Rysunek pomocniczy:

     

Oznaczmy:

$\left|AD\right|=d$  

$\left|AB\right|=\left|BC\right|=\left|AC\right|=a$  

$\left|AE\right|=\left|EB\right|=\left|BG\right|=\left|GC\right|=\left|CF\right|=\left|AF\right|=\frac{1}{2}a$  

$\left|AG\right|=\left|CE\right|=\left|BF\right|=h_p=\frac{a\sqrt{3}}{2}$  

$\left|SO\right|=H$  

$\left|AS\right|=\left|BS\right|=\left|CS\right|=b$  

Zauważmy, że

$\left|\sphericalangle ASF\right|=\left|\sphericalangle FSC\right|=\alpha$  

Korzystając z funkcji trygonometrycznych dla trójkąta AFS otrzymujemy:

$\frac{\left|AF\right|}{\left|AS\right|}=\sin \alpha$  

$\frac{\frac{1}{2}a}{b}=\sin \alpha$  

$\frac{1}{2}a=b\sin \alpha \mid :\sin \alpha$  

$\frac{\frac{1}{2}a}{\sin \alpha }=b$  

$\frac{a}{2\sin \alpha }=b$  

oraz

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

$\frac{1}{2}a=h_b\mathrm{tg}\alpha$  

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

$\frac{a}{2\mathrm{tg}\alpha }=h_b$  

 

Korzystając ze wzoru na pole trójkąta dla trójkąta ABS otrzymujemy:

$P_{ABS}=\frac{1}{2}\cdot a\cdot h_b=\frac{1}{2}\cdot a\cdot \frac{a}{2\mathrm{tg}\alpha }=\frac{a^2}{4\mathrm{tg}\alpha }$  

Z drugiej strony otrzymujemy:

$P_{ABS}=\frac{1}{2}\cdot b\cdot d=\frac{1}{2}\cdot \frac{a}{2\sin \alpha }\cdot d=\frac{ad}{4\sin \alpha }$  

Zatem:

$\frac{a^2}{4\mathrm{tg}\alpha }=\frac{ad}{4\sin \alpha }\mid :\frac{a}{4}$  

$\frac{a}{\mathrm{tg}\alpha }=\frac{d}{\sin \alpha }\mid \cdot \mathrm{tg}\alpha$  

$a=\frac{d}{\sin \alpha }\mathrm{tg}\alpha$  

$a=\frac{d}{\sin \alpha }\cdot \frac{\sin \alpha }{\cos \alpha }$  

$a=\frac{d}{\cos \alpha }$  

 

Korzystając z tw. Pitagorasa dla trójkąta AOS otrzymujemy:

${\left(\frac{2}{3}{h}_p\right)}^2+H^2=b^2$  

${\left(\frac{2}{3}\cdot \frac{a\sqrt{3}}{2}\right)}^2+H^2={\left(\frac{a}{2\sin \alpha }\right)}^2$  

${\left(\frac{a\sqrt{3}}{3}\right)}^2+H^2=\frac{a^2}{4{\sin }^2\alpha }$  

$\frac{3a^2}{9}+H^2=\frac{a^2}{4{\sin }^2\alpha }$  

$\frac{1}{3}{a}^2+H^2=a^2\cdot \frac{1}{4{\sin }^2\alpha }$  

$\frac{1}{3}\cdot {\left(\frac{d}{\cos \alpha }\right)}^2+H^2={\left(\frac{d}{\cos \alpha }\right)}^2\cdot \frac{1}{4{\sin }^2\alpha }$  

$\frac{1}{3}\cdot \frac{d^2}{{\cos }^2\alpha }+H^2=\frac{d^2}{{\cos }^2\alpha }\cdot \frac{1}{4{\sin }^2\alpha }$  

$\frac{d^2}{3{\cos }^2\alpha }+H^2=\frac{d^2}{4{\sin }^2\alpha {\cos }^2\alpha }\mid -\frac{d^2}{3{\cos }^2\alpha }$  

$H^2=\frac{3d^2}{12{\sin }^2\alpha {\cos }^2\alpha }-\frac{4d^2{\sin }^2\alpha }{12{\sin }^2\alpha {\cos }^2\alpha }$  

$H^2=\frac{d^2\left(3-4{\sin }^2\alpha \right)}{12{\sin }^2\alpha {\cos }^2\alpha }$  

$H^2=\frac{d^2\left(9-12{\sin }^2\alpha \right)}{36{\sin }^2\alpha {\cos }^2\alpha }$  

$H=\frac{d}{6\sin \alpha \cos \alpha }\cdot \sqrt{9-12{\sin }^2\alpha }$  

 

Obliczmy objętość tego ostrosłupa.

$V=\frac{1}{3}\cdot P_p\cdot H$  

$V=\frac{1}{3}\cdot \frac{a^2\sqrt{3}}{4}\cdot H$  

$V=\frac{1}{3}\cdot \frac{{\left(\frac{d}{\cos \alpha }\right)}^2\sqrt{3}}{4}\cdot \frac{d}{6\sin \alpha \cos \alpha }\cdot \sqrt{9-12{\sin }^2\alpha }$  

$V=\frac{1}{3}\cdot \frac{d^2\sqrt{3}}{4{\cos }^2\alpha }\cdot \frac{d}{6\sin \alpha \cos \alpha }\cdot \sqrt{9-12{\sin }^2\alpha }$  

$V=\frac{d^3\sqrt{3}\cdot \sqrt{3}\cdot \sqrt{3-4{\sin }^2\alpha }}{3\cdot 24\sin \alpha {\cos }^3\alpha }$  

$V=\frac{d^3\sqrt{3-4{\sin }^2\alpha }}{24\sin \alpha {\cos }^3\alpha }$  

$V=\frac{d^3\sqrt{3{\sin }^2\alpha +3{\cos }^2\alpha -4{\sin }^2\alpha }}{12\cdot 2\sin \alpha \cos \alpha \cdot {\cos }^2\alpha }$  

$V=\frac{d^3\sqrt{3{\cos }^2\alpha -{\sin }^2\alpha }}{12\sin 2\alpha {\cos }^2\alpha }$  

$V=d^3\frac{\sqrt{3{\cos }^2\alpha -\left(1-{\cos }^2\alpha \right)}}{12\sin 2\alpha {\cos }^2\alpha }$  

$V=\frac{d^3\sqrt{3{\cos }^2\alpha -1+{\cos }^2\alpha }}{12\sin 2\alpha {\cos }^2\alpha }$  

$V=\frac{d^3\sqrt{4{\cos }^2\alpha -1}}{12\sin 2\alpha {\cos }^2\alpha }$