= Solution
For $n=1.5$, the <critical angle> is $i_c=\arcsin(1/1.5)\simeq41.8103^\circ$. Here $0\leq r<i_c<\alpha=60^\circ$, so the condition at B is $\alpha-r>i_c$. At the threshold,
$$
r_*=60^\circ-i_c\simeq18.1897^\circ,\qquad
\theta_*=\arcsin(1.5\sin r_*)\simeq27.9205^\circ.
$$
Thus the requested positive-angle interval for strict <total internal reflection> is
$$
\boxed{0<\theta<27.9205^\circ.}
$$
At the upper equality the transmitted ray is grazing; normal incidence $\theta=0$ also gives <total internal reflection> at B if the endpoint is included. The distinction follows the question's positive-angle convention.
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