Question

A man goes 24 metre due east and then 10 metre due north how far is he away from his initial position​

Answer 1

Answer:

Distance from initial position can be found with the help of the Pythagoras theorem.

If, AB= 24m

   BC= 10m

Then AC= ?

$} AC^{2} = AB^{2} + BC^{2} \\\\\\AC^{2} = 24^{2} + 10^{2} \\\\AC = \sqrt{24^{2} + 10^{2}$

$AC = \sqrt{576+100} \\\\AC = 26 m$