Question Number 220560 by mehdee7396 last updated on 15/May/25
$${sin}\alpha=\mathrm{0}.\mathrm{8}\:\Rightarrow\:\frac{{BE}}{{EF}}=? \\ $$
Answered by mehdee7396 last updated on 15/May/25
Answered by mr W last updated on 15/May/25
$${BE}=\mathrm{2}{R}\:\mathrm{sin}\:\alpha−\frac{{R}}{\mathrm{sin}\:\alpha} \\ $$$${EF}={R}−\frac{{R}}{\mathrm{tan}\:\alpha} \\ $$$$\frac{{BE}}{{EF}}=\frac{\mathrm{2}\:\mathrm{sin}\:\alpha−\frac{\mathrm{1}}{\mathrm{sin}\:\alpha}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}}=\frac{\mathrm{2}×\mathrm{0}.\mathrm{8}−\frac{\mathrm{1}}{\mathrm{0}.\mathrm{8}}}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}}=\frac{\mathrm{7}}{\mathrm{5}} \\ $$
Answered by mehdee7396 last updated on 15/May/25
$$\angle{OEC}=\alpha \\ $$$${BE}×{EC}={EF}×\left({R}+{OE}\right) \\ $$$$\Rightarrow\frac{{BE}}{{EF}}=\frac{{R}+{OE}}{{EC}}={sin}\alpha+{cos}\alpha=\frac{\mathrm{7}}{\mathrm{5}} \\ $$$$\: \\ $$