The vector sum of three forces having magnitudes $| \overrightarrow F_1 | = 100 N$, ;& ;$| \overrightarrow F_2 | = 80 N$ &  $| \overrightarrow F_3 | = 60 N$ acting on a particle is zero. the angle between $\overrightarrow F_1$  & $\overrightarrow F_2$ is nearly:-

$\text { For equilibrium:} \\\begin{array}{c}\left|\vec{F}_{1}+\vec{F}_{2}\right|=\left|\vec{F}_{3}\right| \\\end{array}$

$\begin{array}{c}\text { Squaring both side }\\\Rightarrow\left|\vec{F}_{1}+\vec{F}_{2}\right|^{2}=\left|\vec{F}_{3}\right|^{2}\\\Rightarrow\left|\vec{F}_{1}\right|^{2}+\left|\vec{F}_{2}\right|^{2}+2\left|F_{1}\right|\left|F_{2}\right|\cos\theta\\=\left|\vec{F}_{3}\right|^{2} \\\Rightarrow 100^{2}+80^{2}+2\times 100 \times 80\times\cos\theta\\=60^{2}\end{array}$

$\begin{aligned}\Rightarrow \cos\theta&=\frac{-12800}{16000} \\\Rightarrow \cos \theta&=\frac{-4}{5}\end{aligned}$

$\theta=143.13^{\circ}$