Answer
$55.46^{\circ}$
Work Step by Step
Lets assume the height of insect above the water is H and the horizontal distance to the water from the fish is L. Therefore, we can write the equations for the water and position of insect as:
$H=d\sin \varphi ;L=d\cos \varphi \Rightarrow \dfrac {H}{L}=\tan \varphi \left.....( 1\right) $
Lets assume the initial velocity of water is $v_{0}$ and $α$ is the initial angle of water relative to the horizontal line. So,
$H=H_{\max }=\dfrac {v^{2}_{0}\sin ^{2}\alpha }{2g} ..........(2)$ since water hits the insect at the top of its parabolic path and $L=\dfrac {L_{\max }}{2}=\dfrac {2v^{2}_{0}\sin \alpha \cos \alpha }{2g}=\dfrac {v^{2}_{0}\sin \alpha \cos \alpha }{g} .......(3)$
From equations 1,2,3; we get $\dfrac {H}{L}=\dfrac {\tan \alpha }{2}=\tan \varphi \Rightarrow \alpha =\arctan \left( 2\tan \varphi \right) \approx 55.46^{\circ}$
