1. ## Trig Word Problem

Two forest fire towers, A and B are 20.3km apart. The bearing from A to B is N70degreesE. The ranger in each tower observes a fire and radios the fire's bearing from the tower. The bearing from tower A is N25degreesE. From tower B, the bearing is N15degreesW. How far is the fire from each tower?

2. Hello, casey_k!

We need a good sketch for this problem . . .

Two forest fire towers, $\displaystyle A$ and $\displaystyle B$ are 20.3 km apart.
The bearing from $\displaystyle A$ to $\displaystyle B$ is $\displaystyle N70^oE$

The ranger in each tower observes a fire and radios the fire's bearing from the tower.
The bearing from tower $\displaystyle A$ is $\displaystyle N25^oE.$
From tower $\displaystyle B$, the bearing is $\displaystyle N15^oW.$

How far is the fire from each tower?
Code:
      N               P
|         F     |
|         *     |
|        / \    |
|       /40°\15°|
|      /     \  |
|     /       \ |
|    /         \|
|25°/       95° * B
|  /        *   |
| / 45° *    70°|
|/  *  20.3     |
A *               Q
We are given: .$\displaystyle AB \,=\, 20.3,\;\;\angle NAB \:=\: 70^o \:=\: \angle ABQ$

The fire is at $\displaystyle F.$
$\displaystyle \angle NAF = 25^o \quad\Rightarrow\quad \angle FAB = 45^o$

Since $\displaystyle \angle PBF = 15^o,\;\angle ABQ = 70^o$, then $\displaystyle \angle FBA = 95^o$ . . . then: .$\displaystyle \angle F = 40^o$

In $\displaystyle \Delta AFB$. use the Law of Sines:

. . $\displaystyle \frac{AF}{\sin95^o} \:=\:\frac{20.3}{\sin40^o} \quad\Rightarrow\quad AF \:\approx\:31.5\text{ km}$

. . $\displaystyle \frac{BF}{\sin45^o} \:=\:\frac{20.3}{\sin40^o} \quad\Rightarrow\quad BF \:\approx\:22.3\text{ km}$