Question

a man goes 30 km due north and 40 km you East how far away is he from its initial position​

Answer 1

$\huge\frak{\colorbox{orange}{ \red{\dag \:Answer \dag}}}$

50 km.

$\large\bold\color{tan}{\underbrace{Explanation}:-}$

We can apply Pythagoras theorem here as directions are at right angles to each other.

His distance from his initial point is:-

$\small {(distance \:from \:initial\: point )}^{2} = {(30)}^{2} + {(40)}^{2}$

$= > 900 + 1600$

$\small = > {(distance \: from\:initial \:point )}^{2} = 2500$

$\small = >distance\:from\:initial \:point= \sqrt{2500} =$

$=> 50 km$

So, the man is 50 km away from his initial point.

Answer 2

Answer:

\huge\frak{\colorbox{orange}{ \red{\dag \:Answer \dag}}}

50 km.

\large\bold\color{tan}{\underbrace{Explanation}:-}

We can apply Pythagoras theorem here as directions are at right angles to each other.

His distance from his initial point is:-

\small {(distance \:from \:initial\: point )}^{2}  =  {(30)}^{2} +  {(40)}^{2}  

=  > 900 + 1600

\small =  >   {(distance \: from\:initial \:point )}^{2} =  2500

\small =  >distance\:from\:initial \:point \sqrt{2500}  = 50 \: km

So, the man is 50 km away from his initial point.

Step-by-step explanation:

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