Question

a force of 50 n is inclined at 30 degree to the the horizontal find its horizontal component

Answer 1

Fx=Fcos@
put the given values you will get
Fx=25underoot3

Answer 2

$\huge {\rm {\underline {Question}}}$

A force inclined at 50 ° to horizontal , if it's component in this direction is 50 N , find magnitude of force and its vertical components.

$\huge {\rm {\underline {Answer}}}$

Here $\tt {F_x = 50N}$ , $\theta = 50°$

But $\tt {F_x = F cos \theta}$

Therefore ,

$\bf {F = \frac {F_x}{cos \theta} = \frac {50}{cos 50°} = \frac {50}{0.6428} = 77.78N}$

Also ,

$F_y \\ \implies F sin \theta \\ \implies 77.78 \times sin50° \\ \implies 77.78 \times 0.7660 \\ \implies 59.58N$

Hope it helps dear....