Physics - Vectors Question with Solution | TestHub

The magnitude of the resultant of two vectors $\vec{A}$ and $\vec{B}$ is given by $R=\sqrt{A^2+B^2+2AB\cos \theta }$. Given $F_1=6$ N, $F_2=8$ N, and $R=10$ N. Substituting these values, $10=\sqrt{6^2+8^2+2(6)(8)\cos \theta }$.$100=36+64+96\cos \theta ⟹100=100+96\cos \theta ⟹96\cos \theta =0⟹\cos \theta =0$. Therefore, $\theta =90^{\circ }$.