Two forces having magnitude A and (A)/(2) are perpendicular to each...

Using parallelogram law of vector addition, the formula can be written as

${A}_{net} = \sqrt{{{A}_{1}}^{2} + {A}^{2}2 + 2{A}_{1}{A}_{2}cos\theta}$

The data given is

\left|\overset{\to}{{A}_{1}}\right| = A, \left|\overset{\to}{{A}_{2}}\right| = \dfrac{A}{2}$</p><p>Since the vectors are perpendicular$\cos 90 ° = 0$</p><p>$⇒ {F}_{net} = \sqrt{{A}^{2} + \dfrac{{A}^{2}}{4}}= \dfrac{\sqrt{5}A}{2}