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Trignometry

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9 Dec 2011 23:14:41 IST

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Mathematics

Show that for a triangle, a sin (B-C)\(b^2-c^2) = b sin (C-A) \(c^2-a^2) = c sin (A-B) \(a^2-b^2), here a,b & c are sides of triangle.

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lucky_kn

Cool goIITian

Joined: 30 Aug 2008

Posts: 76

26 Dec 2011 11:50:48 IST

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take the first part ...

frm the sine rule .. ie sinA/a =sinB /b =sinC/c = R

write b and c in terms of a,sinB and sinC

solve this and u will obtain the final answer as sinA/a = R

similarly u can show tat the other two parts also equal to R ...

i m sure u will get the answer with dis..

try ..

gud luck.. :D

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