Answer
$F_{R}$ = $\sqrt {F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}CosU}$
Angle X= $tan^{-1}[\frac{F_{1}sin U}{F_{2} + F_{1}CosU}]$
Work Step by Step
By using $Cosine Law:$
$F_{R}$ = $\sqrt {F_{1}^{2}+F_{2}^{2}-2F_{1}F_{2}Cos(180-U)}$
"U" is the angle between $F_{1}$ and $F_{2}$
since $Cos(180-U) = -CosU$
$F_{R}$ = $\sqrt {F_{1}^{2}+F_{2}^{2}-2F_{1}F_{2}(-CosU)}$
$F_{R}$ = $\sqrt {F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}CosU}$ Ans
$X$ is the direction angle made by $F_{R}$
tan$X$= $[\frac{F_{1}sin U}{F_{2} + F_{1}CosU}]$
Angle $X$= $tan^{-1}[\frac{F_{1}sin U}{F_{2} + F_{1}CosU}] Ans$
