A force `3 N` is acting along the `NE` and another force `4 N` is acting the `NW`. Find the angle made by the resultant force with north.

Explanation

Method I (component approach) :
`F_(x) = 3 cos 45^(@) - 4 cos 45^(@) = - cos 45^(@) = -(1)/(sqrt(2))`
`F_(y) = 3 sin 45^(@) + 4 sin 45^(@) = 7 sin 45^(@) = (7)/(sqrt(2))`
`tan theta = (F_(x))/(F_(y)) = (1)/(7)`
`theta = tan^(-1) ((1)/(7))`
The resultant force makes an angle `tan^(-1) ((1)/(7))` with north, left of it.
Method II : Since there are only two forces , we can use the law of parallelogram
Since `4 gt 3` , the resultant will be towards `4 N`.
`tan alpha = (3)/(4) rArr alpha = tan^(-1) ((3)/(4))`
`theta + alpha = 45^(@)`
`theta = 45^(@) - tan^(-1) ((3)/(4))`
(If the value of `tan^(-1) ( 3//4) = 37^(@)` given)
`theta = 45^(@) - 37^(@) = 8^(@)`
The angle between by the resultant force with north (left of it ) is `8^(@)`.