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  1. Geocoding
  2. Romania
  3. Bacău County
  4. Ferestrău-Oituz

Ferestrău-Oituz: latitude and longitude

Ferestrău-Oituz, Bacău County, Romania is at latitude 46.201598 and longitude 26.571734.

Latitude46.201598
Longitude26.571734
Decimal degrees46.201598, 26.571734
Degrees, minutes, seconds46°12'5.75"N 26°34'18.24"E
Geo URIgeo:46.201598,26.571734

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<iframe src="https://geloky.com/geocoding/embed/Ferestr%C4%83u-Oituz+Romania" width="100%" height="350"
  style="border:0" loading="lazy" title="Map of Ferestrău-Oituz Romania"></iframe>
<p>Map of <a href="https://geloky.com/geocoding/place/Ferestr%C4%83u-Oituz+Romania">Ferestrău-Oituz Romania GPS coordinates</a> by Geloky</p>

FAQ

Questions about Ferestrău-Oituz

The latitude of Ferestrău-Oituz is 46.201598 and the longitude is 26.571734. In degrees, minutes and seconds: 46°12'5.75"N 26°34'18.24"E.