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A diver at a depth of 12 m in water $\\mu=\\dfrac{4}{3}$ sees the sky in a cone of semiverticale
angle :

$\sin^{-1}(\dfrac{4}{3})$
$\tan^{-1}(\dfrac{4}{3})$
$\sin^{-1}(\dfrac{3}{4})$
90°
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