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Question Number 45270 by ajfour last updated on 11/Oct/18

Commented by ajfour last updated on 11/Oct/18

$$Find\mathit{\alpha }intermsofRandr.$$

Answered by MrW3 last updated on 11/Oct/18

$$PT=\frac{r}{\tan \alpha }=\frac{r\cos \alpha }{\sin \alpha }$$$$PA=\frac{R}{\sin \alpha }$$$$AT=PA-PT=\frac{R-r\cos \alpha }{\sin \alpha }$$$$AC=R+r$$$${AC}^2={AT}^2+{TC}^2$$$${(R+r)}^2=\frac{{(R-r\cos \alpha )}^2}{{\sin }^2\alpha }+r^2$$$$(R^2+2Rr)(1-{\cos }^2\alpha )=R^2-2Rr\cos \alpha +r^2{\cos }^2\alpha$$$$R^2+2Rr-(R^2+2Rr){\cos }^2\alpha =R^2-2Rr\cos \alpha +r^2{\cos }^2\alpha$$$${(R+r)}^2{\cos }^2\alpha -2Rr\cos \alpha -2Rr=0$$$$\cos \alpha =\frac{2Rr+\sqrt{4R^2{r}^2+8Rr{(R+r)}^2}}{2{(R+r)}^2}$$$$\cos \alpha =\frac{Rr+\sqrt{Rr[{(R+r)}^2+Rr]}}{{(R+r)}^2}$$$$\Rightarrow \alpha ={\cos }^{-1}\frac{Rr+\sqrt{Rr[{(R+r)}^2+Rr]}}{{(R+r)}^2}$$

Commented by ajfour last updated on 11/Oct/18

$$ThankyouSir!$$

Answered by ajfour last updated on 11/Oct/18

$$PM=\frac{R\cos \alpha }{\sin \alpha }=\frac{r}{\sin \alpha }+\sqrt{{(R+r)}^2-R^2}$$$$AP=\frac{R}{\sin \alpha }=\frac{r\cos \alpha }{\sin \alpha }+\sqrt{{(R+r)}^2-r^2}$$$$\Rightarrow \frac{R\cos \alpha -r}{R-r\cos \alpha }=\frac{\sqrt{r(2R+r)}}{\sqrt{R(2r+R)}}=\frac{a}{b}\left(let\right)$$$$\Rightarrow bR\cos \alpha -br=aR-ar\cos \alpha$$$$\cos \alpha =\frac{aR+br}{ar+bR}$$$$\mathit{\alpha }={\cos }^{-1}\left(\frac{R\sqrt{r(2R+r)}+r\sqrt{R(2r+R)}}{r\sqrt{r(2R+r)}+R\sqrt{R(2r+R)}}\right).$$

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