Question Number 191404 by Mingma last updated on 23/Apr/23
Answered by mr W last updated on 23/Apr/23
$${as}\:{Q}\mathrm{191305}\:{but}\:\theta=\mathrm{0}° \\ $$$${r}=\frac{{ab}}{{a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\left({a}+\sqrt{\mathrm{3}}{b}\right)\mathrm{sin}\:\theta+\left(\sqrt{\mathrm{3}}{a}+{b}\right)\mathrm{cos}\:\theta} \\ $$$${with}\:\theta=\mathrm{0}° \\ $$$${r}=\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\left(\mathrm{1}+\sqrt{\mathrm{3}}\right){a}+\mathrm{2}{b}} \\ $$$${with}\:{a}=\mathrm{4},\:{b}=\mathrm{3}: \\ $$$${r}=\frac{\mathrm{12}}{\:\mathrm{15}+\mathrm{4}\sqrt{\mathrm{3}}}\approx\mathrm{0}.\mathrm{547} \\ $$
Commented by mr W last updated on 23/Apr/23
Commented by Mingma last updated on 23/Apr/23
Excellent, sir!