two forces F1=500 N due east and F2=250 N due north have their common initial point. F2-F1 is and angle is..

Note:their is an arrow on F1 and F2..

let

f = f1 + f2

v= f1 - f2

there fore on adding both we get

f+v = 2 f1

and f can be calculated by vetor sum of addition = 559.07N towards north east

and therefore

v = 724.06 N towards 22.5 degree north

since both are in two diff direction so

the angle between them is 90⁰

so l F2vector+(-F1vector) l = √f1²+f2²-2f1f2(cos90).......................cos90=0

by puting the value of f1  f2

we get 

250√5 N and angle is 

Θ=tan^-1(500/250) in north of west

Θ=tan^-1(2)

MAGNITUDE OF RESULTANT FORCE IS CLEARLY 250√5N AND THE ANGLE BETWEEN THE RESULTANT AND F1 IS tan^-1(1/2).