The given problem describes a quadrilateral pyramid $S.ABCD$ where the base $ABCD$ is a square, and $SA$ is perpendicular to the base. We need to determine the angle between the two lines $SC$ and $BD$.

Step-by-step solution:

1. Coordinate System Setup:

   • Assume the square $ABCD$ is in the $xy$-plane.
   • Let $A(0,0,0)$, $B(a,0,0)$, $C(a,a,0)$, and $D(0,a,0)$.
   • Since $SA$ is perpendicular to the base, let $S(0,0,h)$.

2. Finding the Vectors:

   • Vector $\overrightarrow{SC}$: $\overrightarrow{SC} = (a - 0, a - 0, 0 - h) = (a, a, -h)$
   • Vector $\overrightarrow{BD}$: $\overrightarrow{BD} = (0 - a, a - 0, 0 - 0) = (-a, a, 0)$

3. Computing the Dot Product: $\overrightarrow{SC} \cdot \overrightarrow{BD} = (a, a, -h) \cdot (-a, a, 0)$ $= a(-a) + a(a) + (-h)(0) = -a^2 + a^2 + 0 = 0$

4. Conclusion: Since the dot product is $0$, the two vectors are perpendicular, meaning the angle between $SC$ and $BD$ is $90^\circ$.

Thus, the correct answer is B. $90^\circ$.

Would you like more details on any step? Here are some related questions:

1. How do you find angles between lines using vectors?
2. What happens if $SA$ is not perpendicular to the base?
3. How do you generalize this method for other pyramidal shapes?
4. What is the significance of dot products in 3D geometry?
5. Can we use another method (e.g., trigonometry) to verify the angle?

Tip: Always check perpendicularity using the dot product—it’s a quick and reliable method!