Hi there. I have been struggling to understand this question. I got the answer of 221N but i'm not sure if it is correct. Help please

Let F_AB be the force in cable AB and F_BC be the force in cable BC.  Let θ_A be the angle between cable AB and the horizontal and θ_C be the angle between cable BC and the horizontal.

 

Horizontal force balance

F_ABcos(θ_A) = F_BCcos(θ_C)          ...(1)

Vertical force balance

F_ABsin(θ_A) = mg + F_BCsin(θ_C)  ...(2)

m = 2000kg; g = 9.81m/s².

 

We also have that θ_A = tan^-1(5.6/2.2) and θ_C = tan^-1(2.4/5.6)

 

Replace F_BC in (2) by using (1)

F_ABsin(θ_A) = mg + F_ABcos(θ_A)sin(θ_C)/cos(θ_C)

Rearrange to get F_AB

F_AB = mg/(sin(θ_A) - cos(θ_A)sin(θ_C)/cos(θ_C))

$${\mathtt{FAB}} = {\frac{{\mathtt{2\,000}}{\mathtt{\,\times\,}}{\mathtt{9.81}}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{5.6}}}{{\mathtt{2.2}}}}\right)}\right)}{\mathtt{\,-\,}}{\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{5.6}}}{{\mathtt{2.2}}}}\right)}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{2.4}}}{{\mathtt{5.6}}}}\right)}\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{2.4}}}{{\mathtt{5.6}}}}\right)}\right)}}}\right)}} \Rightarrow {\mathtt{FAB}} = {\mathtt{25\,347.418\: \!086\: \!936\: \!201\: \!807\: \!8}}$$

FAB ≈ 25.347 kN

 

Substitute this back into (1) to get F_BC

F_BC = F_ABcos(θ_A)/cos(θ_C)

$${\mathtt{FBC}} = {\frac{{\mathtt{25\,347.418}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{5.6}}}{{\mathtt{2.2}}}}\right)}\right)}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{2.4}}}{{\mathtt{5.6}}}}\right)}\right)}}} \Rightarrow {\mathtt{FBC}} = {\mathtt{10\,083.657\: \!337\: \!858\: \!801\: \!218\: \!7}}$$

FBC ≈ 10.083 kN