Chapter 3 - Equilibrium of a Particle - Section 3.3 - Coplanar Force Systems - Problems - Page 102: 24

$x=1.6m; y=1.98m$

We know that $2T cos\theta=F_2$ $\implies cos\theta=\frac{F_2}{2T}$ We plug in the known values to obtain: $cos\theta=\frac{1000}{2\times 800}=0.62$ $\implies \theta=51.31^{\circ}$ As $tan\theta=\frac{2}{x}$ $\implies x=\frac{2}{tan\theta}$ $\implies x=\frac{2}{tan 51.31^{\circ}}=1.6m$ Now $y=x tan\theta$ We plug in the known values to obtain: $y=1.6\times 1.24=1.98m$