If A = B + C and magnitudes of A , B and C are 5, 4 and 3 units...

Given $\vec{A} = \vec{B} + \vec{C}$

Rearranging the terms, we get:

$\vec{B} = \vec{A} - \vec{C}$

Taking the dot product of $\vec{B}$ with itself:

$|\vec{B}|^2 = |\vec{A} - \vec{C}|^2$

$|\vec{B}|^2 = |\vec{A}|^2 + |\vec{C}|^2 - 2|\vec{A}||\vec{C}|\cos\theta$

where $\theta is the angle between$\vec{A}$and$\vec{C}$.<br><br>Substituting the given magnitudes A = 5$, $B = 4$, and $C = 3$:

4^2 = 5^2 + 3^2 - 2(5)(3)\cos\theta$<br><br>$16 = 25 + 9 - 30\cos\theta$<br><br>$16 = 34 - 30\cos\theta$<br><br>$30\cos\theta = 18$<br><br>$\cos\theta = \dfrac{18}{30} = \dfrac{3}{5}$<br><br>$\theta = \cos^{-1} \left(\dfrac{3}{5}\right)

Answer: \cos^{-1}\left({3}/{5}\right)