Question

select the correct answer.

fire station b is 15 miles due east of fire station a. firefighters at station a spot a fire at \\(n\\ 60^\circ\\ e\\), or \\(30^\circ\\). firefighters at station b spot the same fire at \\(n\\ 40^\circ\\ w\\), or \\(320^\circ\\).

what is the approximate distance between station b and the fire?

11.7 miles
7.6 miles
29.5 miles
6.4 miles

Explanation:

Identify the triangle properties

We have a triangle formed by station A, station B, and the fire. Let:

• \(A\) be the angle at station A: \(A = 30^\circ\)
• \(B\) be the angle at station B: \(B = 50^\circ\)
• \(c\) be the distance between station A and station B: \(c = 15\text{ miles}\)
• \(a\) be the distance between station B and the fire (opposite to angle \(A\))

Calculate the third angle

Using the triangle angle sum theorem:

$$ F = 180^\circ - (30^\circ + 50^\circ) = 100^\circ $$

Apply the Law of Sines

To find the distance \(a\) between station B and the fire:

$$ \frac{a}{\sin(A)} = \frac{c}{\sin(F)} $$

Using Right Triangle Trigonometry principles extended to general triangles:

$$ \frac{a}{\sin(30^\circ)} = \frac{15}{\sin(100^\circ)} $$

Solve for the distance

Isolating \(a\):

$$ a = \frac{15 \cdot \sin(30^\circ)}{\sin(100^\circ)} $$

Since \(\sin(30^\circ) = 0.5\) and \(\sin(100^\circ) \approx 0.9848\):

$$ a \approx \frac{15 \cdot 0.5}{0.9848} \approx 7.615\text{ miles} $$

The distance is approximately \(7.6\text{ miles}\).

Answer:

• 11.7 miles
• 7.6 miles (Correct answer)
• 29.5 miles
• 6.4 miles