A man goes 24m due east and then 10 m due north. how far is he away from his initial position
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By Pythagoras theorem.
AC^2 = AB^2 + BC^2
AC^2 = 24^2 + 10^2
AC^2 = 576 + 100
= 676
AC = 26.
The distance he traveled from his initial position = 26m.
AC^2 = AB^2 + BC^2
AC^2 = 24^2 + 10^2
AC^2 = 576 + 100
= 676
AC = 26.
The distance he traveled from his initial position = 26m.
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