Let's break down the problem.

You are given:

• A 39-foot ladder placed against a vertical wall.
• The base of the ladder slips and moves 15 feet from the wall.
• You need to determine how far down the wall the ladder has slipped.

This forms a right triangle where:

• The ladder is the hypotenuse (39 feet),
• The distance from the base of the wall to the base of the ladder is one leg (15 feet),
• The height from the ground to where the ladder touches the wall is the other leg.

We can use the Pythagorean theorem:

$a^2 + b^2 = c^2$

where:

• $a$ is the distance from the base of the wall to the ladder (15 feet),
• $b$ is the height the ladder reaches on the wall (the unknown),
• $c$ is the length of the ladder (39 feet).

Plugging in the known values:

$15^2 + b^2 = 39^2$

$225 + b^2 = 1521$

Solving for $b$:

$b^2 = 1521 - 225 = 1296$

$b = \sqrt{1296} = 36 \text{ feet}$

So, originally, the ladder reached 36 feet up the wall.

Now, how far does the ladder slip?

If the ladder is flush with the wall, it starts at the full height of 39 feet. After slipping, it is now only 36 feet up the wall.

Thus, the ladder has slipped by:

$39 - 36 = 3 \text{ feet}$

The correct answer is that the ladder slips 3 feet down the wall.

Would you like more details or have any questions?

Here are some additional related questions:

1. What happens to the height as the base of the ladder moves further away?
2. How can you calculate the angle between the ladder and the ground?
3. What if the ladder had been 50 feet long and slipped the same 15 feet at the base?
4. How can we model this situation with trigonometry instead of the Pythagorean theorem?
5. How does this scenario relate to the concept of similar triangles?

Tip: Always verify your calculations using a second method, like approximations or geometric reasoning, to ensure accuracy.