# Thread: How Far is the Ranger

1. ## How Far is the Ranger

A forest ranger is walking on a path inclined at 5 degrees to the horizontal directly toward a 100-foot-tall fire observation tower. The angle of elevation from the path to the top of the tower is 40 degrees. How far is the ranger from the top of the tower at this time?

2. Hello, magentarita!

I assume you made a sketch . . .

A forest ranger is walking on a path inclined at 5° to the horizontal
directly toward a 100-foot fire observation tower.
The angle of elevation from the path to the top of the tower is 40°.
How far is the ranger from the top of the tower at this time?
Code:
                              * A
* |
*   |
* 50° | 100
*       |
d  *         |
*           |
*         95° |
*               * B
* 35°       *  85°|
*       *           |
*   *   5°            |
R * - - - - - - - - - - - * C

The tower is: $\displaystyle AB = 100$
The ranger is at $\displaystyle R.$

$\displaystyle \angle BRC = 5^o;\;\;\angle ARC = 40^o \quad\Rightarrow\quad \angle ARB = 35^o$

In right triangle $\displaystyle BRC\!:\;\;\angle BRC = 5^o \quad\Rightarrow\quad \angle RBC = 85^o \quad\Rightarrow\quad \angle ABR = 95^o$

In $\displaystyle \Delta ABR\!:\;\;\angle ARB = 35^o,\;\angle ABR = 95^o \quad\Rightarrow\quad \angle RAB = 50^o$

Let $\displaystyle d \:=\:RA$

In $\displaystyle \Delta ABR$, the Law of Sines: .$\displaystyle \frac{d}{\sin95^o} \:=\:\frac{100}{\sin35^o}$

Therefore: .$\displaystyle d \;=\;\frac{100\sin95^o}{\sin35^o} \;=\;173.6812454 \;\approx\;\boxed{173.7\text{ feet}}$

3. ## Great...

Originally Posted by Soroban
Hello, magentarita!

I assume you made a sketch . . .

Code:
                              * A
* |
*   |
* 50° | 100
*       |
d  *         |
*           |
*         95° |
*               * B
* 35°       *  85°|
*       *           |
*   *   5°            |
R * - - - - - - - - - - - * C
The tower is: $\displaystyle AB = 100$
The ranger is at $\displaystyle R.$

$\displaystyle \angle BRC = 5^o;\;\;\angle ARC = 40^o \quad\Rightarrow\quad \angle ARB = 35^o$

In right triangle $\displaystyle BRC\!:\;\;\angle BRC = 5^o \quad\Rightarrow\quad \angle RBC = 85^o \quad\Rightarrow\quad \angle ABR = 95^o$

In $\displaystyle \Delta ABR\!:\;\;\angle ARB = 35^o,\;\angle ABR = 95^o \quad\Rightarrow\quad \angle RAB = 50^o$

Let $\displaystyle d \:=\:RA$

In $\displaystyle \Delta ABR$, the Law of Sines: .$\displaystyle \frac{d}{\sin95^o} \:=\:\frac{100}{\sin35^o}$

Therefore: .$\displaystyle d \;=\;\frac{100\sin95^o}{\sin35^o} \;=\;173.6812454 \;\approx\;\boxed{173.7\text{ feet}}$
Great work as always. Tell me, how do you make this pictures?