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Question Number 44017 by ajfour last updated on 20/Sep/18

Commented by ajfour last updated on 20/Sep/18

$$Aconeiscutbyaplaneatan$$$$angle\theta toitsbaseandpasses$$$$throughapointatthebaseof$$$$cone.Findratioof\mathit{u}\mathit{p}\mathit{p}\mathit{e}\mathit{r}\mathit{c}\mathit{o}\mathit{n}\mathit{e}$$$$\mathit{v}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}\mathit{t}\mathit{o}\mathit{l}\mathit{o}\mathit{w}\mathit{e}\mathit{r}\mathit{c}\mathit{o}\mathit{n}\mathit{e}\mathit{v}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}.$$$$(intermsof\theta ,R,h).$$

Commented by MrW3 last updated on 22/Sep/18

$$thankyousir!$$

Commented by ajfour last updated on 22/Sep/18

$$heightsofproblemsolvingsir!$$$${haven}^{\prime }tcomprehendedityet,$$$$however;youare\mathcal{G}reatsir,$$$$thanksforconfirmingagain.$$$$letmetryandunderstand,iwill,$$$$cozyoupresentitthebesttoo.$$$$$$

Answered by MrW3 last updated on 24/Sep/18

$$totalcone:$$$$V=\frac{\pi R^{2}h}{3}$$$$uppercone\left(obliquecone\right):$$$$A_{base}=\pi ab=areaofellipseat\theta$$$$h_{upper}=(h-R\tan \theta )\cos \theta$$$$let\lambda =\frac{h}{R}=\tan \varphi$$$$V_{upper}=\frac{1}{3}{A}_{base}{h}_{upper}=\frac{\pi ab(h-R\tan \theta )\cos \theta }{3}$$$$=\frac{\pi abR(\lambda -\tan \theta )\cos \theta }{3}$$$$\frac{h_1}{\tan \theta }+\frac{h_1}{h}\times R=2R$$$$\Rightarrow h_1=\frac{2Rh}{R+h\cot \theta }=\frac{2R\lambda \tan \theta }{\lambda +\tan \theta }$$$$2a=\frac{h_1}{\sin \theta }=\frac{2h}{(\lambda +\tan \theta )\cos \theta }$$$$\Rightarrow a=\frac{h}{(\lambda +\tan \theta )\cos \theta }$$$$r=R(1-\frac{h_1}{2h})=R(1-\frac{R}{R+h\cot \theta })=\frac{R\lambda }{\lambda +\tan \theta }$$$$e=R-\frac{h_1}{2\tan \theta }=\frac{R\tan \theta }{\lambda +\tan \theta }$$$$b^2=r^2-e^2=(r+e)(r-e)=R^2(\frac{\lambda }{\lambda +\tan \theta }+\frac{\tan \theta }{\lambda +\tan \theta })(\frac{\lambda }{\lambda +\tan \theta }-\frac{\tan \theta }{\lambda +\tan \theta })$$$$=R^2\left(\frac{\lambda -\tan \theta }{\lambda +\tan \theta }\right)$$$$\Rightarrow b=R\sqrt{\frac{\lambda -\tan \theta }{\lambda +\tan \theta }}$$$$V_{upper}=\frac{\pi abR(\lambda -\tan \theta )\cos \theta }{3}$$$$=\frac{\pi R(\lambda -\tan \theta )\cos \theta }{3}\times \frac{h}{(\lambda +\tan \theta )\cos \theta }\times R\sqrt{\frac{\lambda -\tan \theta }{\lambda +\tan \theta }}$$$$=\frac{\pi R^{2}h}{3}\times \frac{\lambda -\tan \theta }{\lambda +\tan \theta }\times \sqrt{\frac{\lambda -\tan \theta }{\lambda +\tan \theta }}$$$$=V\times {\left(\frac{\lambda -\tan \theta }{\lambda +\tan \theta }\right)}^{\frac{3}{2}}$$$$\Rightarrow \frac{V_{upper}}{V}={\left(\frac{\lambda -\tan \theta }{\lambda +\tan \theta }\right)}^{\frac{3}{2}}=\sqrt{{\left(\frac{h-R\tan \theta }{h+R\tan \theta }\right)}^3}$$$$$$$$Tocuttheconeintotwoequalparts:$$$${\left(\frac{\lambda -\tan \theta }{\lambda +\tan \theta }\right)}^{\frac{3}{2}}=\frac{1}{2}$$$$\Rightarrow \tan \theta =\frac{2^{\frac{2}{3}}-1}{2^{\frac{2}{3}}+1}\times \lambda =\frac{{}^3\sqrt{4}-1}{{}^3\sqrt{4}+1}\times \tan \varphi \approx 0.227\tan \varphi$$$$\Rightarrow \theta \approx {\tan }^{-1}(0.227\tan \varphi )$$

Commented by MrW3 last updated on 22/Sep/18

Commented by MrW3 last updated on 22/Sep/18

Commented by ajfour last updated on 17/Nov/18

$$thankyousir,hatsoff!$$

Commented by MrW3 last updated on 23/Sep/18

Commented by ajfour last updated on 17/Nov/18

$$Sir,ihavedifficultyinunder-$$$$standing{b}^2+e^2=r^2;butyou$$$$havealreadyexplained,letme$$$$tryagain,thanks.$$

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