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In a triangle A B C ,/A=60^0a n db : c=(...

In a triangle `A B C ,/_A=60^0a n db : c=(sqrt(3)+1):2,` then find the value of `(/_B-/_C)dot`

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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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CENGAGE PUBLICATION-PROPERTIES AND SOLUTIONS OF TRIANGLE-Illustration
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