could you give me a worked solution please 
thanks! (i tried doing ratio and i couldn't 
)
Paul, I meant 4 over BOTH exams 
Oh right - I misread, but that sounds about right - although it's probably a bit little - I think Thushan lost more and got a 50 :p
if thushan lost 10 on each i'm sure he would stll find a way to get 50 :X
Firstly, I'll apologize for ambiguous this post is most likely going to be, but here goes nothing
Ok so you remember those triangles you've memorised for methods, where you have angles of
and
we we have the same situation here.
So we know the length of AC will be
and the length of AB will be
where
could be any multiple, as the ratio's will always be the same (I'll show you why this doesn't matter next)
Let's say X = 1;
So,
and
Now if we take moments about C;
We have the tensile force of AB acting at a length of
from C in the anticlockwise direction.
We also have the weight force of the light acting at a length of
from C in the clockwise direction.
(F_{AB})-(1)(mg)=0)
(F_{AB})-(1)(200)=0)
Now let's say X = 20
We have the tensile force of AB acting at a length of
from C in the anticlockwise direction.
We also have the weight force of the light acting at a length of
from C in the clockwise direction.
(F_{AB})-(20)(mg)=0)
(F_{AB})-(20)(200)=0)
(200))
So as you can see you only need to worry about the ratio's of the distances not the actual distance
Additional diagram;
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