13.191. Two forces are applied to a material point, the angle between which is $30^{\circ}$. The magnitude of one of the applied forces is $7 \sqrt{3}$ times the magnitude of the other, and the magnitude of the resultant force is $24 \mathrm{N}$ greater than the magnitude of the smaller force. Determine the magnitude of the smaller force and the resultant force.
## Solution.

Let $x \mathrm{H}$ be the modulus of the smaller force (Fig. 13.11). By the cosine rule: $\quad(24+x)^{2}=x^{2}+(7 \sqrt{3} x)^{2}-2 x \cdot 7 \sqrt{3} x \cdot \cos \left(180^{\circ}-30^{\circ}\right)$, from which $x=2 \mathrm{H}$ is the modulus of the smaller force; $24+2=26 \mathrm{H}$ is the modulus of the resultant force.

Answer: 2 and $26 \mathrm{H}$.