Given: Square base ABCD with AD = DC = 6 cm.

Here are the solutions for the remaining parts of question 2.

2. Square-based pyramid ABCD V
Given: Square base ABCD with AD = DC = 6 cm.
Height of pyramid OV = 10 cm, where O is the center of the base.

b) Find:

ii) The angle between VA and ABCD.
The angle between the line VA and the base plane ABCD is the angle between VA and its projection onto the base, which is OA. So, we need to find $\angle VAO$.
Consider the right-angled triangle VOA.
We have:
VO = 10 cm (height of the pyramid)
OA = $3\sqrt{2}$ cm (calculated in part b.i)
We can use the tangent function:

$$\tan(\angle VAO) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{VO}{OA}$$ $$\tan(\angle VAO) = \frac{10}{3\sqrt{2}}$$

To rationalize the denominator:

$$\tan(\angle VAO) = \frac{10\sqrt{2}}{3\sqrt{2}\sqrt{2}} = \frac{10\sqrt{2}}{6} = \frac{5\sqrt{2}}{3}$$

Now, calculate the angle:

$$\angle VAO = \arctan\left(\frac{5\sqrt{2}}{3}\right)$$ $$\angle VAO \approx \arctan\left(\frac{5 \times 1.41421}{3}\right)$$ $$\angle VAO \approx \arctan\left(\frac{7.07105}{3}\right)$$ $$\angle VAO \approx \arctan(2.3570)$$ $$\angle VAO \approx 67.04^\circ$$

The angle between VA and ABCD is $\boxed{67.04^\circ}$ (to 2 decimal places).

iii) The angle between VDC and ABCD.
The angle between two planes (VDC and ABCD) is the angle between two lines, one in each plane, that are perpendicular to their line of intersection. The line of intersection is DC.
Let M be the midpoint of DC. Then OM is perpendicular to DC, and VM is perpendicular to DC.
So, the angle between the planes VDC and ABCD is $\angle VMO$.
Consider the right-angled triangle VOM.
We have:
VO = 10 cm (height of the pyramid)
OM = $\frac{1}{2} \times \text{side length of base} = \frac{1}{2} \times 6 = 3$ cm.
We can use the tangent function:

$$\tan(\angle VMO) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{VO}{OM}$$ $$\tan(\angle VMO) = \frac{10}{3}$$

Now, calculate the angle:

$$\angle VMO = \arctan\left(\frac{10}{3}\right)$$ $$\angle VMO \approx \arctan(3.3333)$$ $$\angle VMO \approx 73.30^\circ$$

The angle between VDC and ABCD is $\boxed{73.30^\circ}$ (to 2 decimal places).

iv) Volume of the pyramid.
The formula for the volume of a pyramid is:

$$\text{Volume} = \frac{1}{3} \times \text{Base Area} \times \text{Height}$$

The base ABCD is a square with side length 6 cm.
Base Area $= \text{side} \times \text{side} = 6 \text{ cm} \times 6 \text{ cm} = 36 \text{ cm}^2$.
The height of the pyramid VO is 10 cm.
Substitute these values into the formula:

$$\text{Volume} = \frac{1}{3} \times 36 \text{ cm}^2 \times 10 \text{ cm}$$ $$\text{Volume} = 12 \text{ cm}^2 \times 10 \text{ cm}$$ $$\text{Volume} = 120 \text{ cm}^3$$

The volume of the pyramid is $\boxed{120 \text{ cm}^3}$.