# Thread: Finding the height of an object without going off campus

1. ## Finding the height of an object without going off campus

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.

2. Originally Posted by SMA777
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.
Read the attachment. Use the angleometer to get the angles. Use a tape measure to get l.

3. 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.

4. Originally Posted by SMA777
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.
But surely you can measure a distance within the campus .... That's what the distance l is .... To get the expression for l in terms of the angles, use the right-triangles.

5. Originally Posted by mr fantastic
But surely you can measure a distance within the campus .... That's what the distance l is .... To get the expression for l in terms of the angles, use the right-triangles.
OH, I get it. So basically you get one angle, move (on campus) and get another. Got it.

Thank you so much! I really appreciate it

6. Hello, SMA777!

This is a classic problem, found in every textbook.
Here is one approach to it . . .

Student $\displaystyle A$ 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 $\displaystyle CD = h.$

The first observation is made at $\displaystyle A.$
The angle of elevation to the top of the tower is: $\displaystyle \alpha = \angle CAD$

Moving 100 feet closer to the tower, another observation is made at $\displaystyle B.$
The angle of elevation to the top of the tower is: $\displaystyle \beta = \angle CBD$
Let $\displaystyle x = BD.$

In right triangle $\displaystyle CDA\!:\;\tan\alpha \:=\:\frac{h}{x + 100} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan\alpha} - 100$ .[1]

In right triangle $\displaystyle CDB\!:\;\tan\beta \:=\:\frac{h}{x} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan\beta}$ .[2]

Equate [1] and [2]: .$\displaystyle \frac{h}{\tan\alpha} - 100 \:=\:\frac{h}{\tan\beta} \quad\Rightarrow\quad \frac{h}{\tan\alpha} - \frac{h}{\tan\beta} \:=\:100$

. . $\displaystyle h\left(\frac{1}{\tan\alpha} - \frac{1}{\tan\beta}\right) \:=\:100 \quad\Rightarrow\quad h\left(\frac{\tan\beta-\tan\alpha}{\tan\alpha\tan\beta}\right) \:=\:100$

Therefore: .$\displaystyle h \;=\;100\left(\frac{\tan\alpha\tan\beta}{\tan\beta-\tan\alpha}\right)$

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.

7. Thank you, Soroban for your extremely useful hep, as always

We will attempt both mentioned methods. Thank you both very, very much.

8. Originally Posted by SMA777
Thank you, Soroban for your extremely useful hep, as always

We will attempt both mentioned methods. Thank you both very, very much.
The methods are the same. What Soroban has done is given more detail as to where the formula comes from and provided an example of its use.