San Angelo coordinates
Texas, United States
Latitude, Longitude: 31.4638, -100.437
Elevation: 1843 (ft)
San Angelo coordinates in different formats
| Format | Latitude, Longitude |
|---|---|
| Simple decimal standard | 31.4638, -100.437 |
| Decimal Degrees (DD) | 31.4638° N, 100.437° W |
| Degrees and Decimal Minutes (DDM) | 31°27.828′ N, 100°26.22′ W |
| Degrees, Minutes and Seconds (DMS) | 31°27′49.7″ N, 100°26′13.2″ W |
Where is San Angelo
San Angelo is located at latitude 31.4638 and longitude -100.437 at an elevation of 1843 feet above sea level.
City information
| City | San Angelo |
| Country | United States |
| State | Texas |
| Population | 100,450 (2017) |
| Language | English |
| Currency | USD (Dollar) |