Prvo rješenje.
Dijagonala baze je d 1 = a 2 d_1=a\sqrt2 d 1 = a 2 d 2 = a 2 + v 2 d_2=\sqrt{a^2+v^2} d 2 = a 2 + v 2
Sada imamo
sin α 2 = d 1 2 d 2 , = a 2 2 a 2 + v 2 \sin\frac\alpha2=\frac{d_1}{2d_2},\qquad=\frac{a\sqrt2}{2\sqrt{a^2+v^2}} sin 2 α = 2 d 2 d 1 , = 2 a 2 + v 2 a 2 odakle kvadriranjem i sređivanjem dobivamo
2 ( a 2 + v 2 ) sin 2 α 2 = a 2 , a 2 ( 1 − 2 sin 2 α 2 ) = 2 v 2 sin 2 α 2 , a 2 cos α = 2 v 2 sin 2 α 2 , a = v 2 sin α 2 cos α . \begin{aligned}2(a^2+v^2)\sin^2\frac\alpha2&=a^2,\\a^2\left(1-2\sin^2\frac\alpha2\right)&=2v^2\sin^2\frac\alpha2,\\a^2\cos\alpha&=2v^2\sin^2\frac\alpha2,\\a&=\frac{v\sqrt2\sin\frac\alpha2}{\sqrt{\cos\alpha}}.\end{aligned} 2 ( a 2 + v 2 ) sin 2 2 α a 2 ( 1 − 2 sin 2 2 α ) a 2 cos α a = a 2 , = 2 v 2 sin 2 2 α , = 2 v 2 sin 2 2 α , = cos α v 2 sin 2 α . Drugo rješenje.
Dijagonala baze je d 1 = a 2 d_1=a\sqrt2 d 1 = a 2 d 2 = a 2 + v 2 d_2=\sqrt{a^2+v^2} d 2 = a 2 + v 2
Primijenimo kosinusov poučak na označeni kut:
cos α = d 2 2 + d 2 2 − d 1 2 2 d 2 d 2 , \cos\alpha=\frac{d_2^2+d_2^2-d_1^2}{2d_2d_2}, cos α = 2 d 2 d 2 d 2 2 + d 2 2 − d 1 2 , tj. 2 d 2 2 cos α = 2 d 2 2 − d 1 2 2d_2^2\cos\alpha=2d_2^2-d_1^2 2 d 2 2 cos α = 2 d 2 2 − d 1 2
2 ( a 2 + v 2 ) cos α = 2 ( a 2 + v 2 ) − 2 a 2 , a 2 = v 2 ( 1 − cos α ) cos α , \begin{aligned}2(a^2+v^2)\cos\alpha&=2(a^2+v^2)-2a^2,\\a^2&=\frac{v^2(1-\cos\alpha)}{\cos\alpha},\end{aligned} 2 ( a 2 + v 2 ) cos α a 2 = 2 ( a 2 + v 2 ) − 2 a 2 , = cos α v 2 ( 1 − cos α ) , tj. a = v 1 − cos α cos α a=v\sqrt{\dfrac{1-\cos\alpha}{\cos\alpha}} a = v cos α 1 − cos α
