The sum of the magnitudes of two forces is $18\text{N}$ and the magnitude of their resultant is $12\text{N}$. If the resultant makes an angle of ${90}^{\circ }$ with the smaller force, then find the magnitude of the forces.

The correct option is A $5\text{N},13\text{N}$
Let us say the magnitudes of the two vectors are $P$ and $Q$, where $P<Q$

As per the question,

$P+Q=18$

$\Rightarrow P=18-Q\dots \dots \left(1\right)$

and, resultant

$R=12=\sqrt{P^2+Q^2+2PQ\cos \theta }\dots \dots \left(2\right)$

If $\varphi$ is the angle between smaller force $P$ and resultant $R$,

$\tan \varphi =\frac{Q\sin \theta }{P+Q\cos \theta }$

Given: $\varphi ={90}^{\circ }$

$\Rightarrow \frac{Q\sin \theta }{P+Q\cos \theta }=\tan {90}^{\circ }=\infty$

$\Rightarrow P+Q\cos \theta =0\dots \dots \left(3\right)$

From eq. $\left(1\right)\&\left(3\right)$, we have

$Q(1-\cos \theta )=18\dots \dots \left(4\right)$

Also, from eqn. $\left(1\right)$ and $\left(2\right)$, we have

$R^2=P^2+Q^2+2PQ\cos \theta$

$\Rightarrow R^2=P^2+Q^2+2PQ+2PQ\cos \theta -2PQ$

$(P+Q{)}^2-R^2=2PQ(1-\cos \theta )$

$2PQ(1-\cos \theta )={18}^2-{12}^2=180$

$PQ(1-\cos \theta )=90\dots \dots \left(5\right)$

Dividing $\left(5\right)$ by $\left(4\right)$, we get

$P=5\text{units}$

and from $\left(1\right)$

$Q=13\text{units}$

Hence, option $\left(\text{A}\right)$ is the correct answer.