Question Number 206659 by cherokeesay last updated on 21/Apr/24
Answered by mr W last updated on 21/Apr/24
Commented by mr W last updated on 21/Apr/24
$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${a}=\mathrm{2}{R}\:\mathrm{sin}\:\alpha=\frac{\mathrm{2}{R}}{\:\sqrt{\mathrm{5}}} \\ $$$$\beta=\frac{\pi}{\mathrm{4}}−\alpha \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{1}−\mathrm{tan}\:\alpha}{\mathrm{1}+\mathrm{tan}\:\alpha}=\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${b}=\mathrm{2}{R}\:\mathrm{sin}\:\beta=\frac{\mathrm{2}{R}}{\:\sqrt{\mathrm{10}}} \\ $$$$\Rightarrow\frac{{a}}{{b}}=\frac{\sqrt{\mathrm{10}}}{\:\sqrt{\mathrm{5}}}=\sqrt{\mathrm{2}} \\ $$
Commented by cherokeesay last updated on 21/Apr/24
$${nice}\:!\:{thank}\:{you}\:! \\ $$
Commented by MM42 last updated on 23/Apr/24
$$\:\underline{\underbrace{\lesseqgtr}} \\ $$
Answered by MM42 last updated on 23/Apr/24
$${sin}\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\:\&\:{cos}\alpha=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\:\&\:\:\theta=\alpha+\beta=\mathrm{45}\: \\ $$$$\frac{{b}}{{sin}\left(\mathrm{90}+\alpha\right)}=\frac{\mathrm{1}}{{sin}\theta}\Rightarrow{b}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{10}}} \\ $$$${a}=\mathrm{4}{sin}\alpha=\frac{\mathrm{4}}{\:\sqrt{\mathrm{5}}} \\ $$$$\Rightarrow\frac{{a}}{{b}}=\sqrt{\mathrm{2}}\:\:\checkmark\:\: \\ $$
Commented by MM42 last updated on 23/Apr/24
