We write the first equation so that the constant term is on the right hand side:

F1 sin 69.3° − F2 sin 71.1° − F3 sin 56.6° = −926

In matrix form, we write the equations as:

`((sin 69.3°,-sin 71.1°,-sin 56.6°),(cos 69.3°,-cos 71.1°,cos 56.6°),(7.80 sin 69.3°,-1.50 sin 71.1°,-5.20 sin 56.6°))((F_1),(F_2),(F_3))`

`=((-926),(0),(0))`

So the solution for the system is:

`((F_1),(F_2),(F_3))=((sin 69.3°,-sin 71.1°,-sin 56.6°),(cos 69.3°,-cos 71.1°,cos 56.6°),(7.80 sin 69.3°,-1.50 sin 71.1°,-5.20 sin 56.6°))^-1((-926),(0),(0))`

`=((425.5),(1079.9),(362.2))`

So

`F_1= 425.5\ "N"`

`F_2= 1079.9\ "N"`

`F_3= 362.2\ "N"`

This is very easy and quick in Scientific Notebook, Matlab or any other computer algebra system!

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