= Solution
The acute <angles> follow from part (ii) and the <Snell's law> relation:
$$
\boxed{\theta=\pi/6=30^\circ,\qquad\phi=\pi/4=45^\circ.}
$$
The path <lengths> below and above the interface are respectively
$$
\ell_1=\frac{\sqrt3}{\cos(\pi/6)}=2,\qquad\ell_2=\frac1{\cos(\pi/4)}=\sqrt2.
$$
Hence the total distance and minimum travel <time> are
$$
\boxed{\ell=2+\sqrt2,\qquad T_{\min}=\frac2{c/2}+\frac{\sqrt2}{c/\sqrt2}=\frac6c.}
$$
For a sketch to scale, choose the interface as $y=0$, with $A=(0,-\sqrt3)$ and $B=(2,1)$. The crossing is $C=(1,0)$ because $d_1\tan\theta=d_2\tan\phi=1$. The figure uses equal axis scales and shows the two normal-relative <angles>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1-q20-fastest-path.png]
{title=The time-minimising two-segment path in Question 20, drawn with equal coordinate scales}
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