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Question Number 39993 by ajfour last updated on 14/Jul/18

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18

$$let\mathrm{\angle }AOT=\theta$$$$OT\parallel toBCso\frac{AO}{AC}=\frac{AT}{AB}=\frac{OT}{BC}$$$$pointT(cos\theta ,sin\theta )andpointC(-1,0)$$$$eqnofOTisy=xtan\theta$$$$eqnofCBisy=(x+1)tan\theta$$$$eqnofcircleis{x}^2+y^2=1$$$$solvey=(x+1)tan\theta and{x}^2+y^2=1$$$$x^2+x^2{tan}^2\theta +2{xtan}^2\theta +{tan}^2\theta -1=0$$$$x^2\left({sec}^2\theta \right)+x\left(2{tan}^2\theta \right)+{tan}^2\theta -1=0$$$$x^2+2{xsin}^2\theta +{sin}^2\theta -{cos}^2\theta =0$$$$x^2+x(1-cos2\theta )-\left(cos2\theta \right)=0$$$$x=\frac{-(1-cos2\theta )\pm \sqrt{1-2cos2\theta +{cos}^{2}2\theta +4cos2\theta }}{2}$$$$x=\frac{-(1-cos2\theta )\pm (1+cos2\theta )}{2}$$$$x=\frac{-1+cos2\theta +1+cos2\theta }{2},\frac{-1+cos2\theta -1-cos2\theta }{2}$$$$x=cos2\theta ,-1$$$$whenx=cos2\theta y=sin2\theta$$$$whenx=-1soy=0$$$$sopointC(-1,0)pointP(cos2\theta ,sin2\theta )$$$$CP=\sqrt{{(cos2\theta +1)}^2+{\left(sin2\theta \right)}^2}$$$$=\sqrt{{co}_{}{s}^{2}2\theta +2cos2\theta +1+{sin}^{2}2\theta }$$$$=\sqrt{2(1+cos2\theta )}$$$$=\sqrt{4{cos}^2\theta }=2cos\theta where\theta =\mathrm{\angle }AOT$$$$wait...$$$$$$$$$$$$$$$$$$$$$$$$$$

Answered by ajfour last updated on 15/Jul/18

$$BC=1+\frac{a}{2}$$$$BP=1-\frac{a}{2}$$$$BC.BP={BT}^2=1-\frac{a^2}{4}$$$${OB}^2=1+{BT}^2={(1+\frac{a}{2})}^2-1$$$$\Rightarrow 1+1-\frac{a^2}{4}=\frac{a^2}{4}+a$$$$\Rightarrow a^2+2a-4=0$$$$or{(a+1)}^2=5$$$$\Rightarrow a=\sqrt{5}-1.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18

$$hwBCandBPfound...plsexplain..$$

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18

$$yesigotit..perpedicuarfromcentrebisectthe$$$$chord...andperpendicularfromcentre$$$$makearectangle...oppositesidearesameequals$$$$to1$$

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