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The forces are acting along the ropes.
Denote their magnitudes as B, C and D in accordance with the points of attachment.
For reasons of symmetry C=D. It is sufficient to get two independent equations.
Project forces on the y-axis:
`B*sin(45°) = 2*C*cos(30°)*sin(60°),`
or `B = C*sqrt(2)*(3/2),`
and on the z-axis:
`2*C*cos(30°)*cos(60°) + B*cos(45°) = mg,`
or `C*sqrt(3)/2 + B*1/sqrt(2) = mg.`
Substitute from the first equation into the second:
`C*(sqrt(3)/2 + 3/2) = mg,` or
`C*2.37 approx 25*9.8=245,` or
`C approx 245/2.37 approx 103.4 (N).`
So `B approx 103.4*2.12 approx 219.2 (N).`
The answer: B=219.2 N, C=D=103.4 N.
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