Answer
(a) $54.1^\circ$
(b) $48.3^\circ$, $32.2^\circ$, $24.5^\circ$, see explanations
Work Step by Step
(a) Given $\beta=10^\circ, n=3$, use the formula to get $\theta=sin^{-1}(\frac{1}{(2\times3+1)tan(10^\circ)})\approx54.1^\circ$
(b) For $\beta=15^\circ, n=2$, $\theta=sin^{-1}(\frac{1}{(2\times2+1)tan(15^\circ)})\approx48.3^\circ$
For $\beta=15^\circ, n=3$, $\theta=sin^{-1}(\frac{1}{(2\times3+1)tan(15^\circ)})\approx32.2^\circ$
For $\beta=15^\circ, n=4$, $\theta=sin^{-1}(\frac{1}{(2\times4+1)tan(15^\circ)})\approx24.5^\circ$
When $n=0$, $\frac{1}{(2\times0+1)tan(15^\circ)}=3.73\gt1$
When $n=1$, $\frac{1}{(2\times1+1)tan(15^\circ)}=1.24\gt1$
Both values are outside the domain of $sin^{-1}$ and can not give any valid $\theta$.
