1. Problem statement: A body of weight 18 gm.wt is placed on a rough horizontal plane and a horizontal force F acts to make it about to move, and the angle between the resultant reaction force and the limiting static friction force is $\cos\theta=4/5$. Find the magnitude of the resultant reaction force. 2. Free-body and variables: Let $N$ be the normal reaction (vertical), $f$ the limiting static friction (horizontal), and $R$ the resultant reaction. Because friction is horizontal and normal is vertical, the components of $R$ are $R\cos\theta=f$ and $R\sin\theta=N$. 3. Formula and rule: The vertical equilibrium gives $N=18$. We use the component relation $$R\sin\theta=N$$ to solve for $R$. 4. Compute $\sin\theta$: Since $\cos\theta=4/5$ and $\sin\theta>0$ we have $$\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-(4/5)^2}=3/5$$. 5. Solve for $R$: From $$R\sin\theta=N$$ we get $$R=\frac{N}{\sin\theta}$$. 6. Substitute and simplify with cancellation: $$R=\frac{18}{3/5}=18\times\frac{5}{3}=\frac{18\times5}{3}$$ $$=\frac{\cancel{18}^{6}\times5}{\cancel{3}}=6\times5=30$$ 7. Final answer: The magnitude of the resultant reaction force is $30$.