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Question Number 39175 by ajfour last updated on 03/Jul/18

Answered by MJS last updated on 03/Jul/18

$$\mathrm{let}d>R+r$$$$2\mathrm{rectangular}\mathrm{triangles}:$$$$d^2-{(R-r)}^2=p^2$$$$d^2-{(R+r)}^2=q^2$$$$\frac{d}{\sin 90°}=\frac{p}{\sin {\beta }_1}=\frac{R-r}{\sin {\gamma }_1}\Rightarrow {\gamma }_1=\arcsin \frac{R-r}{d}$$$$\frac{d}{\sin 90°}=\frac{q}{\sin {\beta }_2}=\frac{R+r}{\sin {\gamma }_2}\Rightarrow {\gamma }_2=\arcsin \frac{R+r}{d}$$$$\alpha ={\gamma }_1+{\gamma }_2=\arcsin \frac{R-r}{d}+\arcsin \frac{R+r}{d}$$$$...\mathrm{but}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{confusing}\mathrm{myself}\mathrm{without}\mathrm{drawing}$$$$\mathrm{scetches},\mathrm{so}\mathrm{please}\mathrm{check}\mathrm{and}\mathrm{correct}$$

Commented by ajfour last updated on 04/Jul/18

$$CorrectanswerSir.$$

Commented by MJS last updated on 04/Jul/18

$$\mathrm{thank}\mathrm{you}$$

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