The resultant of 2 forces of magnitude 1N and 13N is `6sqrt(5)N`. Find the tangent of the angle made by resultant with 1N force.

Explanation

Let P=1N
Q=13N ltbgt `R=6sqrt(5)N`
By Parallelogram rule of vector addition
`R^(2)=P^(2)+Q^(2)+2PQ cos theta`
`(6sqrt(5))^(2)=(1)^(2)+(13)^(2)+2xx1xx13cos theta`

`theta=cos^(-1)((5)/(13))` . . . .(1)
`tan alpha=(Q sin theta)/(P+Q cos theta)=(13 sin theta)/(1+13 cos theta)`
`(13xx(12)/(13))/(1+13xx(5)/(13))=(12)/(6)=2`
`alpha= tan^(-1)2` . . . .(2)
Ans. Angle between given vectors is
`theta=cos^(-1)((5)/(13))`
Angle of resultant with `1N force
`=alpha-tan^(-1)(2)`