1. **State the problem:** Two fire-lookout stations A and B are 140 miles apart, with A directly south of B. The fire is spotted with bearings N55°E from A and S60°E from B. We need to find the distance from the fire to each station. 2. **Set up the coordinate system:** Place station B at the origin $(0,0)$ and station A at $(0,-140)$ since A is directly south of B. 3. **Find the coordinates of the fire:** - From station A, bearing N55°E means the fire is located at an angle 55° east of north from A. - From station B, bearing S60°E means the fire is located at an angle 60° east of south from B. 4. **Express the fire's coordinates from each station:** Let the fire be at point $F=(x,y)$. From A at $(0,-140)$: $$x = d_A \sin 55^\circ$$ $$y = -140 + d_A \cos 55^\circ$$ where $d_A$ is the distance from A to the fire. From B at $(0,0)$: $$x = d_B \sin 60^\circ$$ $$y = -d_B \cos 60^\circ$$ where $d_B$ is the distance from B to the fire. 5. **Set the coordinates equal since both represent the same point $F$:** $$d_A \sin 55^\circ = d_B \sin 60^\circ$$ $$-140 + d_A \cos 55^\circ = -d_B \cos 60^\circ$$ 6. **Solve the system:** From the first equation: $$d_A = d_B \frac{\sin 60^\circ}{\sin 55^\circ}$$ Substitute into the second equation: $$-140 + \left(d_B \frac{\sin 60^\circ}{\sin 55^\circ}\right) \cos 55^\circ = -d_B \cos 60^\circ$$ 7. **Simplify:** $$-140 + d_B \frac{\sin 60^\circ \cos 55^\circ}{\sin 55^\circ} = -d_B \cos 60^\circ$$ Bring all $d_B$ terms to one side: $$-140 = -d_B \cos 60^\circ - d_B \frac{\sin 60^\circ \cos 55^\circ}{\sin 55^\circ}$$ $$-140 = -d_B \left( \cos 60^\circ + \frac{\sin 60^\circ \cos 55^\circ}{\sin 55^\circ} \right)$$ Multiply both sides by $-1$: $$140 = d_B \left( \cos 60^\circ + \frac{\sin 60^\circ \cos 55^\circ}{\sin 55^\circ} \right)$$ 8. **Calculate the trigonometric values:** $$\sin 55^\circ \approx 0.8192$$ $$\cos 55^\circ \approx 0.5736$$ $$\sin 60^\circ \approx 0.8660$$ $$\cos 60^\circ = 0.5$$ 9. **Evaluate the expression:** $$\cos 60^\circ + \frac{\sin 60^\circ \cos 55^\circ}{\sin 55^\circ} = 0.5 + \frac{0.8660 \times 0.5736}{0.8192} = 0.5 + \frac{0.4967}{0.8192} = 0.5 + 0.6063 = 1.1063$$ 10. **Find $d_B$:** $$d_B = \frac{140}{1.1063} \approx 126.5$$ miles 11. **Find $d_A$:** $$d_A = d_B \frac{\sin 60^\circ}{\sin 55^\circ} = 126.5 \times \frac{0.8660}{0.8192} = 126.5 \times 1.057 = 133.7$$ miles **Final answer:** - Distance from station A to fire: approximately $133.7$ miles - Distance from station B to fire: approximately $126.5$ miles