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Question Number 88541 by ajfour last updated on 11/Apr/20

Commented by ajfour last updated on 11/Apr/20

$$Thetwoshownspheres,eachof$$$$radiusraretangenttoeachother$$$$andtothefacesofadihedral$$$$angle\alpha .Findtheradiusofthe$$$$spherethatistangenttothe$$$$planesandthetwospheresaswell.$$$$Ans:$$$$R=r[1+2\tan {}^2\left(\frac{\alpha }{2}\right)\pm \tan \left(\frac{\alpha }{2}\right)\sqrt{3+4\tan {}^2\left(\frac{\alpha }{2}\right)}]$$

Commented by ajfour last updated on 11/Apr/20

Commented by Ar Brandon last updated on 11/Apr/20

$$Whichapppermittedyoutodesign$$$$thisplease?$$

Commented by ajfour last updated on 11/Apr/20

$$LekhDiagram;overtimeiam$$$$friendlywiththisapp..$$

Commented by Ar Brandon last updated on 02/May/20

$$\mathrm{thanks}$$

Answered by mr W last updated on 11/Apr/20

Commented by mr W last updated on 11/Apr/20

$$O^{\prime }A=\frac{r}{\sin \frac{\alpha }{2}}$$$$OB=\frac{R}{\sin \frac{\alpha }{2}}$$$${(R+r)}^2=r^2+{(\frac{R}{\sin \frac{\alpha }{2}}-\frac{r}{\sin \frac{\alpha }{2}})}^2$$$$R^2(1-{\sin }^2\frac{\alpha }{2})-2Rr(1+{\sin }^2\frac{\alpha }{2})+r^2=0$$$$\frac{R}{r}=\frac{1+{\sin }^2\frac{\alpha }{2}\pm \sqrt{{(1+{\sin }^2\frac{\alpha }{2})}^2-(1-{\sin }^2\frac{\alpha }{2})}}{1-{\sin }^2\frac{\alpha }{2}}$$$$\frac{R}{r}=\frac{1+{\sin }^2\frac{\alpha }{2}\pm \sin \frac{\alpha }{2}\sqrt{3+{\sin }^2\frac{\alpha }{2}}}{1-{\sin }^2\frac{\alpha }{2}}$$$$\frac{R}{r}=1+2{\tan }^2\frac{\alpha }{2}\pm \tan \frac{\alpha }{2}\sqrt{3+4{\tan }^2\frac{\alpha }{2}}$$

Commented by ajfour last updated on 11/Apr/20

$$ohyes,thiswastooeasysir,$$$$ihadexpectedittobesomewhat$$$$difficult;thanksSir.$$

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