Question Number 189976 by 073 last updated on 25/Mar/23
Answered by JDamian last updated on 25/Mar/23
$${E} \\ $$
Commented by 073 last updated on 25/Mar/23
$$\mathrm{solution} \\ $$
Answered by HeferH last updated on 26/Mar/23
$$\mathrm{let}\:\mathrm{HD}\:=\:\mathrm{x} \\ $$$$\mathrm{DB}\:=\:\mathrm{x}\sqrt{\mathrm{2}} \\ $$$$\mathrm{cot}\:\alpha\:=\:\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}\: \\ $$$$\mathrm{tan}\:\alpha\:=\:\frac{\mathrm{x}}{\mathrm{x}\sqrt{\mathrm{2}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\:\mathrm{cot}\:\alpha\:=\:\sqrt{\mathrm{2}}\: \\ $$