Let the two forces be $F_1$ and $F_2$, where $F_1$ is the smaller force.

Given: $F_1 + F_2 = 18$ and the magnitude of their resultant is $R = 12$

Use the condition that the resultant is perpendicular to the smaller force.

If the resultant $\vec R = \vec F_1 + \vec F_2$ is perpendicular to $\vec F_1$, then $\vec R \cdot \vec F_1 = 0$

So, $(\vec F_1 + \vec F_2) \cdot \vec F_1 = 0$ $F_1^2 + F_1 F_2 \cos\theta = 0$ $F_1 + F_2\cos\theta = 0 \quad \text{(after dividing by }F_1\text{)}$

Hence, $\cos\theta = -\frac{F_1}{F_2}$

Write the formula for the resultant magnitude.

$R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta$

Substitute $\cos\theta = -\dfrac{F_1}{F_2}$: $R^2 = F_1^2 + F_2^2 + 2F_1F_2\left(-\frac{F_1}{F_2}\right)$ $R^2 = F_1^2 + F_2^2 - 2F_1^2$ $R^2 = F_2^2 - F_1^2$

Since $R=12$, $144 = F_2^2 - F_1^2$ $144 = (F_2 - F_1)(F_2 + F_1)$

Use the given sum $F_1 + F_2 = 18$:

$144 = (F_2 - F_1)(18)$ $F_2 - F_1 = 8$

Solve the two equations:

$F_2 + F_1 = 18$ $F_2 - F_1 = 8$

Adding: $2F_2 = 26 \Rightarrow F_2 = 13$

Then: $F_1 = 18 - 13 = 5$

Therefore, the magnitudes of the two forces are $13N \text{ and } 5N$

Check with options:

This matches Option B: $13N, 5N$.

Comparison with stored correct answer:

Stored correct answer is B, which matches our derived answer.