(w_x = 10 \cos 210^\circ) (\cos 210^\circ = \cos(180^\circ + 30^\circ) = -\cos 30^\circ = -\frac\sqrt32) So (w_x = 10 \cdot \left(-\frac\sqrt32\right) = -5\sqrt3)
$\boxed\approx6.4, \ \alpha\approx68.7^\circ$
Dados los vectores:
$\boxed\approx6.4, \ \alpha\approx68.7^\circ$ ejercicios trigonometria 1 bach vectores