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In a Δ ABC?
In a Δ ABC
[(b²+c²)/bc]CosA +[(a²+b²)/ab]CosC+[(a²+c²)/ac]CosB is
(1) 2abc (2) 3abc (3) 2 (4) 3
2 Answers
- IndicaLv 74 years agoFavorite Answer
[(b²+c²)/bc]CosA +[(a²+b²)/ab]CosC+[(a²+c²)/ac]CosB
By the Cosine Rule cosA=(b²+c²−a²)/(2bc) with similar for cosB and cosC
∴ [(b²+c²)/bc]CosA = (b²+c²)(b²+c²−a²)/(2b²c²) = ½(1/b²+1/c²)(b²+c²−a²)
[(b²+c²)/bc]CosA = ½(2+c²/b²+b²/c²−a²/b²−a²/c²)
Ditto
[(a²+b²)/ab]CosC = ½(2+a²/b²+b²/a²−c²/b²−c²/a²)
[(a²+c²)/ac]CosB = ½(2+c²/a²+a²/c²−b²/a²−b²/c²)
Sum these to give ½(2+2+2) = 3
