American painter (1836-1918) – Archibald Willard was born in Bedford (City in Cuyahoga County, Ohio, United States of America) on August 22nd, 1836 and died in United States of America (Country primarily located in North America) on October 11th, 1918 at the age of 82. Today Archibald Willard would be 187 years old.
Archibald Willard died in 1918 at the age of 82.
Birthday | August 22nd, 1836 (Monday) |
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Place of Birth | Bedford City in Cuyahoga County, Ohio, United States of America |
Death Date | October 11th, 1918 (Friday) |
Death place | United States of America Country primarily located in North America |
Birth sign (Zodiac) | Leo (The Lion) ♌ |
Chinese Zodiac | Monkey 猴 |
Archibald Willard was born on August 22nd, 1836.
Archibald Willard died on October 11th, 1918 at the age of 82 in United States of America (Country primarily located in North America). Today Archibald Willard would be 187 years old.
Bedford (City in Cuyahoga County, Ohio, United States of America).
United States of America (Country primarily located in North America).
Archibald Willard was born in the zodiac sign Leo (The Lion).
Archibald Willard was born in 1836 in the year of the Chinese zodiac Monkey.