Help with really easy Maths question (its killing me!)

cos70=47.5/h

rearranging gives h= 75.002 N in each string.

Atleast that is what i get.

As the problem is symmetrical i only considered half of it. 95/2 = 47.5

The horizontal force provided by one the sections of the string must equal half the horizontal force provided by the archer (the other section providing the other half). This is based on the assumption that the two sections of string are 70 degrees above and below the direction the archer is applying the force.

The answer could be wrong.

Draw a diagram of ONE half of the bowstring and resolve the forces horizontally.
So you should get

$2Tcos70 = 95$

And then rearrange for T. I think that's right, I had to draw a lil diagram in my head. Just ask if you don't understand.

Its not 47.5 = cos70/T, Is 47.5/cos70 = T. That should get you the right answer.

Draw a diagram of ONE half of the bowstring and resolve the forces horizontally.
So you should get

$2Tcos70 = 95$

And then rearrange for T. I think that's right, I had to draw a lil diagram in my head. Just ask if you don't understand.

Draw a diagram of ONE half of the bowstring and resolve the forces horizontally.
So you should get

$2Tcos70 = 95$

And then rearrange for T. I think that's right, I had to draw a lil diagram in my head. Just ask if you don't understand.

By really easy, i though you meant addition, or something along those lines...

Wait mate, your a legend! I just figured it out!!!!

thanks a bunch, and thanks to all of you who answered...
rep+