To determine the distance and bearing of airport B from airport A, we will use vector components and the law of cosines.

Step 1: Break the journey into components

We break the two flight paths into their north-south (y) and east-west (x) components using trigonometry.

First leg: 300 km on a bearing of 348°

• x-component = $300 \sin 348^\circ$
• y-component = $300 \cos 348^\circ$

Using trigonometric values:

• $\sin 348^\circ = -\sin 12^\circ \approx -0.2079$
• $\cos 348^\circ = \cos 12^\circ \approx 0.9781$

So: $x_1 = 300 (-0.2079) = -62.37 \text{ km}$ $y_1 = 300 (0.9781) = 293.43 \text{ km}$

Second leg: 350 km on a bearing of 100°

• x-component = $350 \sin 100^\circ$
• y-component = $350 \cos 100^\circ$

Using trigonometric values:

• $\sin 100^\circ = \cos 10^\circ \approx 0.9848$
• $\cos 100^\circ = -\sin 10^\circ \approx -0.1736$

So: $x_2 = 350 (0.9848) = 344.68 \text{ km}$ $y_2 = 350 (-0.1736) = -60.76 \text{ km}$

Step 2: Find net displacement

Total x-displacement: $X = x_1 + x_2 = -62.37 + 344.68 = 282.31 \text{ km}$

Total y-displacement: $Y = y_1 + y_2 = 293.43 + (-60.76) = 232.67 \text{ km}$

Step 3: Find the distance from A to B

Using the Pythagorean theorem:

$d = \sqrt{X^2 + Y^2}$

$d = \sqrt{(282.31)^2 + (232.67)^2}$

$d = \sqrt{79700.54 + 54138.33} = \sqrt{133838.87} \approx 366 \text{ km}$

Step 4: Find the bearing of B from A

Bearing is given by:

$\theta = \tan^{-1} \left( \frac{X}{Y} \right)$

$\theta = \tan^{-1} \left( \frac{282.31}{232.67} \right)$

$\theta = \tan^{-1} (1.213) \approx 50.2^\circ$

Since the displacement is in the first quadrant (northeast), the final bearing is:

$\mathbf{N 50.2^\circ E} \quad \text{or} \quad \mathbf{50.2^\circ}$

Final Answer

• Distance from A to B: 366 km
• Bearing from A to B: 50.2°

Would you like a step-by-step diagram for visualization? Let me know!

Related Questions:

1. How do you determine the bearing when moving in different quadrants?
2. What is the significance of using trigonometric components in navigation problems?
3. How can we use the Law of Cosines instead of vector components in this case?
4. How would the calculation change if wind speed affected the plane’s movement?
5. How can we determine the shortest path a plane should take to return to A?

Tip: Always check whether your bearing angle makes sense based on the movement in the coordinate plane.