Rysunek pomocniczy:

Z funkcji trygonometrycznych mamy:

$\mathrm{tg}\alpha =\frac{H}{\frac{1}{2}a\sqrt{2}}$  

$\frac{1}{2}a\sqrt{2}\cdot \mathrm{tg}\alpha =H$  

 

Z treści zadania wiemy, że:

$V=\frac{1}{3}{P}_p\cdot H$  

$V=\frac{1}{3}\cdot a^2\cdot H$  

$V=\frac{1}{3}\cdot a^2\cdot \frac{1}{2}a\sqrt{2}\cdot \mathrm{tg}\alpha \mid \cdot 6$  

$6V=a^3\sqrt{2}\cdot \mathrm{tg}\alpha \mid :\sqrt{2}\mathrm{tg}\alpha$  

$\frac{6V}{\sqrt{2}\mathrm{tg}\alpha }=a^3$  

$\sqrt[3]{\frac{6V}{\sqrt{2}\mathrm{tg}\alpha }}=a$  

 

Zatem:

$H=\frac{1}{2}a\sqrt{2}\cdot \mathrm{tg}\alpha$  

$H=\frac{1}{2}\cdot \sqrt[3]{\frac{6V}{\sqrt{2}\mathrm{tg}\alpha }}\cdot \sqrt{2}\cdot \mathrm{tg}\alpha$  

$H=\sqrt[3]{\frac{1}{8}}\cdot \sqrt[3]{\frac{6V}{\sqrt{2}\mathrm{tg}\alpha }}\cdot \sqrt[3]{{\sqrt{2}}^3}\cdot \sqrt[3]{{\mathrm{tg}}^3\alpha }$  

$H=\sqrt[3]{\frac{1}{8}\cdot \frac{6V}{\sqrt{2}\mathrm{tg}\alpha }\cdot 2\sqrt{2}\cdot {\mathrm{tg}}^3\alpha }$  

$H=\sqrt[3]{\frac{3V{\mathrm{tg}}^2\alpha }{2}}$  

$H=\sqrt[3]{\frac{3V{\sin }^2\alpha }{2{\cos }^2\alpha }}$  

 

Wobec tego otrzymujemy:

$\frac{H}{b}=\sin \alpha$  

$H=b\sin \alpha$  

$\sqrt[3]{\frac{3V{\sin }^2\alpha }{2{\cos }^2\alpha }}=b\cdot \sqrt[3]{{\sin }^3\alpha }\mid :\sqrt[3]{{\sin }^3\alpha }$  

$\sqrt[3]{\frac{3V{\sin }^2\alpha }{2{\cos }^2\alpha }\cdot \frac{1}{{\sin }^3\alpha }}=b$  

$\sqrt[3]{\frac{3V}{2{\cos }^2\alpha \sin \alpha }}=b$