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

Answered by MJS last updated on 03/Jul/18

let d>R+r 2 rectangular triangles: d^2 −(R−r)^2 =p^2 d^2 −(R+r)^2 =q^2 (d/(sin 90°))=(p/(sin β_1 ))=((R−r)/(sin γ_1 )) ⇒ γ_1 =arcsin ((R−r)/d) (d/(sin 90°))=(q/(sin β_2 ))=((R+r)/(sin γ_2 )) ⇒ γ_2 =arcsin ((R+r)/d) α=γ_1 +γ_2 =arcsin ((R−r)/d) +arcsin ((R+r)/d) ...but I′m confusing myself without drawing scetches, so please check and correct

$$\mathrm{let}\:{d}>{R}+{r} \\ $$$$\mathrm{2}\:\mathrm{rectangular}\:\mathrm{triangles}: \\ $$$${d}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} \\ $$$${d}^{\mathrm{2}} −\left({R}+{r}\right)^{\mathrm{2}} ={q}^{\mathrm{2}} \\ $$$$\frac{{d}}{\mathrm{sin}\:\mathrm{90}°}=\frac{{p}}{\mathrm{sin}\:\beta_{\mathrm{1}} }=\frac{{R}−{r}}{\mathrm{sin}\:\gamma_{\mathrm{1}} }\:\Rightarrow\:\gamma_{\mathrm{1}} =\mathrm{arcsin}\:\frac{{R}−{r}}{{d}} \\ $$$$\frac{{d}}{\mathrm{sin}\:\mathrm{90}°}=\frac{{q}}{\mathrm{sin}\:\beta_{\mathrm{2}} }=\frac{{R}+{r}}{\mathrm{sin}\:\gamma_{\mathrm{2}} }\:\Rightarrow\:\gamma_{\mathrm{2}} =\mathrm{arcsin}\:\frac{{R}+{r}}{{d}} \\ $$$$\alpha=\gamma_{\mathrm{1}} +\gamma_{\mathrm{2}} =\mathrm{arcsin}\:\frac{{R}−{r}}{{d}}\:+\mathrm{arcsin}\:\frac{{R}+{r}}{{d}} \\ $$$$...\mathrm{but}\:\mathrm{I}'\mathrm{m}\:\mathrm{confusing}\:\mathrm{myself}\:\mathrm{without}\:\mathrm{drawing} \\ $$$$\mathrm{scetches},\:\mathrm{so}\:\mathrm{please}\:\mathrm{check}\:\mathrm{and}\:\mathrm{correct} \\ $$

Commented by ajfour last updated on 04/Jul/18

Correct answer Sir.

$${Correct}\:{answer}\:{Sir}. \\ $$

Commented by MJS last updated on 04/Jul/18

thank you

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

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