By Newton's second and third laws, we have the following net forces at
[tex]\sum F_x = -T_1 \cos(60^\circ) + T_2 \cos(60^\circ) = 0[/tex]
[tex]\sum F_y = T_1 \sin(60^\circ) + T_2 \sin(60^\circ) - 600 \,\mathrm N = 0[/tex]
Together, these equations tell us [tex]\boxed{T_1=T_2\approx 346 \,\mathrm N}[/tex].
[tex]\sum F_x = T_3 \cos(20^\circ) - T_2 \cos(60^\circ) - T_5 = 0[/tex]
[tex]\sum F_y = T_3 \sin(20^\circ) - T_2 \sin(60^\circ) = 0[/tex]
[tex]\sum F_x = -T_4 \cos(20^\circ) + T_1 \cos(60^\circ) + T_5 = 0[/tex]
[tex]\sum F_y = T_4 \cos(20^\circ) - T_1 \sin(60^\circ) = 0[/tex]
Combining the horizontal force equations for points B and C and using the fact that [tex]T_1=T_2[/tex] gives
[tex](T_3 \cos(20^\circ) - T_2 \cos(60^\circ) - T_5) + (-T_4 \cos(20^\circ) + T_1 \cos(60^\circ) + T_5) = 0 + 0[/tex]
[tex]\implies (T_3 - T_4) \cos(20^\circ) = 0 \implies T_3=T_4[/tex]
Then combining the vertical force equations for B and C, we find
[tex](T_3 \sin(20^\circ) - T_2 \sin(60^\circ)) + (T_4 \sin(20^\circ) - T_1 \sin(60^\circ)) = 0 + 0[/tex]
[tex]\implies 2 T_3 \sin(20^\circ) - 2 T_1 \sin(60^\circ) \implies T_3 = \dfrac{\sin(60^\circ)}{\sin(20^\circ)} T_1[/tex]
so that [tex]\boxed{T_3 = T_4 \approx 877 \,\mathrm N}[/tex]
Lastly solve for [tex]T_5[/tex] using either horizontal force equation for B or C.
[tex]-T_4 \cos(20^\circ) + T_1 \cos(60^\circ) + T_5 = 0[/tex]
[tex]\implies T_5 = \left(\dfrac{\sin(60^\circ)}{\sin(20^\circ)} - \cos(60^\circ)\right) T_1 \implies \boxed{T_5 \approx 651\,\mathrm N}[/tex]
