Hello, VDestinV!

Did you make a sketch?

A satellite dish that is 5 m high sits on top of a building.

From a point at the base of the building, the angles of elevation of the botom and top

of the satellite dish are 39.1° and 44.7°.

Determine the height of the building, to one decimal place. Code:

A *
| *
5 | *
| *
B * *
| * *
h | * *
| 39.1° * *
C * - - - - - - - * D
x

$\displaystyle AB$ is the satellite dish: .$\displaystyle AB = 5$

$\displaystyle BC$ is the building: .$\displaystyle BC = h$

$\displaystyle \angle BDC = 39.1^o,\;\;\angle ADC = 44.7^o$

Let $\displaystyle CD = x$

In $\displaystyle \Delta BCD\!:\;\tan39.1^o \:=\: \frac{h}{x} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan39.1^o}\;\;{\color{blue}[1]}$

In $\displaystyle \Delta ACD\!:\;\;\tan44.7^o \:=\:\frac{h+5}{x}\quad\Rightarrow\quad x \:=\:\frac{h+5}{\tan44.7^o}\;\;{\color{blue}[2]}$

Equate [1] and [2]: .$\displaystyle \frac{h}{\tan39.1^o} \:=\:\frac{h+5}{\tan44.7^o} \quad\Rightarrow$ . $\displaystyle h\tan44.7^o \:=\:(h+5)\tan39.1^o$

. . $\displaystyle h\tan44.7^o \:=\:h\tan39.1^o + 5\tan39.1^o \quad\Rightarrow\quad h\tan44.7^o - h\tan39.1^o \:=\:5\tan39.1^o$

. . $\displaystyle h(\tan44.7^o - \tan39.1^o) \:=\:5\tan39.1^o \quad\Rightarrow\quad h \:=\:\frac{5\tan39.1^o}{\tan44.7^o - \tan39.1^o}$

Therefore: .$\displaystyle h \;=\;22.9693949 \;\approx\; 23.0$ m.