; z_2=\dfrac{6}{7}\)