1. **Problem statement:** Calculate angle $\angle B$ in trapezium ABCD where $AB=8$, $DC=2.5$, $AD=3.2$, and $\angle A=60^\circ$.\n\n2. **Known facts and formula:** In trapezium ABCD, $AB \parallel DC$. We want $\angle B$, adjacent to side $AB$. Use the sine and cosine rules in triangle $ABD$.\n\n3. **Step 1: Find length $BD$ using Law of Cosines in $\triangle ABD$:**\n$$BD^2 = AB^2 + AD^2 - 2 \times AB \times AD \times \cos(60^\circ)$$\n$$= 8^2 + 3.2^2 - 2 \times 8 \times 3.2 \times 0.5 = 64 + 10.24 - 25.6 = 48.64$$\n$$BD = \sqrt{48.64} \approx 6.97$$\n\n4. **Step 2: Use Law of Sines to find $\angle B$:**\n$$\frac{\sin(\angle B)}{AD} = \frac{\sin(60^\circ)}{BD}$$\n$$\sin(\angle B) = \frac{AD \times \sin(60^\circ)}{BD} = \frac{3.2 \times \frac{\sqrt{3}}{2}}{6.97} \approx \frac{2.77}{6.97} = 0.397$$\n\n5. **Step 3: Calculate $\angle B$:**\n$$\angle B = \arcsin(0.397) \approx 23.4^\circ$$\n\n**Final answer:** $\angle B \approx 23.4^\circ$ rounded to one decimal place.