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I'm really stuck on this question...

"A truck dealer wants to suspend a 2.5 tonne truck as shown for advertising. the distance b (which is 17.3m) shows a horizontal distance of the suspension point B from the anchor point A. The sum of the lengths of cables AB and BC is 42m. calculate the tensions in the cables."

The diagram shows a 40m horizontal distance from anchor point A to anchor point C.

I can see straight away that hanging point B is 17.3m horizontally from the left (point A) and therefore 22.7m from the right (point C) but also the force verticle would be

2,500kg*9.81ms^-2 =24,525N in a downward direction (Weight of the truck).

I know some theory to use where the F=T1sin[tex]\theta[/tex]1+T2sin[tex]\theta[/tex]2

however to work any other tensions out i require the angles of the cables to verticle but i dont know even 2 sides lol. I know the AB+BC=42 and AC=40

challenging question :( unless I'm missing something simple and fundimental...

plz help

ben