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.