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In a triangle ABC, angle A = 60^(@) and ...

In a triangle ABC, `angle A = 60^(@) and b : c = (sqrt3 + 1) : 2`, then find the value of `(angle B - angle C)`

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`(b)/(c) = (sqrt3 + 1)/(2) rArr (b - c)/(b + c) = (sqrt3 + 1 -2)/(srt3 + 1 + 2) = (sqrt3 -1)/((sqrt3 + 1)) (1)/(sqrt3)`
Now using `tan.(B - C)/(2) = (b - C)/(b + c) cot. (A)/(2)`, we get
`tan.(B - C)/(2) = (sqrt3 - 1)/((sqrt3 + 1)) = 2 - sqrt3 rArr (B - C)/(2) = 15^(@)`
`:. B - C = 30^(@)`
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