Answer
$Θ=21.3^0$
$F_1 = 869 N$
Work Step by Step
* Summing Forces in X direction, We get
$800 Sin 60^0 = F_1 Sin(60^0 + Θ) - \frac{12}{13}(180) $
$F_1Sin(60^0+Θ) = 858.9N$
* Summing Forces in Y direction, We get
$800 Cos 60^0 = F_1 Cos(60^0 + u) + 200 +\frac{5}{13}(180)$
$F_1Cos(60^0+Θ) = 130.76N$
$Finding$ $Θ:$
$\frac{Sin(60^0+Θ)}{Cos(60^0+Θ)}=\frac{858.9/F_1}{130.76/F_1}$
$tan(60^0+Θ)=\frac{858.9}{130.76}$
$60^0 + Θ=81.34^0$
$Θ=21.34$
Then
$F_1=\frac{858.9}{Sin(81.34^0)}$
$F_1 = 869N$
