Question
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Answer
In $\triangle AMO$
$\operatorname{COS} 30^{\circ}=\frac{A O}{M O}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{A O}{6}$
$\Rightarrow A O=5.20\ m $
In $\triangle BNO$
$\sin 47^\circ = \frac{O B}{N O}$
$\Rightarrow 0.73=\frac{O B}{5}$
$\Rightarrow O B=3.6\ m$
$A B=O A+O B$
$\Rightarrow A B=5,20+3.65$
$\Rightarrow A B=8.85 \ m$
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