If you know (((b^2 +c^2 −a^2 )/(2bc)))^2 +(((c^2 +a^2 −b^2 )/(2ca)))^2 +(((a^2 +b^2 −c^2 )/(2ab)))^2 =3, then what′s the value of ((b^2 +c^2 −a^2 )/(2bc))+((c^2 +a^2 −b^2 )/(2ac))+((a^2 +b^2 −c^2 )/(2ab))?


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Question Number 84871 by tw000001 last updated on 17/Mar/20

$If you know$ ${(\frac{b^{2} + c^{2} - a^{2}}{2 b c})}^{2} + {(\frac{c^{2} + a^{2} - b^{2}}{2 c a})}^{2} + {(\frac{a^{2} + b^{2} - c^{2}}{2 a b})}^{2} = 3,$ $then {what}^{'} s the value of$ $\frac{b^{2} + c^{2} - a^{2}}{2 b c} + \frac{c^{2} + a^{2} - b^{2}}{2 a c} + \frac{a^{2} + b^{2} - c^{2}}{2 a b} ?$
Commented by ajfour last updated on 17/Mar/20

$1 ?$
	Answered by MJS last updated on 17/Mar/20

	${(\frac{b^{2} + c^{2} - a^{2}}{2 b c})}^{2} + {(\frac{c^{2} + a^{2} - b^{2}}{2 c a})}^{2} + {(\frac{a^{2} + b^{2} - c^{2}}{2 a b})}^{2} = 3$ $\Leftrightarrow$ $(a^{2} + b^{2} + c^{2}) (a + b + c) (- a + b + c) (a - b + c) (a + b - c) = 0$ $\Leftrightarrow$ $c = - a - b \lor c = - a + b \lor c = a - b \lor c = a + b$ $\Rightarrow$ $\frac{b^{2} + c^{2} - a^{2}}{2 b c} + \frac{c^{2} + a^{2} - b^{2}}{2 c a} + \frac{a^{2} + b^{2} - c^{2}}{2 a b} = x$ $x = - 3 \lor x = 1$ $the red c gives - 3, the others give 1$
	Commented by MJS last updated on 17/Mar/20

	${you}^{'} re welcome$
	Commented by tw000001 last updated on 17/Mar/20

	$Thank you for your solution .$
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