SOLUTION: Find the distance in nautical miles between Manila and San Francisco. Manila is located at 14° 36’N latitude and 121° 05’ E longitude. San Francisco is situated at 37° 48�

Question 1165481: Find the distance in nautical miles between Manila and San Francisco. Manila is located at 14° 36’N latitude and 121° 05’ E
longitude. San Francisco is situated at 37° 48’ N latitude and 122° 24’ W longitude.
a. 7,856.2 nautical miles b. 5,896.2 nautical miles c. 6,326.2 nautical miles d. 6,046.2 nautical miles
Answer by CPhill(2138) (Show Source): You can put this solution on YOUR website!
This is a great-circle distance problem, calculated using the spherical law of cosines. The distance obtained is directly in nautical miles because the Earth's radius is often approximated as 3,440 nautical miles, and distances on a great circle are often measured by the central angle $\theta$ (in minutes, where $1' = 1 \text{ nautical mile}$).
### 1. Convert Coordinates to Decimal Degrees
First, convert the coordinates from degrees/minutes to decimal degrees.
$$1' = \frac{1^\circ}{60}$$
#### Manila (M)
* Latitude ($\phi_M$): $14^\circ 36' \text{ N} = 14 + \frac{36}{60} = 14.60^\circ$
* Longitude ($\lambda_M$): $121^\circ 05' \text{ E} = 121 + \frac{5}{60} \approx 121.0833^\circ$
#### San Francisco (SF)
* Latitude ($\phi_{SF}$): $37^\circ 48' \text{ N} = 37 + \frac{48}{60} = 37.80^\circ$
* Longitude ($\lambda_{SF}$): $122^\circ 24' \text{ W} = - (122 + \frac{24}{60}) = -122.40^\circ$ (West is negative)
### 2. Calculate the Difference in Longitude ($\Delta\lambda$)
$$\Delta\lambda = \lambda_M - \lambda_{SF}$$
$$\Delta\lambda = 121.0833^\circ - (-122.40^\circ) = 121.0833^\circ + 122.40^\circ$$
$$\mathbf{\Delta\lambda = 243.4833^\circ}$$
Since the shortest distance around the globe is sought, we must use the smaller arc. $360^\circ - 243.4833^\circ = 116.5167^\circ$.
$$\mathbf{\Delta\lambda_{min} = 116.5167^\circ}$$
### 3. Use the Spherical Law of Cosines
The formula for the central angle ($\theta$) between two points on a sphere is:
$$\cos(\theta) = \sin(\phi_M)\sin(\phi_{SF}) + \cos(\phi_M)\cos(\phi_{SF})\cos(\Delta\lambda_{min})$$
$$\cos(\theta) = \sin(14.60^\circ)\sin(37.80^\circ) + \cos(14.60^\circ)\cos(37.80^\circ)\cos(116.5167^\circ)$$
Using the trigonometric values:
* $\sin(14.60^\circ) \approx 0.2520$
* $\sin(37.80^\circ) \approx 0.6129$
* $\cos(14.60^\circ) \approx 0.9678$
* $\cos(37.80^\circ) \approx 0.7903$
* $\cos(116.5167^\circ) \approx -0.4466$
$$\cos(\theta) \approx (0.2520)(0.6129) + (0.9678)(0.7903)(-0.4466)$$
$$\cos(\theta) \approx 0.1544 - 0.3418$$
$$\mathbf{\cos(\theta) \approx -0.1874}$$
$$\theta = \arccos(-0.1874) \approx 100.803^\circ$$
### 4. Convert Angle to Nautical Miles (NM)
The great-circle distance ($D$) in nautical miles is found by converting the angle $\theta$ to minutes of arc (since 1 minute of arc along a great circle equals 1 nautical mile).
$$D = \theta \times 60$$
$$D = 100.803^\circ \times 60 \text{ minutes/degree}$$
$$D \approx 6048.18 \text{ nautical miles}$$
The calculated distance is $\mathbf{6,048.2}$ nautical miles, which is closest to option **d**.
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The correct option is **d. 6,046.2 nautical miles**.
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