Question

The horizontal component of a Force of 23 N applied at 38º is.

Answer 1

To be given in question

Force (F)=23N

$\theta=38°$

To be asked in question

F_horizontal=?

We know that

$F_{hor}=Fcos\theta\rightarrow(1)$

equation (1) Put value

$F_{hor}=23cos(38°)$

$F_{hor}=18.1242 N$