To solve this problem, we need to break down the aircraft's journey into two legs and use vector addition to find the resultant distance and bearing from the starting point.

1. **First Leg:**
- Distance: 600 km
- Bearing: 100 degrees

We can convert this bearing into a standard Cartesian coordinate system where 0 degrees is along the positive x-axis (East), 90 degrees is along the positive y-axis (North), etc.

Bearing of 100 degrees means 10 degrees south of east. We can use trigonometry to find the x and y components of this leg:
- $ x_1 = 600 \cos(100^\circ) $
- $ y_1 = 600 \sin(100^\circ) $

2. **Second Leg:**
- Distance: 400 km
- Bearing: 160 degrees

Bearing of 160 degrees means 20 degrees east of south. Again, we use trigonometry to find the x and y components of this leg:
- $ x_2 = 400 \cos(160^\circ) $
- $ y_2 = 400 \sin(160^\circ) $

3. **Total Displacement:**
- Sum the x-components: $ x_{\text{total}} = x_1 + x_2 $
- Sum the y-components: $ y_{\text{total}} = y_1 + y_2 $

4. **Resultant Distance:**
- Use the Pythagorean theorem to find the resultant distance from the origin:
$$
\text{Distance} = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2}
$$

5. **Resultant Bearing:**
- Use the arctangent function to find the angle of the resultant vector:
$$
\theta = \tan^{-1}\left(\frac{y_{\text{total}}}{x_{\text{total}}}\right)
$$
- Adjust the angle to the correct bearing format.

Let's calculate each step:

1. **First Leg Components:**
- $ x_1 = 600 \cos(100^\circ) = 600 \cos(80^\circ) \approx 600 \times 0.1736 \approx 104.16 $ km (East)
- $ y_1 = 600 \sin(100^\circ) = 600 \sin(80^\circ) \approx 600 \times 0.9848 \approx 590.88 $ km (North)

2. **Second Leg Components:**
- $ x_2 = 400 \cos(160^\circ) = 400 \cos(20^\circ) \approx 400 \times (-0.9397) \approx -375.88 $ km (West)
- $ y_2 = 400 \sin(160^\circ) = 400 \sin(20^\circ) \approx 400 \times 0.3420 \approx 136.80 $ km (North)

3. **Total Displacement:**
- $ x_{\text{total}} = 104.16 - 375.88 \approx -271.72 $ km
- $ y_{\text{total}} = 590.88 + 136.80 \approx 727.68 $ km

4. **Resultant Distance:**
$$
\text{Distance} = \sqrt{(-271.72)^2 + (727.68)^2} \approx \sqrt{73832.19 + 529528.22} \approx \sqrt{603360.41} \approx 776.72 \text{ km}
$$

5. **Resultant Bearing:**
$$
\theta = \tan^{-1}\left(\frac{727.68}{-271.72}\right) \approx \tan^{-1}(-2.68) \approx -69.44^\circ
$$
Since the angle is negative and we are in the second quadrant (North-West), we add 180 degrees to get the bearing:
$$
\text{Bearing} = 180^\circ - 69.44^\circ \approx 110.56^\circ
$$

Therefore, the distance of the aircraft from its initial starting point is approximately 776.72 km, and the bearing is approximately 110.56 degrees.

Question

To solve this problem, we need to break down the aircraft's journey into two legs and use vector addition to find the resultant distance and bearing from the starting point.

1. **First Leg:**
   - Distance: 600 km
   - Bearing: 100 degrees

We can convert this bearing into a standard Cartesian coordinate system where 0 degrees is along the positive x-axis (East), 90 degrees is along the positive y-axis (North), etc.

Bearing of 100 degrees means 10 degrees south of east. We can use trigonometry to find the x and y components of this leg:
   - $ x_1 = 600 \cos(100^\circ) $
   - $ y_1 = 600 \sin(100^\circ) $

2. **Second Leg:**
   - Distance: 400 km
   - Bearing: 160 degrees

Bearing of 160 degrees means 20 degrees east of south. Again, we use trigonometry to find the x and y components of this leg:
   - $ x_2 = 400 \cos(160^\circ) $
   - $ y_2 = 400 \sin(160^\circ) $

3. **Total Displacement:**
   - Sum the x-components: $ x_{\text{total}} = x_1 + x_2 $
   - Sum the y-components: $ y_{\text{total}} = y_1 + y_2 $

4. **Resultant Distance:**
   - Use the Pythagorean theorem to find the resultant distance from the origin:
     $$
     \text{Distance} = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2}
     $$

5. **Resultant Bearing:**
   - Use the arctangent function to find the angle of the resultant vector:
     $$
     \theta = \tan^{-1}\left(\frac{y_{\text{total}}}{x_{\text{total}}}\right)
     $$
   - Adjust the angle to the correct bearing format.

Let's calculate each step:

1. **First Leg Components:**
   - $ x_1 = 600 \cos(100^\circ) = 600 \cos(80^\circ) \approx 600 \times 0.1736 \approx 104.16 $ km (East)
   - $ y_1 = 600 \sin(100^\circ) = 600 \sin(80^\circ) \approx 600 \times 0.9848 \approx 590.88 $ km (North)

2. **Second Leg Components:**
   - $ x_2 = 400 \cos(160^\circ) = 400 \cos(20^\circ) \approx 400 \times (-0.9397) \approx -375.88 $ km (West)
   - $ y_2 = 400 \sin(160^\circ) = 400 \sin(20^\circ) \approx 400 \times 0.3420 \approx 136.80 $ km (North)

3. **Total Displacement:**
   - $ x_{\text{total}} = 104.16 - 375.88 \approx -271.72 $ km
   - $ y_{\text{total}} = 590.88 + 136.80 \approx 727.68 $ km

4. **Resultant Distance:**
   $$
   \text{Distance} = \sqrt{(-271.72)^2 + (727.68)^2} \approx \sqrt{73832.19 + 529528.22} \approx \sqrt{603360.41} \approx 776.72 \text{ km}
   $$

5. **Resultant Bearing:**
   $$
   \theta = \tan^{-1}\left(\frac{727.68}{-271.72}\right) \approx \tan^{-1}(-2.68) \approx -69.44^\circ
   $$
   Since the angle is negative and we are in the second quadrant (North-West), we add 180 degrees to get the bearing:
   $$
   \text{Bearing} = 180^\circ - 69.44^\circ \approx 110.56^\circ
   $$

Therefore, the distance of the aircraft from its initial starting point is approximately 776.72 km, and the bearing is approximately 110.56 degrees.

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