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These require extra steps . . . Here is #1 . . .

1. A cell tower 25m tall is on top of a hill.

To an observer standing on level ground near the foot of the hill,

the angle of elevation of the top of the tower is 50°

and the angle of elevation of the bottom of the tower is 32°.

Calculate the height of the top of the tower from the observer's position on the ground. Code:

* B
* |
* |25
* |
* * C
* 18° * :
* * :y
* * 32° :
A * - - - - - - - - - - - * D
x

$\displaystyle BC$ is the 25-foot tower.

The observer is at $\displaystyle A.$

$\displaystyle \angle CAD = 32^o$ . . . $\displaystyle \angle BAD = 50^o$

Let $\displaystyle x = AD,\;y = CD.$

. . We want: $\displaystyle y + 25$

In right triangle $\displaystyle CDA\!:\;\tan32^o \,= \,\frac{y}{x}\quad\Rightarrow\quad x \,=\,\frac{y}{\tan32^o}$ **[1]**

In right triangle $\displaystyle BDA\!:\;\tan50^o\,=\,\frac{y+25}{x}\quad\Rightarro w\quad x\,=\,\frac{y+25}{\tan50^o} $

Equate [1] and [2]: .$\displaystyle \frac{y}{\tan32^o} \:=\:\frac{y+25}{\tan50^o}\quad\Rightarrow\quad y\tan50^o \:=\:\tan32^o(y + 25)$

. . $\displaystyle y\tan50^o \:=\:y\tan32^o + 25\tan32^o\quad\Rightarrow\quad y\tan50^o - y\tan32^o \:=\:25\tan32^o$

Factor: .$\displaystyle y(\tan50^o - \tan32^o) \:=\:25\tan32^o\quad\Rightarrow\quad y \:=\:\frac{25\tan32^o}{\tan50^o - \tan32^o}$

Hence: .$\displaystyle y \:\approx\:27.56\text{ m}\quad\Rightarrow\quad\boxed{y+25 \:=\:52.56\text{ m}}$