Question Number 191403 by Mingma last updated on 23/Apr/23
Answered by mr W last updated on 23/Apr/23
$${as}\:{Q}\mathrm{191305}\:{but}\:\theta=\mathrm{45}° \\ $$$${r}=\frac{{ab}}{{a}+{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\left({a}+\sqrt{\mathrm{3}}{b}\right)\mathrm{sin}\:\theta+\left(\sqrt{\mathrm{3}}{a}+{b}\right)\mathrm{cos}\:\theta} \\ $$$${with}\:\theta=\mathrm{45}° \\ $$$${r}=\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\left(\mathrm{1}+\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}}\right)\left({a}+{b}\right)} \\ $$$${with}\:{a}=\mathrm{12},\:{b}=\mathrm{5}: \\ $$$${r}=\frac{\mathrm{60}}{\:\mathrm{30}+\frac{\mathrm{17}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}{\:\sqrt{\mathrm{2}}}}\approx\mathrm{0}.\mathrm{955} \\ $$
Commented by mr W last updated on 23/Apr/23
Commented by Mingma last updated on 23/Apr/23
Excellent, sir!