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Question Number 39854 by ajfour last updated on 12/Jul/18

Commented by ajfour last updated on 12/Jul/18

$$Choosethecorrectoptions:$$$$\left(i\right)a\approx 1.6\left(ii\right)b\approx 2.7$$$$\left(iii\right)a\approx 1.8\left(iv\right)b\approx 2.4$$$$$$

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Jul/18

$$everyproblemyoupostreallypraiseworthy$$$$youpleaserepostthoseprobemsremain$$$$unsolvedtilldatesothatwegetoption.$$$$tochoose..$$$$$$

Answered by MrW3 last updated on 12/Jul/18

$$letDT=PB=t$$$$\Rightarrow TP=\frac{1}{t}$$$$\Rightarrow TA=t^2$$$$CP=\sqrt{1+\frac{1}{t^2}}$$$$CD=\sqrt{1+t^2}$$$$CB=\sqrt{1+\frac{1}{t^2}}+t$$$$CA=1+t^2$$$${CD}^2+{CB}^2={CA}^2$$$$1+t^2+t^2+1+\frac{1}{t^2}+2t\sqrt{1+\frac{1}{t^2}}=1+t^4+2t^2$$$$1+\frac{1}{t^2}+2t\sqrt{1+\frac{1}{t^2}}=t^4$$$$\Rightarrow t^2+1+2t^2\sqrt{t^2+1}=t^6$$$$\Rightarrow t\approx 1.499$$$$$$$$a=CD=\sqrt{1+t^2}\approx 1.802$$$$b=CB=t+\sqrt{1+\frac{1}{t^2}}=\approx 2.701$$

Commented by ajfour last updated on 12/Jul/18

$$AlrightSir,thanks{}^{\prime }gain.$$

Answered by ajfour last updated on 12/Jul/18

$$Let\mathrm{\angle }CDP=\mathrm{\angle }ACB=\theta$$$$As\mathit{D}\mathit{T}=\mathit{B}\mathit{P}$$$$\Rightarrow a\cos \theta =\frac{a}{\tan \theta }-a\tan \theta$$$$\Rightarrow \sin \theta \cos {}^2\theta =\cos {}^2\theta -\sin {}^2\theta$$$$let\sin \theta =x$$$$\Rightarrow x(1-x^2)=1-2x^2$$$$\Rightarrow {\mathit{x}}^3-2{\mathit{x}}^2-\mathit{x}+1=0$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$letx=t+\frac{2}{3}$$$$\Rightarrow t^3+2t^2+\frac{4t}{3}+\frac{8}{27}-2t^2-\frac{8t}{3}-\frac{24}{27}$$$$-\frac{3t}{3}-\frac{18}{27}+\frac{27}{27}=0$$$$\Rightarrow t^3-\frac{7t}{3}-\frac{7}{27}=0$$$$lett=u+v$$$$\Rightarrow u^3+v^3+(u+v)(3uv-\frac{7}{3})=\frac{7}{27}$$$$letuv=\frac{7}{9}\Rightarrow u^3{v}^3=\frac{343}{729}$$$$and{u}^3+v^3=\frac{7}{27}$$$$u^{3}and{v}^{3}arethenrootsof$$$$z^2-\frac{7z}{27}+\frac{343}{729}=0$$$$z=\frac{\frac{7}{27}\pm \sqrt{\frac{49}{27\times 27}-\frac{1372}{729}}}{2}$$$$t=u+v(appropriatevalue<0)$$$$t={\left(\frac{7+i\sqrt{1323}}{54}\right)}^{1/3}+{\left(\frac{7-i\sqrt{1323}}{54}\right)}^{1/3}$$$$lett=r^{1/3}{(\cos \theta +i\sin \theta )}^{1/3}$$$$+r^{1/3}{(\cos \theta -i\sin \theta )}^{1/3}$$$$wherer=\frac{\sqrt{49+1323}}{54}=\frac{\sqrt{1372}}{54}$$$$r\cos \theta =\frac{7}{54}\Rightarrow \cos (2\pi -\theta )=\frac{7}{\sqrt{1372}}$$$$\Rightarrow x=t+\frac{2}{3}=2r^{1/3}\cos \frac{\theta }{3}+\frac{2}{3}$$$$\mathit{x}=2{\left(\frac{\sqrt{1372}}{54}\right)}^{1/3}\cos (\frac{2\pi }{3}-\frac{1}{3}{\cos }^{-1}\frac{7}{\sqrt{1372}})+\frac{2}{3}$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$\Rightarrow \mathit{x}\approx 0.55496$$$$\mathit{a}=\frac{1}{\sin \mathit{\theta }}=\frac{1}{x}\approx 1.802$$$$\mathit{b}=\mathit{a}\cot \mathit{\theta }=\mathit{a}\sqrt{\frac{1}{\sin {}^2\mathit{\theta }}-1}$$$$\approx 1.802\times \sqrt{\frac{1}{{(0.55496)}^2}-1}$$$$\Rightarrow \mathit{b}\approx 2.701.$$

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