1. **State the problem:** We have two airplanes, A and B, both flying directly toward the same airport. Airplane A is 20 miles from the airport. The pilot of A sees B at a 45° angle to the right. The pilot of B sees A at a 50° angle to the left. We need to find the distance from airplane B to the airport. 2. **Identify the triangle and angles:** Let the airport be point C, airplane A be point A, and airplane B be point B. - Distance AC = 20 miles (distance from A to airport). - Angle at A between line AC and AB is 45° (B is 45° to the right of A's path). - Angle at B between line BC and BA is 50° (A is 50° to the left of B's path). 3. **Find the third angle of triangle ABC:** Sum of angles in triangle = 180° $$\angle C = 180^\circ - 45^\circ - 50^\circ = 85^\circ$$ 4. **Use Law of Sines to find distance BC (distance from B to airport):** Law of Sines states: $$\frac{AC}{\sin \angle B} = \frac{BC}{\sin \angle A} = \frac{AB}{\sin \angle C}$$ We know: - $AC = 20$ - $\angle A = 45^\circ$ - $\angle B = 50^\circ$ - $\angle C = 85^\circ$ Using the ratio: $$\frac{20}{\sin 50^\circ} = \frac{BC}{\sin 45^\circ}$$ 5. **Solve for BC:** $$BC = \frac{20 \times \sin 45^\circ}{\sin 50^\circ}$$ 6. **Calculate the sines:** $$\sin 45^\circ = \frac{\sqrt{2}}{2} \approx 0.7071$$ $$\sin 50^\circ \approx 0.7660$$ 7. **Substitute values:** $$BC = \frac{20 \times 0.7071}{0.7660} = \frac{14.142}{0.7660} \approx 18.47$$ **Final answer:** The distance from airplane B to the airport is approximately **18.47 miles**.