Answer
$F_{AB}=98.6N$
$F_{AC}=267N$
Work Step by Step
We can find the required forces as follows:
$W=mg$
$\implies W=(20)(9.81)=1962.N$
and $\theta=tan^{-1}(\frac{1}{2})=26.6$
$\phi=tan^{-1}(\frac{2.5}{2})=51.3$
Now, the sum of the forces in the x-direction is given as
$\Sigma F_x=0$
$\implies -F_{AB}cos\phi-F_{AC} cos\theta+300=0$.eq(1)
and the sum of the forces in the y-direction is given as
$\Sigma F_y=0$
$\implies F_{AB}sin\phi+F_{AC}sin\theta-196.2=0$..eq(2)
Solving eq(1) and eq(2), we obtain:
$F_{AB}=98.6N$
and $F_{AC}=267N$
