This problem is about calculating the latitude of a vessel based on a given distance traveled due east and the change in longitude. The formula that relates distance, change in longitude, and latitude for a journey along a parallel of latitude on a spherical Earth is:

$d = \Delta \lambda \cdot \cos(\phi) \cdot \frac{111.32 \, \text{km}}{1^\circ}$

Where:

• $d$ is the distance traveled eastward (in nautical miles).
• $\Delta \lambda$ is the change in longitude in degrees.
• $\phi$ is the latitude (which we are solving for).
• 1 degree of longitude is approximately 111.32 km at the equator (we'll convert to nautical miles).
• 1 nautical mile = 1.852 km.

Given:

• Distance $d = 245 \, \text{NM}$
• Change in longitude $\Delta \lambda = 7^\circ 25' = 7 + \frac{25}{60} = 7.4167^\circ$

Step-by-Step Calculation:

1. Convert the distance into kilometers: $245 \, \text{NM} = 245 \times 1.852 = 453.74 \, \text{km}$

2. Now use the formula and rearrange to solve for $\cos(\phi)$: $453.74 = 7.4167 \cdot \cos(\phi) \cdot 111.32$

3. Simplify and solve for $\cos(\phi)$: $\cos(\phi) = \frac{453.74}{7.4167 \cdot 111.32} = \frac{453.74}{825.79} = 0.5494$

4. Find $\phi$: $\phi = \cos^{-1}(0.5494) \approx 56.74^\circ$

Answer:

The vessel was approximately at a latitude of $56.74^\circ$ North (or South depending on the direction).

Would you like any further clarification or details on this calculation?

Here are 5 related questions:

1. How would the formula change if the vessel was traveling along a meridian (north-south)?
2. What is the significance of using nautical miles in navigational problems?
3. How would a change in the Earth's radius affect the calculation of latitude?
4. Can this method be applied to vessels moving westward?
5. How does the curvature of the Earth influence long-distance navigational calculations?

Tip: Always ensure that the units used (nautical miles, kilometers, degrees) are consistent throughout calculations to avoid errors.