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Balanced forces

One question i stumbled upon:

A bow is pulled back such that an angle of 140 degrees is made at the point where the string intersects itself, the force needed to do this is 95N.

Calculate the tension in each string.

Firstly, i tried 95sin(70) but gave me the wrong answer, then i thought for a good while and thought that between the string and the actual bow creates an isosceles triangle, with angles 70, 70 and 40 degrees.

Then, i used the sin rule of 95/sin40=x/sin70
rearranging gets 95sin70/sin40= x = correct answer.

Now, is this what i would have been expected to do or is there an alternate method?
No, you need the components of the tension in the two halves of the string in the direction in which the arrow fires. Opposite to the direction you are pulling.
If the big angle is 140 then half this is 70. The two halves of the string make angles of 70 degs to the forwards direction.
So T cos 70 + T cos 70 = 95 is the equation you need to write.

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