Solve this question in good English and give the reason why we use sinO not conso and explain detaily

The student's partial work indicates that the $16 \text{ N}$ force contributes to an anticlockwise moment. We will follow the student's calculation for this moment:

$$M_{ACW} = 16 \cos 60^\circ \times \frac{a}{2}$$ $$M_{ACW} = 16 \times 0.5 \times \frac{a}{2} = 4a$$

Reason for using $\cos 60^\circ$: This implies that the $16 \text{ N}$ force is acting at a distance $\frac{a}{2}$ from the pivot, and its component perpendicular to the plank is $16 \cos 60^\circ$. This would be the case if the $16 \text{ N}$ force makes an angle of $60^\circ$ with the horizontal, and the plank is vertical ($\theta = 90^\circ$). In this scenario, the horizontal component of the force ($16 \cos 60^\circ$) is perpendicular to the vertical plank.

There are two forces causing clockwise moments:

1. Moment due to the weight of the plank (6 N): The weight acts vertically downwards at the center of the uniform plank, which is at a distance $\frac{a}{2}$ from the pivot. The plank is inclined at an angle $\theta$ to the horizontal. The perpendicular distance from the pivot to the line of action of this vertical force is the horizontal projection of the moment arm: $(\frac{a}{2}) \cos \theta$.

   $$M_{CW1} = 6 \text{ N} \times (\frac{a}{2}) \cos \theta = 3a \cos \theta$$

   Reason for using $\cos \theta$: The force (weight) is vertical. The moment arm is along the plank, which is at an angle $\theta$ to the horizontal. To find the perpendicular distance from the pivot to the line of action of the vertical force, we project the moment arm onto the horizontal axis. This projection is $r \cos \theta$.

2. Moment due to the 4 N force: This force acts at the end of the plank (distance $a$ from the pivot) and is shown acting perpendicular to the plank.

   $$M_{CW2} = 4 \text{ N} \times a = 4a$$

   Reason for not using sin/cos: The force is already given as acting perpendicular to the moment arm (the plank). Therefore, the full force contributes to the moment, and no trigonometric component is needed.

For the system to be in equilibrium, the total anticlockwise moment must equal the total clockwise moment:

$$M_{ACW} = M_{CW1} + M_{CW2}$$ $$4a = 3a \cos \theta + 4a$$

Subtract $4a$ from both sides of the equation:

$$0 = 3a \cos \theta$$

Since the length $a$ cannot be zero, we must have:

$$\cos \theta = 0$$

Therefore, the angle $\theta$ is:

$$\theta = \arccos(0)$$ $$\theta = 90^\circ$$

The angle $\theta$ is $\boxed{90^\circ}$.