A man goes $10\ m$ due south and then $24\ m$ due west. How far is he from the starting point?
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starting from O, let the man goas from O to A and then A to B as shows in the figure.
Then, $\text{OA}=10\text{m},\text{AB}=24\text{m }$ and $\angle\text{OAB}=90^\circ$
Using Pythagoras theorem:
$OB^2 = OA^2+AB^{2}$
$ \Rightarrow OB^2= 10^2 + 24^{2}$
$ \Rightarrow OB^2 = 100 + 576$
$\Rightarrow OB^2 = 676$
$\Rightarrow\text{OB}=\sqrt{676}=26\text{m}$ Hence, the man is $26\ m$ south-west from the starting position.
Then, $\text{OA}=10\text{m},\text{AB}=24\text{m }$ and $\angle\text{OAB}=90^\circ$
Using Pythagoras theorem:
$OB^2 = OA^2+AB^{2}$
$ \Rightarrow OB^2= 10^2 + 24^{2}$
$ \Rightarrow OB^2 = 100 + 576$
$\Rightarrow OB^2 = 676$
$\Rightarrow\text{OB}=\sqrt{676}=26\text{m}$ Hence, the man is $26\ m$ south-west from the starting position.
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