Question

a man goes 15m due west and 8 m north find how far from its starting point

Answer 1

first the man goes to the west 15 m from point A to point B and then it turns north from point B and walks 8 m to point C.

This forms a right angled triangle ABC
So,by pythagoras theorem

$AC^{2} = AB^{2}+ BC^{2}$

SO, 
$AC^{2} = 15^{2} + 8^{2}$
$AC^{2} = 289$
$AC^{2}=225+64$

$AC= \sqrt{289}$
AC = 17 metres.

So the man is now 17metres away from his starting point.

Answer 2

Answer:

first the man goes to the west 15 m from point A to point B and then it turns north from point B and walks 8 m to point C.

This forms a right angled triangle ABC

So,by pythagoras theorem

SO, 

AC = 17 metres.

So the man is now 17metres away from his starting point

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Step-by-step explanation: