In the plane, three common point force vectors F1, F2 and F3 are in equilibrium. It is known that the module of vector F1 is 1n, the module of vector F2 is (radical 6 + radical 2) / 2n, and the angle between vector F1 and vector F2 is 45 degrees 1) Find the size of the vector F3 2) The angle between vector F1 and vector F3

In the plane, three common point force vectors F1, F2 and F3 are in equilibrium. It is known that the module of vector F1 is 1n, the module of vector F2 is (radical 6 + radical 2) / 2n, and the angle between vector F1 and vector F2 is 45 degrees 1) Find the size of the vector F3 2) The angle between vector F1 and vector F3

(F3)^2=(F1)^2+(F2)^2-2F1F2cos45°
=1^2+[(√6+√2)/2]^2-2*1*(√6+√2)/2*√2/2
=1+[(8+4√3)/4]-(√6+√2)√2/2
=1+2+√3-(√3+1)
=1+2+√3-√3-1
=2
F3=√2
cosφ=[(F1)^2+(F3)^2-(F2)^2]/2F1F3
=[1+2-(2+√3)]/2*1*√2
=[1-√3]/2√2
=(√2-√6)/4
φ=105°
Supplementary notes
cos105=cos(45+60)
=cos45cos60-sin45sin60
=（√2-√6)/4