Given right triangular prism $ABCA_1B_1C_1$, the surrounding edge(not sure if this is the right term in English, but the surrounding edge are $AA_1, BB_1, CC_1$) are equal to $\frac{\sqrt{5}}{5}$ and the base edges(again I'm not sure if this is the right term, but those edges are $AB, BC, AC, A_1B_1, B_1C_1, A_1C_1$) are equal to 1.find the $\cos \angle (AB_1; BC_1)$
Solution: Let's take point A_2 such that $AA_2CB$ is parallelogram. Then $AA_2C_1B_1$ is also a parallelogram. Also they are rhombuses. This means that $\angle (AB_1, BC_1) = \angle (A_2C_1; BC_1)$. From here we can find that $ BA_2 = \sqrt{3}, AB_1 = BC_1 = \frac{\sqrt{30}}{5} $ Then $A_2C_1 = \frac{\sqrt{30}}{5}$, now we can apply cosine law for triangle $BA_2C_1$, we get that $cos \angle BC_1A_1 = - \frac{1}{4}$
Here is drawing:
my question is how did they got that $ BA_2 = \sqrt{3} AB_1 = BC_1 = \frac{\sqrt{30}}{5} $
should be
$BO = \dfrac{\sqrt{3}}2$ since $\angle BOA = 90^\circ$ and $BO^2 = BA^2 - OA^2 = 1 - (\dfrac12)^2 = \dfrac34$.
$BA_2 = 2 \times BO = \sqrt{3}$.
$AB_1 = \sqrt{AB^2 + BB_1^2} = \sqrt{1^2+(\frac{\sqrt{5}}{5})^2} = \frac{\sqrt{30}}5$
$BC_1 = \sqrt{BC^2 + CC_1^2} = \sqrt{1^2+(\frac{\sqrt{5}}{5})^2} = \frac{\sqrt{30}}5$