The answers are F1429 Alpha622 degrees Betta110 degrees Gamm

Solution

lets first write the forces in cartesian vector form.

F1 = F1cos(alpha) i + F1cos(beta)j + F1 cos(gamma)k

F2 = 200 (-j) = -200 j lb

F3 = - 400sin30i + 400cos30j

so resultant of all three forces.

FR = F1 + F2 + F3

F1cos(alpha) i + F1cos(beta)j + F1 cos(gamma)k -200 j - 400sin30i + 400cos30j = -350k

(F1cos(alpha) -200)i + (F1cos(beta) + 146.41)j + (F1 cos(gamma) + 350)k = 0

cos(alpha) = 200 / F1;
cos(beta) = -146.41 /F1;
cos(gamma) = -350/F1 ;

and cos^2(alpha) + cos^2(beta) + cos^2(gamma) = 1

(200 / F1)^2 + (-146.41 /F1)^2 + (-350/F1 )^2 = 1

F1^2 = 183935.94

F1 = 428.88 lb

alpha = cos^-1(200/428.88) = 62.20 deg

beta = cos^-1(-146.41/428.88) = 109.96 deg

gamm = cos^-1(-350/428.88) = 145deg