Question

a force is inclined at 60degre to horizontal if its rectangular components in the horizontal directon is 50 newton then magnitude of force in the vertical direction is

Answer 1

You can see figure Let force R is inclined 60° with horizontal then,

horizontal direction component of force = Rcos60° = 50
R × 1/2 = 50
R = 100 N

Vertical direction component of force = Rsin60 = 100 × √3/2 = 50√3 N

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....