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

Explanation

Starting from A, let the man go from A to B and then from B to C, as shown in the figure. Then,
`AB=15m, BC=8 m and angle ABC=90^(@)`
From right `Delta ABC`, we have
` AC^(2)=AB^(2)+BC^(2)`
`={(15)^(2)+(8)^(2)} m^(2)`
`=(225+64)=m^(2)`
`=289 m^(2)`
`:. AC= sqrt(289)m=17 m`.

Hence, the man is 17 m away from the starting poition.