Question Number 201166 by mnjuly1970 last updated on 01/Dec/23
Answered by mr W last updated on 01/Dec/23
$$\frac{{ah}}{\mathrm{2}}=\mathrm{4} \\ $$$$\Rightarrow{h}=\frac{\mathrm{2}×\mathrm{4}}{\:\sqrt{\mathrm{2}}}=\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${h}×\mathrm{cot}\:{B}+{h}×\mathrm{cot}\:{C}={a} \\ $$$$\Rightarrow\mathrm{cot}\:{B}+\mathrm{cot}\:{C}=\frac{{a}}{{h}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by mnjuly1970 last updated on 01/Dec/23
$$\:\:\:{bravo}\:{sir}\:{thanks}\:{alot}\:…\:\underline{\underbrace{\lesseqgtr}} \\ $$
Commented by mr W last updated on 01/Dec/23
Commented by mr W last updated on 01/Dec/23
$${a}_{\mathrm{1}} ={h}\:\mathrm{cot}\:{B} \\ $$$${a}_{\mathrm{2}} ={h}\:\mathrm{cot}\:{C} \\ $$$${a}={a}_{\mathrm{1}} +{a}_{\mathrm{2}} ={h}\left(\mathrm{cot}\:{B}+\mathrm{cot}\:{C}\right) \\ $$$${S}=\frac{{ah}}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{cot}\:{B}+\mathrm{cot}\:{C}\right)} \\ $$$$\Rightarrow\mathrm{cot}\:{B}+\mathrm{cot}\:{C}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}{S}}=\frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by mnjuly1970 last updated on 01/Dec/23
$${so}\:{nice}\:{master}\:.{grateful} \\ $$