c(acosB-bcosA)=a^2-b^2
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Originally Posted by matsci0000 c(acosB-bcosA)=a^2-b^2 Can we have the rest of the question please, is this a triangle with sides a,b,c and angles opposite the sides A, B, C? Try the difference of the cosine rule for a and b. CB
Originally Posted by matsci0000 c(acosB-bcosA)=a^2-b^2 According to cosine rule a^2 = b^2 + c^2 - 2bc*cosA. From this find bc*cosA. Similarly find ac*coaB, And find the difference. Or, by using projection rule write c as c = acosB + bcosA and simplify.
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