Hi everyone, a friend of mine recently explained to me this problem and I could not help, so I was wondering if someone here could steer us in the right direction.
Basically, there is a project where Student A has to find the hight of tower 5/6 miles away without leaving the campus and using the school campus as "0". They have angleometers and since they cannot measure a straight line from the campus to the tower making a 90 degree angle, they cannot use right triangles*. Does anyone know a way they could do this? Thank you very much.
* they cannot use right triangles that we can think of.
And I know it is quite a confusing so if you need clarification on the question, please let me know.
Thank you Mr. Fantastic for this document.
However, I see that H and l make a 90 degree angle, but since they cannot leave the campus how can they know that is going to be 90 degrees?
Additionally, I do not understand how the ￼l = (H/tanb - H/tana) - Sorry, I dunno how to type that.
If anyone could help explain this, or give an alternate method it would be very much appreciated. Thank you.
Hello, SMA777!
This is a classic problem, found in every textbook.
Here is one approach to it . . .
Student has to find the height of tower some distance away
without leaving the campus. .They have angleometers,Code:* C * * | * * | * * | * * | h * * | * * | * α * β | * - - - - - - - * - - - - - - - * A 100 B x D
The tower is
The first observation is made at
The angle of elevation to the top of the tower is:
Moving 100 feet closer to the tower, another observation is made at
The angle of elevation to the top of the tower is:
Let
In right triangle .[1]
In right triangle .[2]
Equate [1] and [2]: .
. .
Therefore: .
We have found the height of the tower without leaving the campus.
. . In fact, we don't need to know the distance to the tower.