# Thread: trig triangle problems

1. ## trig triangle problems

could someone help me with these problems please? they both involve more than 1 right triangle.

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 degrees and the angle of elevation of the bottom of the tower is 32 degrees. calculate the height of hte top of the tower from the observer's position on the ground.
(if you could, please draw a diagram for me?)

2. from the top of a 120m fire tower, a fire ranger observes smoke in 2 locations. 1 has an angle of depression of 6 degrees, and the other has an angle of depression of 3 degrees. calculate the distance between the smoke sightings when they're on the same side of the tower, and in line with the tower.

thanks.

2. Hello, checkmarks!

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

$BC$ is the 25-foot tower.
The observer is at $A.$
$\angle CAD = 32^o$ . . . $\angle BAD = 50^o$
Let $x = AD,\;y = CD.$
. . We want: $y + 25$

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

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

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

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

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

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

3. Originally Posted by checkmarks
could someone help me with these problems please? ...

2. from the top of a 120m fire tower, a fire ranger observes smoke in 2 locations. 1 has an angle of depression of 6 degrees, and the other has an angle of depression of 3 degrees. calculate the distance between the smoke sightings when they're on the same side of the tower, and in line with the tower.

thanks.
Hello,

you have 2 right triangles. The distance between the fire tower and the spot with smoke is marked x resp. y. The distance between the two fires is
d = y - x

Corresponding (and equal) angles are marked in the same colour.

$x=\frac{120\rm{ m}}{\tan{6^\circ}}\approx 1141.7\rm{ m}$

$y=\frac{120\rm{ m}}{\tan{3^\circ}}\approx 2289.7\rm{ m}$

Thus the distance is d = 1148 m

EB

Merry Christmas and a happy New Year!

4. thank you both :] those were really helpful