A trigonometric question: On an ABC triangle , the edges are a, b and c. (b-c)/(b+c)=1/(root3) m(A)=60 degree angle what is angle of m(B) ?

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A trigonometric question: On an ABC triangle , the edges are a, b and c. (b-c)/(b+c)=1/(root3) m(A)=60 degree angle what is angle of m(B) ?

Mathematics
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if its a right triangle, then the m(B) will equal 30 degrees. if its not a right triangle, then a picture would be helpful
There is no given shape.Thanks anyway.
Answer is equal 105 degrees.

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Other answers:

then you could do it the hard way by making a system of equations: b - c = 1 b + c = sqrt(3) and finding your b and c....
then use the law of cosines to find "a" and the law of sines to find all your angles
b = 1+c (1+c) +c = sqrt(3) 1 + 2c = sqrt(3) 2c = sqrt(3)-1 c = (sqrt(3)-1)/2; b = (1 + sqrt(3))/2
I found a=6 then I found , 12/sqrt(3)=(sqrt(3)-1)/sin(B) But couldn't go further
if your a=6 is good; then: sinA sinB) ---- = ----- a b sinB = b sin(A)/a
sinB = [ (1 + sqrt(3))/2 ] [sqrt(3)/2] / 6 sinB = (sqrt(3) +3)/24
sinB = sqrt(3)/24 + 1/8
the sin inverse funtion will give you one angle for that taio; but you have to be aware that 2 angles are possible with a positive sin.
taio means ratio..... somehow :)
Prof ,My problem is how can i turn this to degree. :)
with the sin inverse function on a calculator.
I got it , sin inverse function , i will try thanks :)

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