Q. A force is inclined at $60{^\circ}$ to the horizontal. If its rectangular component in the horizontal direction is 50 N, then magnitude of the force in the vertical direction is
A) 25 N
B) 75 N
C) 87 N
D) 100 N
Explanation
(c) $A_{x} = 50,\theta = 60{^\circ}$
Then $tan\theta = A_{y}/A_{x}$ or $A_{y} = A_{x\ }tan\theta$
Or $A_{y} = 50\ tan60{^\circ} = 50 \times \sqrt{3} = 87\ $N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.