2. Find the resultant of the following forces : F, = 600 N at 40 in the first quadrant , F2 = 800 N at 20 . in the second quadrant and F3 = 200 N at 60 in the 4" quadrant A. 522.67 N at 63.43 . in the 2nd quadrant * B. 453.45 N at 54.34 . in the 1 st quadrant C . 523.65 N at 56.43 . in the 3"d quadrant D. 321.45 N at 34.43 . in the 4th quadrant y-axis Solution : EF = 600cos 40 - 800cos20 + 200cos60 800 600 M - 192.12 N (+ ) EF = 600sin40 + 800sin20 - 200sin60 20 .x-axis 60 = 486.08 N (T) R = ( EF * ) + ( EF )z 200 N (-192.12 )"+ (486.08 ) R = 522.7 N GR = tan- IF = tan-1 486.08 192.12 J, 48608 N " = 68.430 IF, = - 192.12 N Alternate Solution : ( By Calculator in made : complex ) R = 600240+ 8002160+ 200