find distance from each fire station to the forest fire

[Milyka09]

The Question is:
A fire is blazing in the woods. The two different fire stations H and G are 55 mies away from each other. The ranger at station H sites the fire at an angle of 35degrees and the ranger at station G sites the fire at an angle of 65degrees. What is the distance from EACH station to the Fire?

DISTANCE FROM H____________________________________- DISTANCE FROM G ___________________________________________________

 

[Loren]

The wording in the problem leaves something to be desired. But, try this.
Draw a sketch.
Draw a vertical line that represents a north-south line.
On the line label a point G.
On the line and below point G label another point H.
From point G draw a line that is approximately 65° from north.
From point H draw a line that is approximately 35° from north.
You now have a triangle with two angles and the included side.
With that information you should be able to use the law of cosines to find the length of the other two sides.
Note that 65° is not one of the angles of the triangle.

 

[soroban]

Hello, Milyka09!

The description is not complete . . .

A fire is blazing in the woods. The two fire stations \(\displaystyle H\) and \(\displaystyle G\) are 55 mies apart.
Station \(\displaystyle H\) sights the fire at an angle of \(\displaystyle 35^o\) and station \(\displaystyle G\) sights the fire at an angle of \(\displaystyle 65^o.\)
What is the distance from each station to the fire?

I will assume that the fire is between the two fire stations.

                        F
                        *
                     *80d *
                  *         *
               *              *
            *                   *
         * 35d                65d *
      *  *  *  *  *  *  *  *  *  *  *
      H             55              G

\(\displaystyle \text{We have: }\angle H = 35^o,\;\angle G = 65^o \quad\Rightarrow\quad \angle F = 80^o\)


Law of Sines

\(\displaystyle \frac{FH}{\sin65^o} \:=\:\frac{55}{\sin80^o}\quad\Rightarrow\quad FH \:=\:\frac{55\sin65^o}{\sin80^o}\;\approx\;50.6\text{ miles}\)

\(\displaystyle \frac{FG}{\sin35^o}\:=\:\frac{55}{\sin80^o} \quad\Rightarrow\quad FG \:=\:\frac{55\sin35^o}{\sin80^o} \;\approx\;32.0\text{ miles}\)