A6 The length of the base edge $a$ of a regular quadrilateral pyramid is $3 \sqrt{2} \mathrm{~cm}$, and the length of the height $v_{1}$ of the side face is $\frac{1}{2} \sqrt{30} \mathrm{~cm}$. The size of the angle between the side edge and the base plane is:
(A) $45^{\circ}$
(B) $22,5^{\circ}$
(C) $60^{\circ}$
(D) $120^{\circ}$
(E) $30^{\circ}$
A6. By Pythagoras' theorem, we calculate the length of the side edge $s=\sqrt{\left(\frac{a}{2}\right)^{2}+v_{1}^{2}}=$

$\sqrt{\left(\frac{3 \sqrt{2}}{2}\right)^{2}+\left(\frac{1}{2} \sqrt{30}\right)^{2}}=\sqrt{\frac{9}{2}+\frac{15}{2}}=\sqrt{12}=2 \sqrt{3}$. We calculate the length of the diagonal of the base square $d=a \sqrt{2}=3 \sqrt{2} \cdot \sqrt{2}=6$. Using trigonometric functions, we calculate $\cos \alpha=\frac{\frac{d}{2}}{s}=\frac{3}{2 \sqrt{3}}=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}$, thus the angle $\alpha=30^{\circ}$.

## Section B