1. ## Trig Problem

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 degrees and 44.7 degrees. Determine the height of the building, to one decimal place.

I'm completely lost.

2. Originally Posted by VDestinV
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 degrees and 44.7 degrees. Determine the height of the building, to one decimal place.

I'm completely lost.
Always draw a diagram. Here is my "artistic rendition" of the situtation

We can now use the tangent function to find h by setting up a system of equations

$\tan(39.1^\circ)=\frac{h}{x}$ and $\tan(44.7^\circ)=\frac{h+6}{x}$

From here just solve the system for h.

Good luck.

3. Originally Posted by We can now use the tangent function to find h by setting up a system of equations

[tex
\tan(39.1^\circ)=\frac{h}{x}[/tex] and $\tan(44.7^\circ)=\frac{h+6}{x}$

From here just solve the system for h.
but how do I figure out the x then?

4. 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
$AB$ is the satellite dish: . $AB = 5$
$BC$ is the building: . $BC = h$
$\angle BDC = 39.1^o,\;\;\angle ADC = 44.7^o$
Let $CD = x$

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

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

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

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

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

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