Question

solving a real-world problem
a forest ranger in the west observation tower spots a fire $41^{\circ}$ north of east. fifteen miles directly east, the forest ranger in the east tower spots the same fire at $56^{\circ}$ north of west. how far away is the ranger who is closest to the fire? approximate the distance by rounding to the nearest hundredth of a mile.
9.91 mi
11.87 mi
12.53 mi
18.95 mi
west tower
east tower

Explanation:

Step1: Find triangle's third angle

Let the fire be point $F$, west tower $W$, east tower $E$.
Sum of angles in $\triangle WFE$ is $180^\circ$.
$\angle F = 180^\circ - 41^\circ - 56^\circ = 83^\circ$

Step2: Apply Law of Sines to find $WE=15$

Law of Sines: $\frac{WF}{\sin 56^\circ} = \frac{EF}{\sin 41^\circ} = \frac{15}{\sin 83^\circ}$

Step3: Calculate distance from west tower

$WF = \frac{15 \times \sin 56^\circ}{\sin 83^\circ}$
$\sin 56^\circ \approx 0.8290$, $\sin 83^\circ \approx 0.9925$
$WF \approx \frac{15 \times 0.8290}{0.9925} \approx 12.53$ mi

Step4: Calculate distance from east tower

$EF = \frac{15 \times \sin 41^\circ}{\sin 83^\circ}$
$\sin 41^\circ \approx 0.6561$
$EF \approx \frac{15 \times 0.6561}{0.9925} \approx 9.91$ mi

Step5: Identify the closer distance

Compare $12.53$ mi and $9.91$ mi; $9.91$ is smaller.

Answer:

9.91 mi