Question

A force 150N is inclined at 50° to the horizontal direction. Find it's component in the horizontal and vertical direction

Answer 1

From the right triangle above:

$$F_x = F\cdot \cos 50\degree = 150N\cdot \cos 50\degree \approx 96.4N\\ F_y = F\cdot \sin 50\degree = 150N\cdot \sin 50\degree \approx 115N$$

Answer. In horizontal direction 96.4N, in vertical direction 115N.