Question Number 219339 by Spillover last updated on 23/Apr/25
Answered by mr W last updated on 23/Apr/25
$$\alpha=\mathrm{45}°+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{3} \\ $$
Answered by mr W last updated on 23/Apr/25
Commented by mr W last updated on 23/Apr/25
$$\frac{{FK}}{{KH}}=\frac{{FC}}{{EH}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{FK}}{{EF}}=\frac{{FK}}{{FH}}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{cot}\:\alpha \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\mathrm{3} \\ $$
Answered by Spillover last updated on 04/May/25
Answered by Spillover last updated on 04/May/25
