Let's break down the problem step by step and **draw the situation** to help visualize it. --- ## **Step 1: Draw the Diagram** Let’s summarize the bearings: - **A to C**: S 58° E (i.e., 58° east of due south from A) - **A to B**: N 86° E (i.e., 86° east of due north from A) - **B to C**: S 32° W (i.e., 32° west of due south from B) Let’s draw point A somewhere, then: - From A, draw a line **N 86° E** for AB. This is almost due east (since due east is N 90° E). - From A, draw a line **S 58° E** for AC. This is 58° east of due south (or 32° south of due east). - From B, the bearing to C is **S 32° W**, so from B, draw a line 32° west of due south to C. Let’s label the points as A, B, and C. --- ## **Step 2: Find the Distance AB** Given: - Plane speed = 250 mph - Time from A to B = 2.6 hr \[ AB = 250 \times 2.6 = 650 \text{ miles} \] --- ## **Step 3: Find the Angles in Triangle ABC** **Bearings:** - **A to C:** S 58° E - Measured clockwise from north: S = 180°, plus 58° east, so bearing from north = 180° - 58° = 122° - **A to B:** N 86° E - Measured clockwise from north: 86° - **B to C:** S 32° W - Measured clockwise from north: 180° + 32° = 212° **Angle at A (between AB and AC):** - AB is N 86° E from A (almost due east). - AC is S 58° E from A. Let’s find the angle between these two legs at A. - The angle **from north to AB** = 86° - The angle **from north to AC** = 180° - 58° = 122° \[ \text{Angle at A} = 122^\circ - 86^\circ = 36^\circ \] --- ## **Step 4: Use Law of Sines** We have: - Side \( AB = 650 \) mi - Angle at A = 36° - We need to find \( AC \) But to use Law of Sines, we need another side or angle in triangle ABC. Let’s find the **angle at B**. **At B:** - The bearing from B to C is S 32° W (212° from north) - The bearing from A to B is N 86° E (86° from north) - The bearing from B to A is the reverse of A to B: Add 180°, so 86° + 180° = 266°. This is the bearing from B to A. The angle at B in triangle ABC is the **difference between the bearings from B to C and from B to A**: \[ \text{Angle at B} = 266^\circ - 212^\circ = 54^\circ \] --- ## **Step 5: Find Angle at C** The sum of angles in a triangle is 180°. \[ \text{Angle at C} = 180^\circ - 36^\circ - 54^\circ = 90^\circ \] --- ## **Step 6: Use Law of Sines** Let: - \( AB = 650 \) mi - Angle at C = 90° - Angle at A = 36° - Angle at B = 54° - \( AC = x \) \[ \frac{AB}{\sin C} = \frac{AC}{\sin B} \] \[ \frac{650}{\sin 90^\circ} = \frac{x}{\sin 54^\circ} \] \[ 650 = \frac{x}{\sin 54^\circ} \] \[ x = 650 \times \sin 54^\circ \] \[ \sin 54^\circ \approx 0.8090 \] \[ x = 650 \times 0.8090 \approx 526 \text{ mi} \] --- ## **Final Answer** \[ \boxed{526 \text{ mi}} \] --- ## **Summary Table** | Side | Opposite Angle | |-------------|---------------| | AB = 650 mi | C = 90° | | AC = ? | B = 54° | | BC = ? | A = 36° | --- ## **Diagram** ``` N | | | A---------B \ / \ / \ / C ``` - **A to B**: almost due east (N 86° E) - **A to C**: S 58° E, 36° down from AB - **Angle at A**: 36° - **Angle at B**: 54° - **Angle at C**: 90° --- ### **Boxed Answer** The distance from **A to C** is \[ \boxed{526\ \text{mi}} \]