explain how to conpitmpute the angle $60\backslash$deg becoause not given the question idonI don't kknow

the beam shown, it carries a trapezoidal load which varies from $2k\frac{N}{m}$ at $A$ to a max of $3k\frac{N}{m}$ at $C,$ then decreases to $2k\frac{N}{m}$ at the right support $B.$ A clockwise moment of $600kN.m$ is apptied at point $C.$

Determine the reaction at B$.$

Determine the vertical component of the reaction at $A.$

Determine the horizontal component of the reaction at A

Solution:

Reaction it B$.$

$P=\frac{(2+3)}{2}(12)(2)$

$P=5(12)$

$P=60kN$

$2M_4=0$

$R_6$Cos$60(24)=60(12)+600$

$R_0=110kN$