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Question

Find the value of $∠ r$ for the case shown in figure

${sin}^{- 1} (0.5)$
$75^{\circ} - {sin}^{- 1} (0.5)$
$90^{\circ}$
$60^{\circ}$

A

{sin}^{- 1} (0.5)

B

75^{\circ} - {sin}^{- 1} (0.5)

C

90^{\circ}

D

60^{\circ}

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Solution

Verified by Toppr

$∠ i = 45^{0}; ∠ r = ?$
$μ = \frac{s i n i}{s i n r}$
$s i n r = \frac{s i n i}{μ} = \frac{0.707}{\sqrt{2}} = 0.5$
$r = s i n^{- 1} (0.5)$
Hence, option A is correct.

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