a) 6.1 N
Given a force of magnitude F, the horizontal (x) component of the force can be found using the equation:
![F_x = F cos \theta](https://img.qammunity.org/2020/formulas/physics/middle-school/8e1wxx1cqymp8nk5yvxad3050l706labca.png)
where
F is the magnitude of the force
is the angle between the direction of the force and the horizontal
In this problem, we have
F = 6.6 N
![\theta=21.6^(\circ)](https://img.qammunity.org/2020/formulas/physics/high-school/n2vrz1sh5vp4831ammwgrs1jraaey9614y.png)
Therefore, the horizontal component is
![F_x = (6.6)(cos 21.6^(\circ))=6.1 N](https://img.qammunity.org/2020/formulas/physics/high-school/dol2dsgf5cktzwhxjmx4j2lq926vbl52eg.png)
b) 2.4 N
The vertical (y) component of the force can be found using the equation:
![F_y = F sin \theta](https://img.qammunity.org/2020/formulas/physics/college/sfkbwe0xbdqpzjydf6p7y3jnwgtkoqd0q7.png)
where
F is the magnitude of the force
is the angle between the direction of the force and the horizontal
Here we have
F = 6.6 N
![\theta=21.6^(\circ)](https://img.qammunity.org/2020/formulas/physics/high-school/n2vrz1sh5vp4831ammwgrs1jraaey9614y.png)
So, the vertical component is
![F_y = (6.6)(sin 21.6^(\circ))=2.4 N](https://img.qammunity.org/2020/formulas/physics/high-school/o166qodxn5c2kcdr16t3g5bi23f0qgcytv.png)