Two forces, while acting on a particle in opposite; directions, have the resultant of $10N$. If they act at right angles to each other, the resultant is found to be $50N$. Find the two forces.

$\text{Step 1: Resultant force when the forces are opposite}$ $\text{[Ref. Fig. 1]}$

Let $F_1$ and $F_2$ are the magnitudes of two forces

$R_1=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \theta }$

$\Rightarrow 10N=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \theta {180}^0}$

$\Rightarrow 10=\sqrt{(F_1{)}^2+F_2^2-2F_1{F}_2}$

$\Rightarrow 10=F_1-F_2$ $.....\left(1\right)$

$\text{Step 2: Resultant force when the forces are at right angle}$ $\text{[Ref. Fig. 2]}$

$R_2=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos {90}^0}$

$\Rightarrow 50=\sqrt{F_1^2+F_2^2}$

$\Rightarrow 2500=F_1^2+F_2^2$ $....\left(2\right)$

$\text{Step 3: Solving equations}$

From eq $\left(1\right)$ substitute the value of $F_1$ in eq $\left(2\right)$

$2500=(F_2+10{)}^2+(F_2^2)$

$\Rightarrow F_2^2+10F_2-1200=0$

$\Rightarrow (F_2+40)(F_2-30)=0$

$F_2=30N$ [Since the magnitude of force can't be negative , so we will not consider negative value]

Substituting this value of $F_2$ in eq (1)

$10=F_1-30\Rightarrow F_1=40N$

Hence, the Magnitudes of two forces are $40N$ and $30N$.