Question

part 1 of 2 two fire - lookout stations are 260 miles apart, with station a directly south of station b. both stations spot a fire. the bearing of the fire from station a is \\(\text{n}55^{\circ}\text{e}\\) and the bearing of the fire from station b is \\(\text{s}60^{\circ}\text{e}\\). how far, to the nearest tenth of a mile, is the fire from each lookout station? the distance from station b to the fire is \\(\square\\) miles. (round to the nearest tenth)

Explanation:

Step1: Define triangle angles

Let station A be at point $A$, station B at point $B$, fire at point $C$.

• $AB = 260$ miles, $A$ is south of $B$.
• Bearing from $A$: $N55^\circ E$ → $\angle CAB = 90^\circ - 55^\circ = 35^\circ$
• Bearing from $B$: $S60^\circ E$ → $\angle CBA = 90^\circ - 60^\circ = 30^\circ$
• $\angle ACB = 180^\circ - 35^\circ - 30^\circ = 115^\circ$

Step2: Apply Law of Sines

Law of Sines: $\frac{BC}{\sin\angle CAB} = \frac{AB}{\sin\angle ACB}$
Solve for $BC$ (distance from B to fire):
$BC = \frac{AB \cdot \sin\angle CAB}{\sin\angle ACB}$

Step3: Substitute values

$BC = \frac{260 \cdot \sin(35^\circ)}{\sin(115^\circ)}$
$\sin(35^\circ) \approx 0.5736$, $\sin(115^\circ) \approx 0.9063$
$BC \approx \frac{260 \cdot 0.5736}{0.9063} \approx \frac{149.136}{0.9063} \approx 164.6$

Answer:

164.6