Sacu

From Wikipedia, the free encyclopedia
  (Redirected from Tincova)
Jump to: navigation, search
For other uses, see Sacu (disambiguation).
Sacu
Commune
Location in Caraș-Severin County
Location in Caraș-Severin County
Sacu is located in Romania
Sacu
Sacu
Location in Romania
Coordinates: 45°34′N 22°07′E / 45.567°N 22.117°E / 45.567; 22.117
Country  Romania
County Caraş-Severin County
Population (2002)[1]
 • Total 1,681
Time zone EET (UTC+2)
 • Summer (DST) EEST (UTC+3)

Sacu (Hungarian: Szákul) is a commune in Caraş-Severin County, western Romania with a population of 1681 people. It is composed of three villages: Sacu, Sălbăgelu Nou (Gyulatelep) and Tincova (Tinkova).

References[edit]

  1. ^ (Romanian) "Sacu", at the Erdélyi Magyar Adatbank's Recensământ 2002; Retrieved on August 22, 2009

Coordinates: 45°34′N 22°07′E / 45.567°N 22.117°E / 45.567; 22.117