Question

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

Explanation:

Step1: Find triangle's third angle

Let the fire be point $F$, west tower be $W$, east tower be $E$.
At $W$: angle inside $\triangle WFE$ is $90^\circ - 41^\circ = 49^\circ$
At $E$: angle inside $\triangle WFE$ is $90^\circ - 56^\circ = 34^\circ$
Third angle at $F$: $180^\circ - 49^\circ - 34^\circ = 97^\circ$

Step2: Apply Law of Sines

Law of Sines: $\frac{WE}{\sin\angle F} = \frac{WF}{\sin\angle E} = \frac{EF}{\sin\angle W}$
$WE = 15$ mi, so first solve for $WF$:
$\frac{15}{\sin97^\circ} = \frac{WF}{\sin34^\circ}$
$WF = \frac{15 \times \sin34^\circ}{\sin97^\circ}$

Step3: Calculate $WF$ value

$\sin34^\circ \approx 0.5592$, $\sin97^\circ \approx 0.9925$
$WF \approx \frac{15 \times 0.5592}{0.9925} \approx 8.47$ (Note: correction, recalculating precisely: $15*0.5592=8.388; 8.388/0.9925≈8.45$? No, wait, correct internal angles: actually, the angles from the towers are measured from east/west, so the angle at $W$ is $41^\circ$ from east towards north, so internal angle of triangle is $41^\circ$? No, re-express:
Correct triangle angles:

• The line between towers is east-west (15 mi).
• West tower sees fire $41^\circ$ north of east: so the angle between $WE$ (east direction) and $WF$ is $41^\circ$, so internal angle at $W$ is $41^\circ$.
• East tower sees fire $56^\circ$ north of west: angle between $EW$ (west direction) and $EF$ is $56^\circ$, so internal angle at $E$ is $56^\circ$.

Then angle at $F$: $180 - 41 - 56 = 83^\circ$. This is the correct angle set.

Step1 (Corrected): Find angle at $F$

$\angle F = 180^\circ - 41^\circ - 56^\circ = 83^\circ$

Step2 (Corrected): Law of Sines for $EF$

$\frac{EF}{\sin41^\circ} = \frac{15}{\sin83^\circ}$
$EF = \frac{15 \times \sin41^\circ}{\sin83^\circ}$

Step3 (Corrected): Calculate $EF$

$\sin41^\circ \approx 0.6561$, $\sin83^\circ \approx 0.9925$
$EF \approx \frac{15 \times 0.6561}{0.9925} \approx \frac{9.8415}{0.9925} \approx 9.91$ mi

Step4 (Corrected): Calculate $WF$

$\frac{WF}{\sin56^\circ} = \frac{15}{\sin83^\circ}$
$\sin56^\circ \approx 0.8290$
$WF \approx \frac{15 \times 0.8290}{0.9925} \approx \frac{12.435}{0.9925} \approx 12.53$ mi

Step5: Identify closest distance

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

Answer:

A. 9.91 mi