To solve this problem, we can use trigonometry, specifically the tangent function, which relates the height ($h$) to the given angles and distance across the canyon.

Given Information:

• The distance across the canyon (the length of the bridge) is $5025$ feet.
• The angles of elevation to the ends of the bridge from the deepest point in the canyon are $75^\circ$ and $73^\circ$.

Steps:

Let's label the midpoint directly below the bridge as $O$ and the points where the angles are taken from as $A$ and $B$ at the ends of the bridge.

1. Divide the bridge into two right triangles where:

   • One triangle has an angle of $75^\circ$ and half the bridge distance, $\frac{5025}{2} = 2512.5$ feet.
   • The other triangle has an angle of $73^\circ$ and the same horizontal distance of $2512.5$ feet.

2. Apply the tangent function:

   • For the triangle with the $75^\circ$ angle: $\tan(75^\circ) = \frac{h}{2512.5}$ Solving for $h$: $h = 2512.5 \times \tan(75^\circ)$

   • For the triangle with the $73^\circ$ angle: $\tan(73^\circ) = \frac{h}{2512.5}$ Solving for $h$: $h = 2512.5 \times \tan(73^\circ)$

3. Calculate the depths: Plugging in the values for $\tan(75^\circ)$ and $\tan(73^\circ)$ using a calculator:

   • $\tan(75^\circ) \approx 3.732$
   • $\tan(73^\circ) \approx 3.271$

   Then:

   • For $75^\circ$: $h \approx 2512.5 \times 3.732 = 9375.63$ feet
   • For $73^\circ$: $h \approx 2512.5 \times 3.271 = 8222.24$ feet

Final Answer

The canyon is approximately $8222$ feet deep (rounded to the nearest foot).

Would you like further details or clarification on any step?

Relative Questions

1. How would the solution change if the angles of elevation were different?
2. What trigonometric identities are used to derive the tangent function?
3. How can you verify if the depth calculation is correct?
4. Can this method be used if the bridge were not straight across?
5. How would the answer change if the length of the bridge increased?

Tip

In trigonometric problems involving elevation angles, it’s helpful to split complex shapes into right triangles to use tangent or other trigonometric ratios effectively.