a man goes 10 m due east and then 24m due north find the distance from the starting point
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AB = 10 m , BC = 24 m and AC = X m.
By Pythagoras theroem ,
AC² = AB² + BC²
AC² = (10)² + (24)²
AC² = 676
AC = √676 = 26 m.
By Pythagoras theroem ,
AC² = AB² + BC²
AC² = (10)² + (24)²
AC² = 676
AC = √676 = 26 m.
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