A prism of angle 70° is made of glass of refractive index 1.5…

Step 1. The critical angle. $\sin C = \dfrac{1}{1.5} = 0.6667$, so $C = 41.8^\circ$.

Step 2. The largest prism angle. The angle $r_1$ cannot be larger than $C$ either, because $\sin r_1 = \dfrac{\sin i}{n}$ and $\sin i \le 1$. So $r_1 \le C$ and $r_2 \le C$. Since $r_1 + r_2 = A$, we need $A \le 2C = 83.6^\circ$.

Step 3. The smallest angle of incidence. The ray just leaves when $r_2 = C$. Then $r_1 = A - C = 70^\circ - 41.8^\circ = 28.2^\circ$. Snell's law at the first face gives $\sin i = n\sin r_1 = 1.5 \times 0.4726 = 0.709$, so $i = 45.1^\circ$.

Step 4. The deviation there. At that moment $e = 90^\circ$, so $\delta = i + e - A = 45.1^\circ + 90^\circ - 70^\circ = 65.1^\circ$.

Step 5. The minimum deviation. $n = \dfrac{\sin\frac{A + \delta_m}{2}}{\sin\frac{A}{2}}$, so $\sin\dfrac{A + \delta_m}{2} = 1.5 \times \sin 35^\circ = 0.8604$. Then $\dfrac{A + \delta_m}{2} = 59.4^\circ$, so $\delta_m = 48.7^\circ$.