The resultant of two forces has magnitude 20 N. One of the forces is of magnitude `20sqrt(3)` N and makes an angle of `90^(@)` with the resultant. Then, the other force must be of magnitude

Options

A. 40N
B. 20N
C. `20sqrt(3)N`
D. `10sqrt(3)N`

Answer

A

Explanation

`tan beta=(B tan)/(A+B cos theta)`………1 `:.oo=(B tan theta)/(A+Bcos theta):.A+Bcos theta=0`
`implies20sqrt(3)=-Bcos theta` ……..
`R^(2)=A^(2)+B^(2)+2ABcos theta` and substituting 2 in 1 we get
`400-1200+2400=B^(2)impliesB=40N`