Physics - Vectors Question with Solution | TestHub

Let the two forces be $F_1=3P$ and $F_2=\sqrt{3}P$.

The resultant force is given as $R=\sqrt{15}P$.

Using the formula for the magnitude of the resultant of two vectors:

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

Substituting the given values:

$(\sqrt{15}P{)}^2=(3P{)}^2+(\sqrt{3}P{)}^2+2(3P)(\sqrt{3}P)\cos \theta$

$15P^2=9P^2+3P^2+6\sqrt{3}{P}^2\cos \theta$

$15P^2=12P^2+6\sqrt{3}{P}^2\cos \theta$

$3P^2=6\sqrt{3}{P}^2\cos \theta$

$\cos \theta =\frac{3}{6\sqrt{3}}=\frac{1}{2\sqrt{3}}$

$\theta ={\cos }^{-1}\frac{1}{2\sqrt{3}}$

Answer: ${\cos }^{-1}\frac{1}{2\sqrt{3}}$