Theodore Foster
(1752-1828)

American politician (1752-1828) – Theodore Foster was born in Brookfield (Town in Massachusetts, USA) on April 29th, 1752 and died in Providence (Capital city of Rhode Island, United States) on January 13th, 1828 at the age of 75. Today Theodore Foster would be 272 years old.

Age

how old was Theodore Foster when he died?

75

Theodore Foster died in 1828 at the age of 75.


Biographical data

Birthday April 29th, 1752 (Saturday)
Place of Birth Brookfield
Town in Massachusetts, USA
Death Date January 13th, 1828 (Sunday)
Death place Providence
Capital city of Rhode Island, United States
Birth sign (Zodiac) Taurus (The Bull) ♉
Chinese Zodiac Monkey 猴

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Also born in 1752

Theodore Foster

Birthday

When was Theodore Foster born?

Theodore Foster was born on April 29th, 1752.

Death

When did Theodore Foster die?

Theodore Foster died on January 13th, 1828 at the age of 75 in Providence (Capital city of Rhode Island, United States). Today Theodore Foster would be 272 years old.

Place of Birth

Where was Theodore Foster born?

Brookfield (Town in Massachusetts, USA).

Death place

Where did Theodore Foster die?

Providence (Capital city of Rhode Island, United States).

Zodiac

What is the zodiac sign of Theodore Foster?

Theodore Foster was born in the zodiac sign Taurus (The Bull).

Chinese Zodiac

Which Chinese zodiac sign does Theodore Foster have?

Theodore Foster was born in 1752 in the year of the Chinese zodiac Monkey.

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Died on January 13th, 1828:
Anniversaries of death

American politician (1752-1828)

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