Question

A man goes 3km due north and then 4 km due
east. How far is he away from his initial position?​

Answer 1

Step-by-step explanation :5km

let his initial position is A

and 3km to North point B and then 4 km to east point C and the distance of A to C is hypotenuse of the right angled triangle formed

so

${h}^{2} = {p}^{2} + {b}^{2} \\ {h}^{2} = {4}^{2} + {3}^{2} \\ {h}^{2} = 16 + 9 \\ {h}^{2} = 25 \\ h = \sqrt{25} \\ h = 5$

So the distance is 5km .

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