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Let's break down and solve each part of the question related to the pyramid \( VABCD \), where the base is a square with side 10 cm and the height of the pyramid is 15 cm.

### 1. The angle between AV and the base:
- \( AV \) is the slant height of the pyramid, and we need to find the angle between this edge and the base plane.
- The height of the pyramid is given as 15 cm, and the distance from the center of the base to \( A \) is half the diagonal of the square.
- The diagonal of the square \( ABCD \) is \( \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2} \) cm, so the distance from the center \( O \) to any vertex, including \( A \), is half of this: \( 5\sqrt{2} \) cm.

Using trigonometry, the angle \( \theta \) between \( AV \) and the base can be found using:
\[
\tan(\theta) = \frac{\text{Height of pyramid}}{\text{Distance from center to } A} = \frac{15}{5\sqrt{2}} = \frac{15}{7.07}
\]
\[
\theta = \tan^{-1} \left( \frac{15}{7.07} \right) \approx \tan^{-1} (2.12) \approx 64.5^\circ
\]

### 2. The angle between the plane VAB and the base:
- The plane \( VAB \) includes the slant height \( AV \), the height \( VO \), and the base edge \( AB \).
- We are looking for the dihedral angle between the plane \( VAB \) and the base plane.
- The perpendicular height \( VO = 15 \) cm and the slant height \( VA \) is \( \sqrt{15^2 + (5\sqrt{2})^2} = \sqrt{225 + 50} = \sqrt{275} \approx 16.58 \) cm.

Using the dot product, the cosine of the angle \( \alpha \) between the two planes is:
\[
\cos(\alpha) = \frac{\text{Height of pyramid}}{\text{Slant height}} = \frac{15}{16.58} \approx 0.905
\]
\[
\alpha = \cos^{-1}(0.905) \approx 25.8^\circ
\]

### 3. The angle between the planes VAD and VCD:
- Both \( VAD \) and \( VCD \) are triangular planes that share the edge \( VD \). To find the angle between them, consider that these planes meet along \( VD \) and make an angle at this edge.
- Since \( VABCD \) is a square pyramid, the angle between any two adjacent planes through the vertex \( V \) (such as \( VAD \) and \( VCD \)) is symmetrical.

Using geometric properties of pyramids and adjacent triangular planes, the angle between these planes is found from the angle between the normals to the planes. For regular square pyramids, the angle between adjacent triangular faces is approximately:
\[
\gamma = \cos^{-1}\left(\frac{\text{side length}}{\sqrt{\text{side length}^2 + \text{height}^2}}\right) \approx \cos^{-1}\left(\frac{10}{\sqrt{10^2 + 15^2}}\right) = \cos^{-1}\left(\frac{10}{18.03}\right) \approx 56.3^\circ
\]

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Would you like further details on any specific part of the solution, or do you have any additional questions?

### Related Questions:
1. How do you calculate the slant height of a pyramid?
2. What is a dihedral angle, and how is it measured in 3D geometry?
3. How do you use trigonometry to calculate angles in 3D shapes like pyramids?
4. How does the shape of the base influence the angles in a pyramid?
5. What are the properties of the angles in a regular square pyramid?

### Tip:
In geometry problems involving pyramids, always start by sketching the base and height relations. It simplifies visualization for slant heights, angles, and planes.

The pyramid VABCD has a square base ABCD of side 10cm and a vertex V. The vertical height of the pyramid is 15cm. Find:
1. The angle between AV and the base
2. The angle between the plane VAB and the base
3. The angle between the planes VAD and VCD.

This problem covers the calculation of angles between slant heights and base planes in a square pyramid with a 10cm side and 15cm height. The solution involves the use of trigonometric ratios and geometric properties to find the angles between AV, VAB, and VAD planes.

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