Math Problem Statement

Jim is a licensed surveyor who works at a high-rise building project. During the final round of inspections, he noticed the angle of elevation of the top of a perpendicular building is 25 degrees. Then he moved 110 m nearer to the building and found the angle of the elevation is now 55 degrees. Determine the height of the building at Jim’s worksite (round your answers to two decimal places).

Solution

Let's denote the following:

  • hh = height of the building
  • dd = initial distance from Jim to the building

Given the angles of elevation and the distances moved, we can set up two right triangles and use trigonometric relations to solve for hh and dd.

Step-by-Step Solution:

  1. Initial Setup:

    When the angle of elevation is 25 degrees: tan(25)=hd\tan(25^\circ) = \frac{h}{d}

    When the angle of elevation is 55 degrees after moving 110 meters closer: tan(55)=hd110\tan(55^\circ) = \frac{h}{d - 110}

  2. Express the height hh in terms of dd:

    From the first equation: h=dtan(25)h = d \tan(25^\circ)

  3. Substitute hh into the second equation:

    tan(55)=dtan(25)d110\tan(55^\circ) = \frac{d \tan(25^\circ)}{d - 110}

  4. Solve for dd:

    (d110)tan(55)=dtan(25)(d - 110) \tan(55^\circ) = d \tan(25^\circ)

    dtan(55)110tan(55)=dtan(25)d \tan(55^\circ) - 110 \tan(55^\circ) = d \tan(25^\circ)

    d(tan(55)tan(25))=110tan(55)d (\tan(55^\circ) - \tan(25^\circ)) = 110 \tan(55^\circ)

    d=110tan(55)tan(55)tan(25)d = \frac{110 \tan(55^\circ)}{\tan(55^\circ) - \tan(25^\circ)}

  5. Calculate the values of the tangents:

    tan(25)0.4663\tan(25^\circ) \approx 0.4663 tan(55)1.4281\tan(55^\circ) \approx 1.4281

  6. Substitute these values into the equation for dd:

    d=110×1.42811.42810.4663d = \frac{110 \times 1.4281}{1.4281 - 0.4663}

    d=157.0910.9618d = \frac{157.091}{0.9618}

    d163.34 metersd \approx 163.34 \text{ meters}

  7. Calculate the height hh:

    Using h=dtan(25)h = d \tan(25^\circ):

    h=163.34×0.4663h = 163.34 \times 0.4663

    h76.18 metersh \approx 76.18 \text{ meters}

Final Answer:

The height of the building at Jim's worksite is approximately 76.1876.18 meters.

Would you like more details or have any questions about the solution?

Relative Questions:

  1. How do we derive the tangent function for an angle in a right triangle?
  2. What are the general properties of trigonometric functions?
  3. Can you explain how the law of sines or cosines might be applied in surveying?
  4. What are other methods to measure the height of a building?
  5. How do angles of elevation and depression relate to real-world surveying?
  6. What role does trigonometry play in navigation and astronomy?
  7. How can you use trigonometry to solve real-life problems involving heights and distances?
  8. What are the limitations of using trigonometric methods in surveying?

Tip:

When solving trigonometric problems, always make sure to check the angles and distances to ensure the setup of your equations is accurate.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometry
Right Triangles
Angles of Elevation
Trigonometric Functions

Formulas

Trigonometric ratios (sine, cosine, tangent)

Theorems

-

Suitable Grade Level

Grades 10-12