Question Number 187953 by Mingma last updated on 24/Feb/23 Answered by som(math1967) last updated on 24/Feb/23 $$\bigtriangleup{AOB}=\bigtriangleup{AOC}=\bigtriangleup{AOC}=\frac{\mathrm{1}}{\mathrm{3}}\bigtriangleup{ABC} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{48}=\mathrm{16}{cm}^{\mathrm{2}} \\ $$$$\bigtriangleup{AOM}=\frac{\mathrm{1}}{\mathrm{2}}\bigtriangleup{AOB}=\mathrm{8}{cm}^{\mathrm{2}} \\ $$$$\left[{OM}\:{is}\:{median}\:{of}\bigtriangleup{AOB}\right] \\ $$$$\bigtriangleup{AOK}=\frac{\mathrm{1}}{\mathrm{2}}\bigtriangleup{AOC}=\mathrm{8}{cm}^{\mathrm{2}}…
Question Number 187946 by Mingma last updated on 24/Feb/23 Answered by mr W last updated on 24/Feb/23 $${B}={blue}\:{area} \\ $$$${R}={red}\:{area} \\ $$$${Y}={yellow}\:{area} \\ $$$$\frac{{B}}{{Y}}=\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{9}}{\mathrm{12}} \\…
Question Number 187932 by Rupesh123 last updated on 23/Feb/23 Answered by a.lgnaoui last updated on 24/Feb/23 $${O}_{\mathrm{1}} :{centre}\:{du}\:{cercle}\left({C}_{\mathrm{1}} \right)=\:\:\:{origine}\:{Ref} \\ $$$${Cercles}\left(\:{C}_{\mathrm{1}} \:\right){et}\left({C}_{\mathrm{2}} \right){d}\:{equations}\: \\ $$$$\begin{cases}{{x}^{\mathrm{2}}…
Question Number 187921 by Mingma last updated on 23/Feb/23 Answered by HeferH last updated on 23/Feb/23 $${V}_{\mathrm{1}} −{V}_{\mathrm{2}} \:=\:{V}_{\mathrm{3}} \\ $$ Answered by mr W…
Question Number 187904 by Mingma last updated on 23/Feb/23 Commented by MathGuy last updated on 24/Feb/23 Commented by Mingma last updated on 24/Feb/23 Nice solution, sir! Terms…
Question Number 187901 by Rupesh123 last updated on 23/Feb/23 Commented by Rupesh123 last updated on 23/Feb/23 Find R and r Commented by BaliramKumar last updated on 23/Feb/23 $${R}\:=\:\frac{\mathrm{3}\sqrt{\mathrm{6}}}{\mathrm{4}}…
Question Number 187903 by Mingma last updated on 23/Feb/23 Answered by HeferH last updated on 23/Feb/23 Commented by HeferH last updated on 23/Feb/23 $${i}.\:\:\frac{{a}+{b}}{{c}}=\:\frac{{d}}{{e}}\:\Rightarrow\:\frac{{e}}{{d}}\:=\:\frac{{c}}{{a}+{b}} \\…
Question Number 187891 by Rupesh123 last updated on 23/Feb/23 Answered by mr W last updated on 24/Feb/23 Commented by mr W last updated on 25/Feb/23…
Question Number 187826 by Rupesh123 last updated on 22/Feb/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 187827 by Rupesh123 last updated on 22/Feb/23 Answered by cortano12 last updated on 23/Feb/23 $$\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}}…