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A man walking along a straight line with a velocity $6 \mathrm{~km} / \mathrm{h}$ encounters rain falling vertically down with a velocity $6 \sqrt{3} \mathrm{~km} / \mathrm{h}$. At what angle the man should hold his umbrella so that he can protect himself from rain
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Verified Answer
The correct answer is:
$30^{\circ}$ with respect to vertical
$$
\mathrm{v}_{\mathrm{V}}=6 \sqrt{3} \mathrm{~km} / \mathrm{h}
$$
$$
\begin{aligned}
& \tan \alpha=\frac{6}{6 \sqrt{3}}=\frac{1}{\sqrt{3}} \\
& \tan \alpha=\tan 30^{\circ}
\end{aligned}
$$
$\alpha=30^{\circ}$ with respect to vertical
\mathrm{v}_{\mathrm{V}}=6 \sqrt{3} \mathrm{~km} / \mathrm{h}
$$
$$
\begin{aligned}
& \tan \alpha=\frac{6}{6 \sqrt{3}}=\frac{1}{\sqrt{3}} \\
& \tan \alpha=\tan 30^{\circ}
\end{aligned}
$$
$\alpha=30^{\circ}$ with respect to vertical
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