Question Number 212721 by Spillover last updated on 22/Oct/24 Answered by A5T last updated on 22/Oct/24 Commented by A5T last updated on 22/Oct/24 $${In}\left[{ABC}\right];\:{tan}\theta=\frac{{R}}{{R}+\mathrm{2}{r}};{AC}=\sqrt{\mathrm{2}{R}^{\mathrm{2}} +\mathrm{4}{r}^{\mathrm{2}}…
Question Number 212720 by Spillover last updated on 22/Oct/24 Answered by A5T last updated on 22/Oct/24 $${AB}=\sqrt{\mathrm{2}\left(\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{6};\:{Let}\:\angle{OAB}=\theta \\ $$$$\frac{{sin}\theta}{{r}}=\frac{{sin}\left(\mathrm{180}−\mathrm{2}\theta\right)}{{AB}}\Rightarrow{cos}\theta=\frac{\mathrm{3}}{{r}}\Rightarrow{sin}\mathrm{2}\theta=\frac{\mathrm{6}\sqrt{{r}^{\mathrm{2}} −\mathrm{9}}}{{r}^{\mathrm{2}} } \\ $$$$\left(\mathrm{4}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}}…
Question Number 212723 by ajfour last updated on 22/Oct/24 Commented by ajfour last updated on 22/Oct/24 $${Find}\:\:{R}_{{min}} \\ $$ Answered by mr W last updated…
Question Number 212719 by Spillover last updated on 22/Oct/24 Answered by A5T last updated on 22/Oct/24 Commented by A5T last updated on 22/Oct/24 $${a}=\mathrm{2}\left(\frac{\pi}{\mathrm{6}}−\frac{{sin}\mathrm{60}°}{\mathrm{2}}\right)=\mathrm{2}\left(\frac{\pi}{\mathrm{6}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right)=\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\…
Question Number 212690 by Spillover last updated on 21/Oct/24 Answered by mr W last updated on 21/Oct/24 $$\mathrm{10}×\left(\mathrm{2}{n}\right)=\left(\mathrm{2}\sqrt{\mathrm{15}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{n}=\mathrm{3} \\ $$$$\Rightarrow{R}={n}+\frac{\mathrm{10}}{\mathrm{2}}=\mathrm{8} \\ $$$${painted}\:{area}\:=\pi\left({R}^{\mathrm{2}} −{n}^{\mathrm{2}}…
Question Number 212669 by Spillover last updated on 20/Oct/24 Commented by Spillover last updated on 21/Oct/24 $${ans}\:{X}=\:\mathrm{35}\sqrt{\mathrm{2}}\: \\ $$ Answered by mr W last updated…
Question Number 212668 by Spillover last updated on 20/Oct/24 Answered by efronzo1 last updated on 21/Oct/24 $$\:\:\mathrm{BC}^{\mathrm{2}} =\:\mathrm{64}+\mathrm{25}−\mathrm{40}\:=\:\mathrm{49} \\ $$$$\:\:\mathrm{BC}^{\mathrm{2}} =\:\mathrm{2r}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \Rightarrow\mathrm{r}=\sqrt{\frac{\mathrm{BC}^{\mathrm{2}} }{\mathrm{3}}}=\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{3}} \\…
Question Number 212533 by emilagazade last updated on 16/Oct/24 $${given}\:{isoscele}\:{triangle}\:{with}\:{sides}\:\mathrm{10}\:{and}\:{inradius}\:\mathrm{3}.\:{how}\:{find}\:{base}? \\ $$ Answered by A5T last updated on 16/Oct/24 $${Let}\:{base}={b}\:;{angle}\:{between}\:{non}-{equal}\:{sides}=\theta \\ $$$$\left[\bigtriangleup\right]=\frac{\mathrm{10}×\mathrm{10}{sin}\left(\mathrm{180}−\mathrm{2}\theta\right)}{\mathrm{2}}=\frac{{b}×\mathrm{10}×{sin}\theta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{50}×\mathrm{2}{cos}\theta=\mathrm{5}{b}\Rightarrow{cos}\theta=\frac{{b}}{\mathrm{20}}\Rightarrow{sin}\theta=\frac{\sqrt{\mathrm{400}−{b}^{\mathrm{2}} }}{\mathrm{20}}…
Question Number 212417 by mr W last updated on 13/Oct/24 Commented by mr W last updated on 13/Oct/24 $${unsolved}\:{old}\:{question}\:\mathrm{212291} \\ $$ Answered by mr W…
Question Number 212214 by CrispyXYZ last updated on 06/Oct/24 $$\mathrm{Find}\:\mathrm{tan}\theta. \\ $$ Commented by CrispyXYZ last updated on 06/Oct/24 Answered by a.lgnaoui last updated on…