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Question Number 39404 by ajfour last updated on 06/Jul/18

Commented by ajfour last updated on 06/Jul/18

$$Q.39365solution..$$

Commented by MrW3 last updated on 06/Jul/18

$${let}^{\prime }ssaytheanglesofisoscelestriangles$$$$are\theta ,inourcase\theta =\frac{\pi }{6}.$$$$Arethereanyothervaluesfor\theta such$$$$that\mathrm{\Delta }GJNisequilateral?$$

Commented by behi83417@gmail.com last updated on 06/Jul/18

$$AG=\frac{b}{2}sec\theta ,AN=\frac{c}{2}sec\theta$$$${NG}^2=\frac{b^2}{4}{sec}^2\theta +\frac{c^2}{4}{sec}^2\theta -\frac{bc}{2}{sec}^2\theta .cos(60+2\theta )=$$$$=\frac{1}{4}{sec}^2\theta [b^2+c^2-2bc.cos(60+2\theta )]$$$${NJ}^2=\frac{1}{4}{sec}^2\theta [a^2+c^2-2ac.cos(60+2\theta )]$$$$\Rightarrow a^2+c^2-2ac.cos(60+2\theta )=b^2+c^2-2bc.cos(60+2\theta )$$$$\Rightarrow a^2-b^2=2c(a-b)cos(60+2\theta )$$$$\stackrel{a\ne b}{\Rightarrow }cos(60+2\theta )=\frac{a+b}{2c}$$$$\Rightarrow \theta =\frac{1}{2}{cos}^{-1}\left[\frac{a+b}{2c}\right]-30.$$

Commented by MrW3 last updated on 06/Jul/18

$$AG=\frac{b}{2\cos \theta }$$$$AN=\frac{c}{2\cos \theta }$$$${NG}^2={\left(\frac{b}{2\cos \theta }\right)}^2+{\left(\frac{c}{2\cos \theta }\right)}^2-2\left(\frac{b}{2\cos \theta }\right)\left(\frac{c}{2\cos \theta }\right)\cos (A+2\theta )$$$$=\frac{b^2+c^2-2bc(\cos A\cos 2\theta -\sin A\sin 2\theta )}{4{\cos }^2\theta }$$$$=\frac{b^2+c^2-2bc\cos A\cos 2\theta +2bc\sin A\sin 2\theta }{4{\cos }^2\theta }$$$$=\frac{b^2+c^2-(b^2+c^2-a^2)\cos 2\theta +4\mathrm{\Delta }\sin 2\theta }{4{\cos }^2\theta }$$$$=\frac{a^2\cos 2\theta +(b^2+c^2)(1-\cos 2\theta )+4\mathrm{\Delta }\cos 2\theta }{4{\cos }^2\theta }$$$$\cos 2\theta =1-\cos 2\theta$$$$\Rightarrow \cos 2\theta =\frac{1}{2}$$$$\Rightarrow \theta =\frac{\pi }{6}$$$$\Rightarrow thereisonlyonevaluefor\theta .$$

Commented by ajfour last updated on 06/Jul/18

$$marvelousSir!$$

Commented by behi83417@gmail.com last updated on 06/Jul/18

$$dearmaster!forwhat?$$$$cos2\theta =1-cos2\theta$$

Commented by MrW3 last updated on 06/Jul/18

$$theterm{a}^2\cos 2\theta +(b^2+c^2)(1-\cos 2\theta )$$$$mustbethesameforallthreesides$$$$including$$$$b^2\cos 2\theta +(c^2+a^2)(1-\cos 2\theta )$$$$c^2\cos 2\theta +(a^2+b^2)(1-\cos 2\theta )$$$$butthisistrueonlywhen$$$$\cos 2\theta =1-\cos 2\theta$$$$then$$$$a^2\cos 2\theta +(b^2+c^2)(1-\cos 2\theta )$$$$=b^2\cos 2\theta +(c^2+a^2)(1-\cos 2\theta )$$$$=c^2\cos 2\theta +(a^2+b^2)(1-\cos 2\theta )$$$$=(a^2+b^2+c^2)\cos 2\theta$$$$and{it}^{\prime }sthesameforallthreesides.$$

Answered by ajfour last updated on 06/Jul/18

$$AG=\frac{b}{2}\sec \frac{\pi }{6}=\frac{b}{\sqrt{3}}$$$$AN=\frac{c}{2}\sec \frac{\pi }{6}=\frac{c}{\sqrt{3}}$$$${NG}^2={AG}^2+{AN}^2-2\left(AG\right)\left(AN\right)\cos (A+\frac{\pi }{3})$$$$=\frac{b^2+c^2}{3}-\frac{2bc}{3}(\frac{1}{2}\cos A-\frac{\sqrt{3}}{2}\sin A)$$$$=\frac{b^2+c^2-bc\cos A}{3}+\frac{bc\sin A}{\sqrt{3}}$$$$Nowsince\sin A=\frac{a}{2R}$$$$(Rbeingcircumradiusof△ABC)$$$$and{a}^2=b^2+c^2-2bc\cos A$$$${NG}^2=\frac{2b^2+2c^2-2bc\cos A}{6}+\frac{abc}{2\sqrt{3}R}$$$$=\frac{a^2+b^2+c^2}{6}+\frac{abc}{2\sqrt{3}R}$$$$similarlyNJ{}^{2}and{GJ}^{2}result$$$$inthesameexpression.$$$$Hence△NJGisequilateral.$$

Commented by behi83417@gmail.com last updated on 06/Jul/18

$$niceproofsirAjfour.thanksfor$$$$drawingtheperfectfigure.$$

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