4. In the tetrahedron $V-ABC$, it is known that the base $ABC$ is an isosceles right triangle with $\angle B$ as the right angle, the plane $VAC \perp$ plane $ABC$, $AC=4$, $VA=\sqrt{14}$, and the tangent value of the dihedral angle $V-AB-C$ is $\frac{\sqrt{10}}{3}$. Then the angle formed by $VB$ and the base $ABC$ is equal to $\qquad$ (express the angle using degrees or inverse trigonometric functions).
4. $45^{\circ}$.

As shown in Figure 2, construct VH
$\perp$ plane $ABC$, $HD \perp$
$AB$. Then $AB \perp VD$.
Therefore, $\angle VDH$ is

the plane angle of the dihedral angle $V-AB-C$, i.e.,
$$
\tan \angle VDH = \frac{\sqrt{10}}{3} \text{.}
$$

Figure 2
Let $VH = \sqrt{10} x$, $HD = 3x$, $AH = 3\sqrt{2} x$.
In the right triangle $\triangle VAH$, we have
$(3\sqrt{2} x)^{2} + (\sqrt{10} x)^{2} = (\sqrt{14})^{2} \Rightarrow x = \frac{\sqrt{2}}{2}$
$\Rightarrow VH = \sqrt{5}, AH = 3$.
In $\triangle DBH$, we have
$HB = \sqrt{(3x)^{2} + (2\sqrt{2} - 3x)^{2}} = \sqrt{5}$.
Thus, $HB = VH$.
Therefore, $\angle VBH = 45^{\circ}$, i.e., the angle between $VB$ and the base plane $ABC$ is $45^{\circ}$.