Let AB = 10m and AC = 24m
In ∆CAB, using Pythagoras Theorem,
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
⇒ (AC)2 + (AB)2 = (BC)2
⇒ (24)2 + (10)2 = (BC)2
⇒ (BC)2 = 576 + 100
⇒ (BC)2 = 676
⇒ BC = ±26
⇒ BC = 26 [taking positive square root]
Hence, the man is 26m far from the starting point.