# A man goes 80 m due east and then 150 m due north. How far is he from the starting point?  Let AB = 80m and AC = 150m

In ∆CAB, using Pythagoras Theorem,

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

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

(150)2 + (80)2 = (BC)2

(BC)2 = 22500 + 6400

(BC)2 = 28900

BC = ±170

BC = 170 [taking positive square root]

Hence, the man is 170 m far from the starting point.

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