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Question-193293




Question Number 193293 by TUN last updated on 09/Jun/23
Answered by AST last updated on 09/Jun/23
h=a+b+c  ∣AH∣^2 =∣b+c∣^2 =(((b+c)^2 )/(bc))=(b+c)(b^− +c^− )  ∣BC∣^2 =∣c−b∣^2 =(c−b)(c^− −b^− )=(c−b)(((b−c)/(bc)))=((−(b−c)^2 )/(bc))  ((∣AH∣^2 )/(∣BC∣^2 ))=((−(b+c)^2 )/((b−c)^2 ))=3 (where c=be^(i(π/3)) )  x^2 R^2 =9R^2 ⇒x^2 =9⇒x=3
$${h}={a}+{b}+{c} \\ $$$$\mid{AH}\mid^{\mathrm{2}} =\mid{b}+{c}\mid^{\mathrm{2}} =\frac{\left({b}+{c}\right)^{\mathrm{2}} }{{bc}}=\left({b}+{c}\right)\left(\overset{−} {{b}}+\overset{−} {{c}}\right) \\ $$$$\mid{BC}\mid^{\mathrm{2}} =\mid{c}−{b}\mid^{\mathrm{2}} =\left({c}−{b}\right)\left(\overset{−} {{c}}−\overset{−} {{b}}\right)=\left({c}−{b}\right)\left(\frac{{b}−{c}}{{bc}}\right)=\frac{−\left({b}−{c}\right)^{\mathrm{2}} }{{bc}} \\ $$$$\frac{\mid{AH}\mid^{\mathrm{2}} }{\mid{BC}\mid^{\mathrm{2}} }=\frac{−\left({b}+{c}\right)^{\mathrm{2}} }{\left({b}−{c}\right)^{\mathrm{2}} }=\mathrm{3}\:\left({where}\:{c}={be}^{{i}\frac{\pi}{\mathrm{3}}} \right) \\ $$$${x}^{\mathrm{2}} {R}^{\mathrm{2}} =\mathrm{9}{R}^{\mathrm{2}} \Rightarrow{x}^{\mathrm{2}} =\mathrm{9}\Rightarrow{x}=\mathrm{3} \\ $$

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