Math Problem Statement

A bridge is being built across a canyon. The length of the bridge is 5055 ft. From the deepest point in the canyon, the angles of elevation of the ends of the bridge are 78° and 71°. How deep is the canyon?

Solution

To solve this problem, we need to find the depth of the canyon, hh, using trigonometry. Here’s the setup based on the information provided:

  1. Given:

    • Length of the bridge: 50555055 ft.
    • Angles of elevation from the deepest point of the canyon to the ends of the bridge: 7878^\circ and 7171^\circ.
  2. Objective:

    • Find the depth hh of the canyon.

Solution Outline

The triangle in the diagram can be divided into two right triangles. Let’s designate:

  • AA and BB as the endpoints of the bridge.
  • CC as the point directly below the bridge at the deepest point of the canyon.

The depth hh is the vertical distance from point CC to the line segment ABAB (the bridge).

Using Trigonometry

Since we have two right triangles, we can use the tangent function for each angle:

Step 1: Set up the equations for each side of the bridge

  1. For the left side (angle 7878^\circ): tan(78)=hx\tan(78^\circ) = \frac{h}{x} where xx is the horizontal distance from point CC to the left end of the bridge AA.

    Rearranging, we get: x=htan(78)x = \frac{h}{\tan(78^\circ)}

  2. For the right side (angle 7171^\circ): tan(71)=hy\tan(71^\circ) = \frac{h}{y} where yy is the horizontal distance from point CC to the right end of the bridge BB.

    Rearranging, we get: y=htan(71)y = \frac{h}{\tan(71^\circ)}

Step 2: Relate xx and yy to the total length of the bridge

Since x+y=5055x + y = 5055 ft, we substitute from the equations above: htan(78)+htan(71)=5055\frac{h}{\tan(78^\circ)} + \frac{h}{\tan(71^\circ)} = 5055

Step 3: Solve for hh

  1. Calculate tan(78)\tan(78^\circ) and tan(71)\tan(71^\circ):

    • tan(78)4.7046\tan(78^\circ) \approx 4.7046
    • tan(71)2.9042\tan(71^\circ) \approx 2.9042
  2. Substitute these values into the equation: h4.7046+h2.9042=5055\frac{h}{4.7046} + \frac{h}{2.9042} = 5055

  3. Combine terms: h(14.7046+12.9042)=5055h \left( \frac{1}{4.7046} + \frac{1}{2.9042} \right) = 5055

  4. Calculate the combined term: h×(0.2126+0.3444)=5055h \times (0.2126 + 0.3444) = 5055 h×0.5570=5055h \times 0.5570 = 5055

  5. Solve for hh: h=50550.55709078 fth = \frac{5055}{0.5570} \approx 9078 \text{ ft}

Conclusion

The depth of the canyon is approximately 9078 feet.

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


Here are some related questions that could deepen your understanding:

  1. How would the depth change if the length of the bridge was shorter?
  2. What if the angles were both smaller, say 6060^\circ and 5555^\circ?
  3. How does the tangent function help determine heights in right triangles?
  4. Could you solve this problem using the sine or cosine functions instead?
  5. How would we set up the problem if the bridge was over a different shape, like an arc?

Tip: In trigonometry, the tangent of an angle is often used for finding heights when horizontal distances and angles are known, especially in right triangles.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Right Triangles
Tangent Function

Formulas

tan(angle) = opposite / adjacent
x + y = total length of bridge
Depth (h) = 5055 / (1/tan(78°) + 1/tan(71°))

Theorems

Right Triangle Trigonometry

Suitable Grade Level

Grades 10-12