A force is inclined at `60^(@)` to the horozontal. If its rectangular component in the horizontal direction is `50 N`, then magnitude of the verticle components of force is approximately

Options

A. 25 N
B. 75 N
C. 87 N
D. 100 N

Answer

C

Explanation

Given, `A_(x)=50N and theta =60^(@)`
Then, Tan `theta=(A_(y))/(A_(x))`
`Ay=A_(x)tan theta`
`A_(y)=50 tan 60^(@)`
`A_(y)=50xxsqrt3=87N`