Finding applied force on 50 kg block accelerating up a 20° incline

[ssh222]

Homework Statement

A block of mass 50 kg is free to slide on a frictionless inclined plane that makes an angle of 20° with respect to horizontal. A force F is applied as shown in the drawing pulling parallel to the incline. As a result, the block accelerates up the plane at 4 m/s2.

known:
m= 50 kg
a= 4 m/s^2
Angle = 20 degrees

Homework Equations

F=ma
Fcos(theta)

The Attempt at a Solution

F= 50kg* 4m/s2 = 200 N
200cos(20)=187.9= F x-direction

* The explanation shows me F- 167 N = 200N
F=367N
But I don't know which equation this comes from or where the 167 comes from.

 

[azizlwl]

3. The Attempt at a Solution
F= 50kg* 4m/s2 = 200 N
200cos(20)=187.9= F x-direction
......
200cos(20)=187.9= F x-direction
What force is this?

 

[ssh222]

This was my attempt at finding the applied force..

 

[azizlwl]

Can you tell me why you multiply 200 with Cos(20)?

If you just put an object on frictionless inclined plane, in which direction the object moves?. What makes it move in that direction?