Answer
$\theta_1=74.1^o$
Work Step by Step
$n_1\sin(\theta_1)=n_2\sin(\theta_2)$
$\theta_2=90-\frac{\phi}{2}$
$\theta_1=\arcsin(\frac{n_2\sin(\theta_2)}{n_1})$
$=\arcsin(1.58\sin(90-(90-\frac{75^o}{2})))=74.1^o$
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