1. **Problem statement:** Given a body under three parallel forces at a point with magnitudes 0, 4, and 5 Newtons, find the angle between the second and third forces. 2. **Understanding the problem:** The forces are parallel and act at a point. The problem asks for the angle between the second and third forces, which are 4 N and 5 N respectively. 3. **Key concept:** For three forces in equilibrium acting at a point, the vector sum is zero. The forces can be represented as vectors forming a closed triangle. 4. **Using the Law of Cosines:** If the forces are represented as vectors forming a triangle, the angle between the second and third forces can be found by: $$c^2 = a^2 + b^2 - 2ab \cos(\theta)$$ where $a=4$, $b=5$, and $c=0$ (since the first force is 0 N, the resultant is zero). 5. **Calculate the angle:** $$0^2 = 4^2 + 5^2 - 2 \times 4 \times 5 \times \cos(\theta)$$ $$0 = 16 + 25 - 40 \cos(\theta)$$ $$40 \cos(\theta) = 41$$ $$\cos(\theta) = \frac{41}{40}$$ This is impossible since cosine cannot be greater than 1, so the problem likely means the forces are 3, 4, and 5 Newtons (a common Pythagorean triple). 6. **Assuming forces are 3, 4, and 5 N:** $$5^2 = 3^2 + 4^2 - 2 \times 3 \times 4 \times \cos(\theta)$$ $$25 = 9 + 16 - 24 \cos(\theta)$$ $$25 = 25 - 24 \cos(\theta)$$ $$24 \cos(\theta) = 0$$ $$\cos(\theta) = 0$$ $$\theta = 90^\circ$$ 7. **Answer:** The angle between the second and third forces is $90^\circ$. **Final answer:** (ب) 90