How Do You Calculate the Height of a Building Using Trigonometry?

  • Thread starter Thread starter illmatic899
  • Start date Start date
  • Tags Tags
    Building Height
Click For Summary
SUMMARY

The discussion focuses on calculating the height of a building using trigonometric principles. A surveyor measures the angle of elevation to the top of the building at two different distances: initially at an angle of π/3 and then at π/4 after moving 40 feet further away. By applying the tangent function, the equations tan(π/3) = h/x and tan(π/4) = h/(x + 40) are established. The solution involves solving these equations to find the height (h) and the distance (x) from the building.

PREREQUISITES
  • Understanding of basic trigonometric functions, specifically tangent.
  • Familiarity with solving simultaneous equations.
  • Knowledge of angles in radians, particularly π/3 and π/4.
  • Ability to apply trigonometric principles to real-world problems.
NEXT STEPS
  • Study the properties of tangent functions in trigonometry.
  • Learn how to solve systems of equations involving trigonometric functions.
  • Explore applications of trigonometry in surveying and architecture.
  • Investigate further problems involving angles of elevation and depression.
USEFUL FOR

This discussion is beneficial for surveyors, mathematics students, and professionals in fields requiring practical applications of trigonometry, such as architecture and engineering.

illmatic899
Messages
1
Reaction score
0
A surveyor stands on flat ground at an unknown distance from a tall building. She measures the angle from the horizonal ground to the top of the building; this angle is pi/3. next she paces 40ft further away from the building. the angle from the ground to the top of the building is now measured to be pi/4.
a)how tall is the building
b) If the surveyor moves 20 feet further from the building what will the angle from the horizontal to the building's roof be.


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
Let the height of the building, in feet, be h, the initial distance from the building be x. Then you have [itex]tan(\pi/3)= \sqrt{3}= x/h[/itex]. 40 ft further away, the building is still h feet high and the distance from the building is now x+ 40 feet. Now you have [itex]tan(\pi/4)= 1= (x+40)/h[/itex]. You now have two equations to solve for x and h.

For (b), let [itex]\theta[/itex]. You have already solved for x so you know the distance from the building is x+ 40+ 20. And, of course, you have solved for h. Now, [itex]tan(\theta)= (x+ 60)/h[/itex]. Solve that equation for [itex]\theta[/itex].
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
Replies
11
Views
6K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 18 ·
Replies
18
Views
4K
  • · Replies 8 ·
Replies
8
Views
5K
Replies
40
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
3K