Menu Close

Category: Mensuration

Question-187932

Question Number 187932 by Rupesh123 last updated on 23/Feb/23 Answered by a.lgnaoui last updated on 24/Feb/23 $${O}_{\mathrm{1}} :{centre}\:{du}\:{cercle}\left({C}_{\mathrm{1}} \right)=\:\:\:{origine}\:{Ref} \\ $$$${Cercles}\left(\:{C}_{\mathrm{1}} \:\right){et}\left({C}_{\mathrm{2}} \right){d}\:{equations}\: \\ $$$$\begin{cases}{{x}^{\mathrm{2}}…

Question-187827

Question Number 187827 by Rupesh123 last updated on 22/Feb/23 Answered by cortano12 last updated on 23/Feb/23 $$\left(\mathrm{1}\right)\mathrm{tan}\:\mathrm{30}°=\frac{\mathrm{4}}{\mathrm{a}}\Rightarrow\mathrm{a}=\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{tan}\:\mathrm{22},\mathrm{5}^{\mathrm{o}} \:=\frac{\mathrm{4}}{\mathrm{b}}\:;\:\mathrm{tan}\:\mathrm{45}°=\frac{\mathrm{2tan}\:\mathrm{22},\mathrm{5}^{\mathrm{o}} }{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{22},\mathrm{5}^{\mathrm{o}} } \\ $$$$\Rightarrow\mathrm{tan}\:^{\mathrm{2}}…