Answer
$F_{R}$=12.5KN
Directional Angle = $64.1^{0}$
Work Step by Step
Scalar Notation:
Summing the force components along x and y axes algebraically
$F_{Rx}$ = 4 + 5cos 45° - 8sin 15° = 5.465 kN
$F_{Ry}$= 5sin 45° + 8cos 15° = 11.263 kN
Magnitude:
$F_{R}$ = $\sqrt (F_{Rx})^{2}+(F_{Ry})^{2}$
$F_{R}$ = $\sqrt (5.465)^{2}+(11.263)^{2}$
$F_{R}$ = 12.5KN
And the directional angle of $F_{R}$ measured counterclockwise from the positive x-axis is
= $tan^{-1}$ [$\frac{F_{Ry}}{F_{Rx}}$]
= $tan^{-1}$ [$\frac{11.263}{5.465}$]
= $64.1^{0}$
