How to Calculate Forces in a Beam Using the Method of Sections

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Arty7
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IHi guys with part b of this question I am having trouble. I have cut it it at the specified members and used the top half. Now i take the moment at A to get the force in BC. (CW+)M@A=(6*3)+(6cos(30)*3)=0, But when you solve this it doesn't give you the answer its supposed to be, I am probably blind & missing something. Please check out the attachments i provided some more info on it.

Thankyou in Advance.
 
Attachments
  • Truss diagram of a pin-jointed frame with supports at A and D, 6 kN loads at joints B and C, and bay lengths of 3 m, 3 m and 3 m, with handwritten support reactions VD = 6.0 kN up, VA = 6.0 kN up, HA = 0, and BC = 8.49 kN (C).
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Where do you get 30 degrees from? Check your triangles.