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`Delta ABC` में सिद्ध कीजिए कि `a cos.(B-C)/(2)=(b+c)sin.(A)/(2)` |
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Answer» हम जानते है कि `(sinA)/(A)=(sinB)/(b)=(sinC)/(c)=(1)/(k)` `rArr a=k sin A, b=k sin B, c= sin C` अब `(b+c)/(a)=(ksinB+ksinC)/(ksinA)` `=(sinB+sinC)/(sinA)=(2sin((B+C)/(2))cos((B-C)/(2)))/(2sin.(A)/(2)cos.(A)/(2))` `=(sin(90^@-(A)/(2))cos((B-C)/(2)))/(sin.(A)/(2)cos.(A)/(2))=(cos.(A)/(2)cos((B-C)/(2)))/(sin.(A)/(2)cos.(A)/(2))=(cos.(B-C)/(2))/(sin.(A)/(2))` इसलिए `a cos.(B-C)/(2)=(b+c)sin.(A)/(2)` . |
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