Question Number 45980 by ajfour last updated on 19/Oct/18
Commented by ajfour last updated on 19/Oct/18
$${Determine}\:{sides}\:{of}\:\bigtriangleup,\: \\ $$$$\boldsymbol{{a}},\:{and}\:\boldsymbol{{b}}\:=\:\boldsymbol{{c}}\:,\:{in}\:{terms}\:{of}\:\boldsymbol{{R}},\:\boldsymbol{{r}},\:\boldsymbol{{d}}. \\ $$
Answered by MrW3 last updated on 19/Oct/18
$${let}\:\theta=\angle{B}=\angle{C} \\ $$$$\mathrm{cos}\:\theta=\frac{{R}−{r}}{{d}} \\ $$$${a}=\mathrm{2}×\frac{{R}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}=\mathrm{2}{R}\sqrt{\frac{\mathrm{1}+\mathrm{cos}\:\theta}{\mathrm{1}−\mathrm{cos}\:\theta}} \\ $$$$\Rightarrow{a}=\mathrm{2}{R}\sqrt{\frac{\mathrm{1}+\frac{{R}−{r}}{{d}}}{\mathrm{1}−\frac{{R}−{r}}{{d}}}}=\mathrm{2}{R}\sqrt{\frac{{d}+{R}−{r}}{{d}+{r}−{R}}} \\ $$$$\Rightarrow{b}={c}=\frac{{a}}{\mathrm{2}\:\mathrm{cos}\:\theta}={R}\frac{{d}}{{R}−{r}}\sqrt{\frac{{d}+{R}−{r}}{{d}+{r}−{R}}} \\ $$
Commented by ajfour last updated on 19/Oct/18
$${wonderfully}\:{solved},\:{Sir}\:! \\ $$
Commented by MrW3 last updated on 19/Oct/18
$${thanks}! \\ $$