Example 13 (10th "Hope Cup" Invitational Competition Question) In the regular quadrilateral pyramid $O-A B C D$, $\angle A O B=30^{\circ}$, the dihedral angle formed by faces $O A B$ and $O B C$ is $\theta$, and $\cos \theta=a \sqrt{b}-c$, where $a, b, c \in \mathbf{N}$, and $b$ is not divisible by the square of any prime number, then $a+b+c=$ $\qquad$
Solution: Fill in 14. Reason: As shown in Figure 7-12, let $OB=k$, then $BC^{2}=$ $k^{2}+k^{2}-2k^{2} \cdot \cos 30^{\circ}=(2-\sqrt{3})k^{2}$. In the plane $AOB$, draw $AH \perp OB$ at $H$, connect $HC$, then by the symmetry of the regular quadrilateral pyramid, we know $HC \perp OB$, thus $\angle AHC$ is the plane angle of the dihedral angle $A-OB-C$, so $\angle AHC=\theta$. From $\angle AOB=30^{\circ}$, we know $AH=\frac{1}{2}AO=\frac{k}{2}$, in $\triangle AHC$, by the cosine rule, we get $\cos \theta=$ $\frac{AH^{2}+HC^{2}-AC^{2}}{2AH \cdot HC}=\frac{2AH^{2}-AC^{2}}{2AH^{2}}=4\sqrt{3}-7$. Since the problem states $\cos \theta=a\sqrt{b}-c, a, b, c \in \mathbf{N}$ and $b$ is not divisible by the square of any prime number, so $a=4, b=3, c=7$.