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A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point?
Given:
A man goes 15 metres due west and then 8 metres due north.
To do:
The distance from the starting point.
Solution:
Let the starting position of the man be $O$. The position when he moves $15\ m$ due West be $A$ and the final position when he moves $8\ m$ due North be $B$.
In $∆ABO$,
By Pythagoras theorem,
$BO^2 = AB^2 + AO^2$
$BO^2 = 8^2 + 15^2$
$BO^2 = 64 + 225$
$BO= \sqrt{289}$
$BO = 17\ m$
Therefore, the man is $17\ m$ far from the starting point.
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