Answer
$F_{R} = 9.93 KN$
$F = 1.2 KN$
Work Step by Step
First we will calculate the sum of the two given forces (Q)
$Q = \sqrt (6^{2} + 8^{2}) = 10 KN$
$\theta = \tan^{-1}(\frac{8}{6}) = 53.13°$
For the resultant force to be as minimum as possible, it has to be perpendicular to F; therefore
$F_{R} = 10 \times \sin(53.13° + 30°) = 9.928 KN \approx 9.93 KN$
$F = 10 \times \cos(53.13° + 30°) = 1.196 KN = 1.2 KN$
