Let AB = 80 m and AC = 150 m
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.