I assume you know mechanics, free-body diagram (FBD), moments, etc.

The weight on the drawing is 600N. In the text it is 900N. Which one?

Let us say it is 900N.

First get the vertical, Fy, and horizontal , Fx, components of the two tensions.

Say T is the tension at BE, and U is the tension at CF.

For the T:

T is acting along BE, so get the length of BE.

BE = sqrt[(200^2 +80^2) = 215.4066 mm

By proportion,

Tx = T(200/215.4066) = (0.92848)T

Ty = T(80/215.4066) = (0.37139)T

For the U:

CF = sqrt(100^2 +80^2) = 128.0625 mm

So,

Ux = U(100/128.0625) = (0.78087)U

Uy = U(80/128.0625) = (0.62469)U

Now, to solve for T, cut the system vertically between points C and D to create a FBD. Take moments at point C. Break the T into its Tx and Ty components.

The Tx has no moment at point C because its line of direction passes through point C.

Ty(100) = 900(200)

TY = 1800 N

So, T = 1800 / 0.37139 = 4847 N tension --------------answer.

For the U,

in the same FBD above, break the U into its Ux and Uy components.

Take moments at point B. Ux has no moment there

Uy(100) = 900(100)

Uy = 900 N

So,

U = 900 / 0.62469 = 1441 N tension ----------answer.

For the reaction at point D:

Since the beam is supported by a roller there, there will be no vertical restraint there. It will all be horizontal.

It will be Ux.

It will be 1441(0.78087) = 1125 N compression -----answer.(Oppss, my mistake here. Let me correct it.)

I forgot the the Tx.

The reaction at D should be R = Tx +Ux

R = 4847(0.92848) +1125 = 5625 N compression ---------corrected answer.

Which cable will break first?

Cable BE, because it carries more stress than cable CF. ------answer.

Or, because the T is greater than the U.