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The problem asks for the width of the canyon, given a bridge that is inclined upward at an angle of 20.0° to the horizontal, with a total bridge length of 50.0 meters. 

We can use trigonometry to solve this. The length of the bridge is the hypotenuse of a right triangle, and the width of the canyon corresponds to the horizontal leg.

### Step-by-step solution:

The relationship between the horizontal distance \( W \) (the width of the canyon), the hypotenuse \( L \) (the length of the bridge), and the angle \( \theta \) (the inclination) is given by:

\[
W = L \cdot \cos(\theta)
\]

Where:
- \( L = 50.0 \, \text{m} \)
- \( \theta = 20.0^\circ \)

Let's calculate the width of the canyon:

\[
W = 50.0 \, \text{m} \cdot \cos(20.0^\circ)
\]

I'll now compute this value.The width of the canyon is approximately \( 46.98 \, \text{m} \), which rounds to \( 47 \, \text{m} \), matching the given solution.

Would you like further details on this, or do you have any questions?

Here are five related questions to expand on this concept:
1. How would the width change if the angle of inclination was steeper, say 30°?
2. What is the vertical height difference between the two ends of the bridge?
3. How would the calculation change if the bridge length was increased to 60 meters but kept the same angle?
4. What trigonometric function would you use to find the vertical height directly from the length of the bridge and the angle?
5. If the bridge was horizontal, what would be the width of the canyon?

**Tip:** When dealing with inclined structures, understanding how angles affect horizontal and vertical distances is key to solving problems like this.

A bridge 50.0 m long crosses a canyon. If the bridge is inclined (upward) at an angle of 20.0° to the horizontal, how wide is the canyon?

Solve a trigonometry problem involving a 50.0 m long bridge inclined at 20.0° to the horizontal. Learn how to calculate the width of a canyon using cosine function, suitable for students in Grades 9-12.

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