Question Number 210508 by hardmath last updated on 11/Aug/24
Answered by mr W last updated on 11/Aug/24
Commented by mr W last updated on 11/Aug/24
![ΔABE=((c^2 tan B)/2) x=(c/(cos B))−a y=(x/(tan B)) ΔCDE=((xy)/2)=(1/(2 tan B))((c/(cos B))−a)^2 [ABCD]=ΔABC−ΔCDE =((c^2 tan B)/2)−(1/(2 tan B))((c/(cos B))−a)^2 =((c^2 tan B)/2)−((1/(2 tan B))+((tan B)/2))c^2 −(a^2 /(2 tan B))+((ac)/(sin B)) =((ac)/(sin B))−((a^2 +c^2 )/(2 tan B)) ✓](https://www.tinkutara.com/question/Q210513.png)
$$\Delta{ABE}=\frac{{c}^{\mathrm{2}} \mathrm{tan}\:{B}}{\mathrm{2}} \\ $$$${x}=\frac{{c}}{\mathrm{cos}\:{B}}−{a} \\ $$$${y}=\frac{{x}}{\mathrm{tan}\:{B}} \\ $$$$\Delta{CDE}=\frac{{xy}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{tan}\:{B}}\left(\frac{{c}}{\mathrm{cos}\:{B}}−{a}\right)^{\mathrm{2}} \\ $$$$\left[{ABCD}\right]=\Delta{ABC}−\Delta{CDE} \\ $$$$\:\:=\frac{{c}^{\mathrm{2}} \mathrm{tan}\:{B}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{tan}\:{B}}\left(\frac{{c}}{\mathrm{cos}\:{B}}−{a}\right)^{\mathrm{2}} \\ $$$$\:\:=\frac{{c}^{\mathrm{2}} \mathrm{tan}\:{B}}{\mathrm{2}}−\left(\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{tan}\:{B}}+\frac{\mathrm{tan}\:{B}}{\mathrm{2}}\right){c}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{2}\:\mathrm{tan}\:{B}}+\frac{{ac}}{\mathrm{sin}\:{B}} \\ $$$$\:\:=\frac{{ac}}{\mathrm{sin}\:{B}}−\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}\:\mathrm{tan}\:{B}}\:\checkmark \\ $$
Commented by hardmath last updated on 11/Aug/24
$$ \\ $$Perfect solution as always, thanks dear professor