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Question Number 162100 by amin96 last updated on 26/Dec/21

Commented by mr W last updated on 27/Dec/21

$$\mathbf{tan}\mathit{\alpha }=\frac{2(\mathit{R}-\mathit{r})}{\mathit{R}+\mathit{r}}$$

Answered by mr W last updated on 27/Dec/21

Commented by mr W last updated on 27/Dec/21

$$2p+2r=2R\Rightarrow p=R-r$$$$BC=R-r$$$$CD=R-p=R-(R-r)=r$$$$BE=r+q$$$$CE=R-q$$$$DE=p+q=R-r+q$$$$\frac{{BC}^2+{CE}^2-{BE}^2}{2\times BC\times CE}=-\frac{{CD}^2+{CE}^2-{DE}^2}{2\times CD\times CE}$$$$\frac{{(R-r)}^2+{(R-q)}^2-{(r+q)}^2}{R-r}=-\frac{r^2+{(R-q)}^2-{(R-r+q)}^2}{r}$$$$\frac{R^2-Rr-(R+r)q}{R-r}=-\frac{Rr-(2R-r)q}{r}$$$$(R^2-Rr+r^2)q=(R-r)Rr$$$$\Rightarrow q=\frac{(R-r)Rr}{R^2-Rr+r^2}$$$$BC+CE+EB=(R-r)+(R-q)+(r+q)=2R$$$$-BC+CE+EB=-(R-r)+(R-q)+(r+q)=2r$$$$BC-CE+EB=(R-r)-(R-q)+(r+q)=2q$$$$BC+CE-EB=(R-r)+(R-q)-(r+q)=2(R-r-q)$$$${\mathrm{\Delta }}_{BCE}=\frac{\sqrt{2R\times 2r\times 2q\times 2(R-r-q)}}{4}=\sqrt{Rrq(R-r-q)}$$$${\mathrm{\Delta }}_{BCE}=\frac{BC\times h}{2}$$$$\Rightarrow h=\frac{2\sqrt{Rrq(R-r-q)}}{R-r}=\frac{2(R-r)Rr}{R^2-Rr+r^2}=2q$$$$\sqrt{{BE}^2-h^2}=\sqrt{{(r+q)}^2-4q^2}=\sqrt{(r+3q)(r-q)}$$$$\tan \alpha =\frac{h}{AB+\sqrt{{BE}^2-h^2}}$$$$=\frac{2q}{r+\sqrt{(r+3q)(r-q)}}$$$$=\frac{2q}{r+\sqrt{(r+\frac{3(R-r)Rr}{R^2-Rr+r^2})(r-\frac{(R-r)Rr}{R^2-Rr+r^2})}}$$$$=\frac{2}{r+\frac{r^2(2R-r)}{R^2-Rr+r^2}}\times \frac{(R-r)Rr}{R^2-Rr+r^2}$$$$=\frac{2(R-r)}{R+r}$$$$checkforspecialcaseR=2r:$$$$\tan \alpha =\frac{2(2r-r)}{2r+r}=\frac{2}{3}$$$$$$$${(r+q)}^2=r^2+{(2r-q)}^2$$$$q=\frac{2r}{3}$$$$\tan \alpha =\frac{2r-q}{2r}=\frac{2r-\frac{2r}{3}}{2r}=\frac{2}{3}✓$$

Commented by Tawa11 last updated on 27/Dec/21

$$\mathrm{Great}\mathrm{sir}.\mathrm{Am}\mathrm{always}\mathrm{gaining}\mathrm{from}\mathrm{your}\mathrm{geometry}\mathrm{solution}.$$

Commented by Tawa11 last updated on 27/Dec/21

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{more}\mathrm{sir}$$

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