Step 1: Conceptual Foundation:
When three forces acting at a single point are in equilibrium, their vector sum is zero. This implies that the three force vectors can form a closed triangle. The angles within this force triangle can be determined using the law of cosines. Alternatively, the equilibrium condition can be stated as the resultant of any two forces being equal in magnitude and opposite in direction to the third force.
Step 2: Applicable Formula/Methodology:
Let the three forces be denoted by \( \vec{F}_1, \vec{F}_2, \vec{F}_3 \). For equilibrium, the vector sum must be zero: \( \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0 \).
Rearranging this equation, we get \( \vec{F}_1 + \vec{F}_2 = -\vec{F}_3 \).
Taking the magnitude of both sides yields \( |\vec{F}_1 + \vec{F}_2| = |-\vec{F}_3| = F_3 \).
The magnitude of the resultant of two vectors, \( \vec{F}_1 \) and \( \vec{F}_2 \), is given by the formula \( R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta \), where \( \theta \) is the angle between \( \vec{F}_1 \) and \( \vec{F}_2 \).
Therefore, for equilibrium, \( F_3^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta \). This equation can be solved for \( \cos\theta \).
Step 3: Detailed Analysis and Calculation:
Consider the forces with magnitudes \( F_3 = 8 \) N, \( F_1 = 5 \) N, and \( F_2 = 4 \) N.
The condition \( \vec{F}_1 + \vec{F}_2 = -\vec{F}_3 \) implies that the resultant of \( \vec{F}_1 \) and \( \vec{F}_2 \) must have a magnitude equal to \( F_3 \).
Using the resultant magnitude formula: \( |\vec{F}_1 + \vec{F}_2|^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta \).
Substituting the given magnitudes and the magnitude of the resultant (\( F_3 \)):
\[ 8^2 = 5^2 + 4^2 + 2(5)(4)\cos\theta \]
\[ 64 = 25 + 16 + 40\cos\theta \]
\[ 64 = 41 + 40\cos\theta \]
\[ 64 - 41 = 40\cos\theta \]
\[ 23 = 40\cos\theta \]
\[ \cos\theta = \frac{23}{40} \]
This value of \( \cos\theta \) represents the cosine of the angle between \( \vec{F}_1 \) and \( \vec{F}_2 \) when placed tail-to-tail. However, the provided options suggest that the question might be interpreted differently.
Revisiting the Law of Cosines applied directly to the force triangle, where the sides of the triangle represent the magnitudes of the forces (8, 5, and 4). Let \( \gamma \) be the angle *inside* the triangle opposite the side of length 8. The Law of Cosines states:
\[ 8^2 = 5^2 + 4^2 - 2(5)(4)\cos\gamma \]
\[ 64 = 25 + 16 - 40\cos\gamma \]
\[ 64 = 41 - 40\cos\gamma \]
\[ 23 = -40\cos\gamma \]
\[ \cos\gamma = -\frac{23}{40} \]
The angle \( \theta \) between the vectors when placed tail-to-tail is related to the internal angle \( \gamma \) of the triangle by \( \theta = 180^\circ - \gamma \). Therefore, \( \cos\theta = \cos(180^\circ - \gamma) = -\cos\gamma = -(-\frac{23}{40}) = \frac{23}{40} \).
Given that the provided options have negative cosines, it is highly probable that the question intends for "the angle between the forces" to refer to the internal angle of the force triangle as determined by the Law of Cosines, where the angle is opposite the resultant force. In this context, we are looking for \( \gamma \).
Thus, \( \cos\gamma = -\frac{23}{40} \).
Step 4: Conclusive Result:
Based on the interpretation that the question seeks the internal angle of the force triangle opposite the force of magnitude 8 N, the cosine of this angle is \( -\frac{23}{40} \). The angle is therefore \( \cos^{-1}\left(-\frac{23}{40}\right) \).