Question Number 35446 by ajfour last updated on 19/May/18
Commented by ajfour last updated on 19/May/18
$${Solution}\:{to}\:\:{Q}.\mathrm{35439} \\ $$
Answered by ajfour last updated on 19/May/18
$${d}\mathrm{cos}\:\theta+{R}=\left({a}−{d}\mathrm{sin}\:\theta\right)\mathrm{tan}\:\theta \\ $$$$\Rightarrow\:{d}=\frac{{a}\mathrm{tan}\:\theta−{R}}{\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\mathrm{tan}\:\theta} \\ $$$$\:\:\:\:\:\:\:\:=\:{a}\mathrm{sin}\:\theta−{R}\mathrm{cos}\:\theta \\ $$$$\boldsymbol{{r}}=\sqrt{\boldsymbol{{R}}^{\mathrm{2}} −\boldsymbol{{d}}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\:{r}=\sqrt{{R}^{\mathrm{2}} −\left({a}\mathrm{sin}\:\theta−{R}\mathrm{cos}\:\theta\right)^{\mathrm{2}} }\:. \\ $$