Math Problem Statement

Solve

A man, M, positions himself on a river bank as in the diagram alongside, so he can observe two poles A and B of equal height on the opposite bank of the river. He finds the angle of elevation to the top of pole A is 22°, and the angle of elevation to the top of pole B is 19°. Show how he could use these facts to determine the width of the river, if he knows that A and B are 100 m apart.

Solution

To determine the width of the river, let's analyze the given data step-by-step. Here's the setup based on the information provided:

  1. Known Distances and Angles:

    • The distance between poles AA and BB is 100m100 \, \text{m}.
    • The angle of elevation to the top of pole AA from point MM is 2222^\circ.
    • The angle of elevation to the top of pole BB from point MM is 1919^\circ.
    • Let DD represent the width of the river (i.e., the horizontal distance from MM to AA or BB, since the poles are directly across the river from MM).
  2. Using Right Triangles: Let:

    • hh represent the height of the poles AA and BB.
    • We can set up two right triangles, one with angle 2222^\circ for pole AA and one with angle 1919^\circ for pole BB.
  3. Setting Up Tangent Equations:

    • For pole AA: tan(22)=hD\tan(22^\circ) = \frac{h}{D} So, h=Dtan(22)h = D \cdot \tan(22^\circ)

    • For pole BB: tan(19)=hD+100\tan(19^\circ) = \frac{h}{D + 100} So, h=(D+100)tan(19)h = (D + 100) \cdot \tan(19^\circ)

  4. Equating the Two Expressions for hh: Since both expressions represent hh, we can set them equal to each other: Dtan(22)=(D+100)tan(19)D \cdot \tan(22^\circ) = (D + 100) \cdot \tan(19^\circ)

  5. Solving for DD: Expanding and rearranging terms: Dtan(22)=Dtan(19)+100tan(19)D \cdot \tan(22^\circ) = D \cdot \tan(19^\circ) + 100 \cdot \tan(19^\circ) D(tan(22)tan(19))=100tan(19)D \cdot (\tan(22^\circ) - \tan(19^\circ)) = 100 \cdot \tan(19^\circ) D=100tan(19)tan(22)tan(19)D = \frac{100 \cdot \tan(19^\circ)}{\tan(22^\circ) - \tan(19^\circ)}

  6. Calculating the Value of DD: Using approximate values for the tangents:

    • tan(22)0.4040\tan(22^\circ) \approx 0.4040
    • tan(19)0.3443\tan(19^\circ) \approx 0.3443

    Substitute these values: D=1000.34430.40400.3443D = \frac{100 \cdot 0.3443}{0.4040 - 0.3443} D34.430.0597D \approx \frac{34.43}{0.0597} D576.57mD \approx 576.57 \, \text{m}

Therefore, the width of the river is approximately 576.57m576.57 \, \text{m}.

Would you like more details on any of these steps or further clarification?


Related Questions

  1. How can we verify this calculation using a different trigonometric approach?
  2. What assumptions are made in this solution, and how would errors in angle measurement affect the result?
  3. How would the calculation change if the poles were of different heights?
  4. Can we use sine or cosine rules to solve similar problems in different configurations?
  5. How could you use similar principles to measure the height of a building using two points on the ground?

Tip

When solving trigonometric problems, drawing a clear diagram helps in visualizing relationships and setting up equations correctly.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Angle of Elevation
Right Triangle Trigonometry

Formulas

tan(θ) = opposite / adjacent
D = (100 * tan(19°)) / (tan(22°) - tan(19°))

Theorems

Trigonometric Ratios in Right Triangles

Suitable Grade Level

Grades 10-12