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Question Number 35446 by ajfour last updated on 19/May/18

Commented by ajfour last updated on 19/May/18

$$SolutiontoQ.35439$$

Answered by ajfour last updated on 19/May/18

$$d\cos \theta +R=(a-d\sin \theta )\tan \theta$$$$\Rightarrow d=\frac{a\tan \theta -R}{\cos \theta +\sin \theta \tan \theta }$$$$=a\sin \theta -R\cos \theta$$$$\mathit{r}=\sqrt{{\mathit{R}}^2-{\mathit{d}}^2}$$$$\Rightarrow r=\sqrt{R^2-{(a\sin \theta -R\cos \theta )}^2}.$$

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