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To solve for the **Dlat** (difference in latitude) and **Dep** (departure) made good from point A to point B, we can use trigonometry based on the given course and distance.

### Given:
- Course: 124°T (True)
- Distance: 210 Nautical Miles (NM)

We can calculate Dlat and Dep using the following trigonometric formulas:

\[
\text{Dlat} = \text{Distance} \times \cos(\text{Course})
\]
\[
\text{Dep} = \text{Distance} \times \sin(\text{Course})
\]

Both angles should be calculated in degrees.

### Step-by-Step Solution:

1. **Convert the course to radians** (if using a calculator that works in radians):
   \[
   \theta = 124^\circ
   \]
   
2. **Calculate Dlat**:
   \[
   \text{Dlat} = 210 \times \cos(124^\circ)
   \]

3. **Calculate Dep**:
   \[
   \text{Dep} = 210 \times \sin(124^\circ)
   \]

I'll now compute these values.

### Calculations:

1. **Dlat**:  
\[
Dlat = 210 \times \cos(124^\circ) \approx 210 \times (-0.5592) \approx -117.43 \, \text{NM}
\]

2. **Dep**:  
\[
Dep = 210 \times \sin(124^\circ) \approx 210 \times 0.8290 \approx 174.09 \, \text{NM}
\]

### Conclusion:
- **Dlat** ≈ -117.43 NM (moving southward, hence negative)
- **Dep** ≈ 174.09 NM

Would you like more details or have further questions?

### Related Questions:
1. How do you interpret negative latitude and departure values in navigation?
2. How does the course angle affect the Dlat and Dep values?
3. How would these calculations change if the course were in a different quadrant?
4. What is the relationship between course angle and the cardinal directions (N, E, S, W)?
5. How can these results be applied to plotting a course on a navigational chart?

### Tip:
Always ensure your calculator is set to the correct mode (degrees or radians) depending on the angle format provided in navigation problems.

Find the Dlat and Dep made good from A to B if the course is 124°T and the distance runs is 210 NM.

Learn how to calculate the difference in latitude (Dlat) and departure (Dep) using trigonometry in nautical navigation. Given a course of 124°T and a distance of 210 NM, this solution involves using the cosine and sine functions to find the respective values. Suitable for students in Grades 10-12.

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