How Far is the Ranger

**magentarita** · September 12th 2008, 06:19 PM

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?

**Soroban** · September 12th 2008, 06:55 PM

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: $AB = 100$
The ranger is at $R.$

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

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

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

Let $d \:=\:RA$

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

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

**magentarita** · September 13th 2008, 04:27 AM

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: $AB = 100$
The ranger is at $R.$

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

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

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

Let $d \:=\:RA$

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

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

Great work as always. Tell me, how do you make this pictures?

Thread: How Far is the Ranger

LinkBack

Thread Tools

Display

How Far is the Ranger

Great...

Search Tags