1. **Problem statement:** We need to find the angle $v$ between two sides of lengths 3.7 cm and 4.8 cm such that the area of the triangle is 6.5 cm². 2. **Formula for the area of a triangle given two sides and the included angle:** $$\text{Area} = \frac{1}{2}ab\sin(v)$$ where $a=3.7$, $b=4.8$, and $v$ is the angle between them. 3. **Substitute the known values:** $$6.5 = \frac{1}{2} \times 3.7 \times 4.8 \times \sin(v)$$ 4. **Simplify the multiplication:** $$6.5 = \frac{1}{2} \times 17.76 \times \sin(v)$$ 5. **Calculate half of 17.76:** $$6.5 = 8.88 \times \sin(v)$$ 6. **Isolate $\sin(v)$:** $$\sin(v) = \frac{6.5}{8.88}$$ 7. **Simplify the fraction using cancellation:** $$\sin(v) = \frac{\cancel{6.5}}{\cancel{8.88}} \approx 0.7315$$ 8. **Find the angle $v$ by taking the inverse sine:** $$v = \arcsin(0.7315)$$ 9. **Calculate $v$ in degrees:** $$v \approx 47.1^\circ$$ 10. **Conclusion:** The angle $v$ should be approximately $47.1^\circ$ to get an area of 6.5 cm².