Figure 1 shows a right pyramid with a rectangular base such that VC = 25 cm, AD = 12 cm and AB = 16 cm. If VO is the vertical height of the pyramid, calculate the angle between the planes ADCB and VAB.

The base ABCD is a rectangle with $AD = 12$ cm and $AB = 16$ cm.
Since it's a rectangle, $BC = AD = 12$ cm.
O is the center of the base, and VO is the vertical height of the pyramid.
First, find the length of the diagonal AC of the base using the Pythagorean theorem in $\triangle ABC$:

$$AC^2 = AB^2 + BC^2$$ $$AC^2 = 16^2 + 12^2$$ $$AC^2 = 256 + 144$$ $$AC^2 = 400$$ $$AC = \sqrt{400} = 20 \text{ cm}$$

Since O is the center of the rectangle, $OC$ is half of $AC$:

$$OC = \frac{1}{2} AC = \frac{1}{2}(20) = 10 \text{ cm}$$

Now, find the vertical height VO using the Pythagorean theorem in $\triangle VOC$:

$$VC^2 = VO^2 + OC^2$$

Given $VC = 26$ cm:

$$26^2 = VO^2 + 10^2$$ $$676 = VO^2 + 100$$ $$VO^2 = 676 - 100$$ $$VO^2 = 576$$ $$VO = \sqrt{576} = 24 \text{ cm}$$

The line of intersection of the two planes is AB.
To find the angle between the planes, we need to find a line in each plane that is perpendicular to the line of intersection (AB) at the same point.
Let M be the midpoint of AB.
In the base plane ADCB, the line segment OM is perpendicular to AB (since O is the center and M is the midpoint of a side, OM is parallel to AD and BC, which are perpendicular to AB).
The length of OM is half the length of AD:

$$OM = \frac{1}{2} AD = \frac{1}{2}(12) = 6 \text{ cm}$$

In the plane VAB, the line segment VM is the slant height of the face VAB. Since $\triangle VAB$ is an isosceles triangle (VA = VB in a right pyramid), VM is perpendicular to AB.
Therefore, the angle between the planes ADCB and VAB is $\angle VMO$.

Consider the right-angled triangle VMO. We have:
Opposite side to $\angle VMO$: $VO = 24$ cm
Adjacent side to $\angle VMO$: $OM = 6$ cm
Using the tangent function:

$$\tan(\angle VMO) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{VO}{OM}$$ $$\tan(\angle VMO) = \frac{24}{6}$$ $$\tan(\angle VMO) = 4$$ $$\angle VMO = \arctan(4)$$ $$\angle VMO \approx 75.9637^\circ$$

Rounding to one decimal place:

$$\angle VMO \approx 76.0^\circ$$

The angle between the planes ADCB and VAB is $*76.0^\circ*$.