SOLUTION: Airport B is 320 miles from airport A on a bearing of S40°E. A pilot wishes to fly from A to B, but to avoid a storm must first fly due East at a speed of 210 mph for an hour, and

Question 1173735: Airport B is 320 miles from airport A on a bearing of S40°E. A pilot wishes to fly from A to B, but to avoid a storm must first fly due East at a speed of 210 mph for an hour, and then from this point (call it C) turns to fly to B. Find the distance, to the nearest mile, and the bearing, to the nearest degree, that the pilot must fly to airport B?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this problem step-by-step.
**1. Visualize the Problem**
* **A to B:** 320 miles, bearing S40°E
* **A to C:** Due East, 210 mph for 1 hour
* **C to B:** We need to find the distance and bearing.
**2. Calculate the Distance from A to C**
* Distance = Speed * Time
* Distance AC = 210 mph * 1 hour = 210 miles
**3. Set Up a Coordinate System**
* Let A be the origin (0, 0).
* Since AC is due East, point C is (210, 0).
* To find the coordinates of B, we need to use the distance and bearing from A to B.
**4. Find the Coordinates of B**
* Bearing S40°E means 40 degrees east of south.
* We can use trigonometry to find the coordinates:
* x-coordinate (Eastward): 320 * sin(40°)
* y-coordinate (Southward): 320 * cos(40°)
* Calculate:
* x = 320 * sin(40°) ≈ 205.67 miles
* y = 320 * cos(40°) ≈ 245.13 miles
* Since it's S40°E, the coordinates of B are (205.67, -245.13).
**5. Find the Distance from C to B**
* Use the distance formula:
* CB = √[(x2 - x1)² + (y2 - y1)²]
* CB = √[(205.67 - 210)² + (-245.13 - 0)²]
* CB = √[(-4.33)² + (-245.13)²]
* CB = √[18.75 + 60088.66]
* CB = √60107.41
* CB ≈ 245.17 miles
* Rounded to the nearest mile, CB ≈ 245 miles.
**6. Find the Bearing from C to B**
* We need to find the angle between the East direction from C and the line CB.
* Use the tangent function:
* tan(θ) = |y-coordinate| / |x-coordinate|
* tan(θ) = 245.13 / 4.33
* tan(θ) ≈ 56.61
* θ = arctan(56.61) ≈ 88.99 degrees
* Since B is south and slightly west of C, the bearing is approximately S89°W.
**Final Answers**
* **Distance (CB):** Approximately 245 miles
* **Bearing (C to B):** Approximately S89°W

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