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Author: Tinku Tara

Question-212720

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-212719

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-212745

Question Number 212745 by RoseAli last updated on 22/Oct/24 Answered by mehdee7396 last updated on 22/Oct/24 $${lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{\mathrm{1}+{tanx}}{\mathrm{1}−{tanx}}−\mathrm{1}\right)×\frac{\mathrm{1}}{{sinx}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{\mathrm{2}{tanx}}{\mathrm{1}−{tanx}}\right)×\frac{\mathrm{1}}{{sinx}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{\mathrm{2}}{\mathrm{1}−{tanx}}\right)×\frac{\mathrm{1}}{{cosx}}=\mathrm{2} \\…

Question-212736

Question Number 212736 by RojaTaniya last updated on 22/Oct/24 Answered by efronzo1 last updated on 22/Oct/24 $$\:\:\:\:\frac{\mathrm{x}}{\mathrm{4}}\:=\:\frac{\mathrm{9}}{\mathrm{15}}\:\Rightarrow\mathrm{x}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\:\:\:\:\frac{\mathrm{y}}{\mathrm{5}}=\:\frac{\mathrm{9}}{\mathrm{15}}\Rightarrow\mathrm{y}=\:\mathrm{3} \\ $$$$\:\:\:\:\:\frac{\mathrm{z}}{\mathrm{6}}=\:\frac{\mathrm{9}}{\mathrm{15}}\:\Rightarrow\mathrm{z}=\frac{\mathrm{18}}{\mathrm{5}} \\ $$ Commented by…