Q. A vector is inclined at 30$^\circ$ to the horizontal. If its rectangular component in the horizontal direction be 50 unit, the magnitude of its vertical component is
Solution:
Here, $\theta = 30^\circ$
Let F= magnitude of force.
Given, F cos $\theta $ = 50 or F = $\frac{50}{cos\, 30^\circ} = \frac{50 \times 2 }{\sqrt{3}} = \frac{100}{\sqrt{3}} = 57.74 \, N$
Vertical component = F sin $\theta$ = 57.74 $ \times $
sin 30$^\circ$ = 57.74 $\times \frac{1}{2} $ = 28.87 N
