Question Number 81322 by ajfour last updated on 11/Feb/20
Commented by ajfour last updated on 11/Feb/20
$${Find}\:{radius}\:{of}\:{quatercircle} \\ $$$${in}\:{terms}\:{of}\:{a},{b},{c}. \\ $$
Answered by mr W last updated on 11/Feb/20
Commented by mr W last updated on 11/Feb/20
$$\mathrm{cos}\:\beta=\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ac}}\geqslant\mathrm{0},\:{i}.{e}.\:{a}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant{b}^{\mathrm{2}} \\ $$$$\mathrm{cos}\:\gamma=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$$ \\ $$$${BF}=\frac{{r}}{\mathrm{tan}\:\beta}={r}\:\mathrm{cot}\:\beta \\ $$$${FC}={a}−{r}\:\mathrm{cot}\:\beta \\ $$$${FC}×\mathrm{sin}\:\gamma={r} \\ $$$$\left({a}−{r}\:\mathrm{cot}\:\beta\right)\:\mathrm{sin}\:\gamma={r} \\ $$$$\Rightarrow{r}=\frac{{a}}{\mathrm{cosec}\:\gamma+\mathrm{cot}\:\beta} \\ $$
Commented by ajfour last updated on 12/Feb/20
$${thanks}\:{sir}. \\ $$