Rays of the Sun are seen to make a \(\displaystyle{\left(\angle{r}={43}^{{\circ}}\right)}\) angle to the vertical beneaththe water.

Let incident angle be \(\displaystyle{\left(\angle{i}\right)}\)

Apply --------Snell's Law

refractive index of water = \(\displaystyle\frac{{\sin{{i}}}}{{\sin{{r}}}}\)

so, \(\displaystyle{\sin{{i}}}={1.33}\cdot{\sin{{r}}}\)

\(\displaystyle{i}={65.1}^{{\circ}}\)

Hence ; The angle above the horizon = \(\displaystyle{90}^{{\circ}}-{65.1}^{{\circ}}\)

\(\displaystyle={24.89}^{{\circ}}\)

Let incident angle be \(\displaystyle{\left(\angle{i}\right)}\)

Apply --------Snell's Law

refractive index of water = \(\displaystyle\frac{{\sin{{i}}}}{{\sin{{r}}}}\)

so, \(\displaystyle{\sin{{i}}}={1.33}\cdot{\sin{{r}}}\)

\(\displaystyle{i}={65.1}^{{\circ}}\)

Hence ; The angle above the horizon = \(\displaystyle{90}^{{\circ}}-{65.1}^{{\circ}}\)

\(\displaystyle={24.89}^{{\circ}}\)