# A man goes 15 m due west and then 8 m due north. How far is he from the starting point?

Let AB = 15m and AC = 8m

In ∆CAB, using Pythagoras Theorm,

(Perpendicular)2 + (Base)2 = (Hypotenuse)2

(AC)2 + (AB)2 = (BC)2

(8)2 + (15)2 = (BC)2

(BC)2 = 64 + 225

(BC)2 = 289

BC = ±17

BC = 17 [taking positive square root]

Hence, the man is 17m far from the starting point.

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