Question

fire station b is 15 miles due east of fire station a. firefighters at station a spot a fire at n 60° e, or 30°. firefighters at station b spot the same fire at n 40° w, or 320°. what is the approximate distance between station b and the fire? 6.4 miles 7.6 miles 11.7 miles 29.5 miles

Explanation:

Step1: Find the third - angle of the triangle

The sum of angles in a triangle is $180^{\circ}$. Let the third angle be $\theta$. So, $\theta=180^{\circ}-(30^{\circ}+50^{\circ}) = 100^{\circ}$.

Step2: Apply the Law of Sines

The Law of Sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. Let the distance between station B and the fire be $x$. We know that the side between the two stations has length $15$ miles. So, $\frac{x}{\sin30^{\circ}}=\frac{15}{\sin100^{\circ}}$.

Step3: Solve for $x$

Cross - multiply to get $x=\frac{15\times\sin30^{\circ}}{\sin100^{\circ}}$. Since $\sin30^{\circ}=\frac{1}{2}=0.5$ and $\sin100^{\circ}\approx0.9848$, then $x=\frac{15\times0.5}{0.9848}=\frac{7.5}{0.9848}\approx7.6$ miles.

Answer:

$7.6$ miles