\(\sum_{}^{}\)Fy=0
TBCY+TABY=W
TBCsin\(\theta\)+TABsin\(\theta\)=W
Pero ya se que TBC=TAB
TBCsin\(\theta\)+TBCsin\(\theta\)=W
2TBCsin\(\theta\)=W
sin\(\theta\)=\(\frac{W}{2TBC}\)=\(\frac{3433.5}{13340}\)
\(\theta\)=sin-1(\(\frac{3433.5N}{13340N}\))
=15°