leaning ladders- a nasty quartic
Two ladders leaning in opposite directionsbetween two walls.Each ladder touching the base of the opposite wall Ladders are 40 ft and 60 ft respt. cross at hy of 10 ft , h, what is the distance tween walls w ?
To solve this problem, we can use the concept of similar triangles. Let's consider the following diagram:
```
A B
/ /
/ /
/ /
/ /
/ /
/ /
P /
| /
|/
Q
```
Assume that the smaller ladder (40 ft) is at wall A, while the larger ladder (60 ft) is at wall B. Let P be the point where the smaller ladder touches wall B, and let Q be the point where the larger ladder touches wall A. The height where the ladders cross is point C. The distance between the walls, which we are trying to find, is denoted as W.
From the given information, we know that the height of the point C is 10 ft.
Let's use the concept of similar triangles to solve this problem. The triangles ACP and BCQ are similar since they share the angle at C and have corresponding angles at A and B.
By setting up a proportion, we can relate the lengths of the sides of these triangles. Let's consider the ratios of the sides:
AC/AP = CB/CQ
In other words:
AC / 40 = 60 / (W - 40)
Simplifying this equation, we get:
AC = (40 * 60) / (W - 40)
We also know that the triangle ACP is a right triangle, so we can apply the Pythagorean theorem:
AC^2 + PC^2 = 40^2
Substituting the value of AC from the earlier equation, we get:
[(40 * 60) / (W - 40)]^2 + 10^2 = 40^2
Now, we can solve this equation to find the value of W.