Answer
$F_x = 1.3\ kN$
$F_y = 2.6\ kN$
$F_z = 0.8\ kN$
Work Step by Step
We determine the magnitude of its three components as follows:
$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$
$\cos^2 \alpha + \cos^2 30° + \cos^2 75° = 1$
$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$
We rearrange and calculate:
$\alpha = 64.67°$
$F_x = 3\cos64.67° = 1.3\ kN$
$F_y = 3\cos30° = 2.6\ kN$
$F_z = 3\cos75° = 0.8\ kN$
