We can find equals to the given components
[tex]f1= 200cos(60), 200sin(60)[/tex]
Then you have to find equal for f2= 100cos(170) ,100sin(170) which is [tex]\ \textless \ -98.48, 17.36\ \textgreater \ [/tex]
Substract the vectors to find F3 coordinates [tex]= \ \textless \ 1.52, 190.56\ \textgreater \ [/tex][tex]F3= \ \textless \ 100-98.48\ \textgreater \ , \ \textless \ 173.2+17.36\ \textgreater \ [/tex]
With theese new coordinates it is possible to find the direction and magnitude by using arc tan and square rooting. You should make [tex]X= 1.52 and Y=190.56[/tex] and plug it to the arc tan function tan-[tex]1(190.56/1.52)[/tex]. You will obtain direction 89.54, then find magnitude of the square root of 36315.4 --[tex](1.52)^2 + (190.56)^2[/tex]
