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Question-156671




Question Number 156671 by cortano last updated on 14/Oct/21
Answered by som(math1967) last updated on 14/Oct/21
Commented by som(math1967) last updated on 14/Oct/21
cosα=((10)/(12))=(5/6)  sinα=((√(11))/6)  sin2α=2×((√(11))/6)×(5/6)=((5(√(11)))/(18))  again sin2α=((LM)/(AL))    ((10)/(AL))=((5(√(11)))/(18))⇒AL=((36)/( (√(11))))=((36(√(11)))/(11))  DL^2 =(((36(√(11)))/(11)))^2 −10^2   DL=((14)/( (√(11))))=((14(√(11)))/(11))  LC=(10−((14(√(11)))/(11)))  BN=(√(12^2 −10^2 ))=2(√(11))  CN=10−2(√(11))  perimeter=AL+LC+CN+AN  =((36(√(11)))/(11)) +(10−((14(√(11)))/(11)))+10−2(√(11))+12  =32+((36(√(11)))/(11)) −((36(√(11)))/(11))=32unit
$${cos}\alpha=\frac{\mathrm{10}}{\mathrm{12}}=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$${sin}\alpha=\frac{\sqrt{\mathrm{11}}}{\mathrm{6}} \\ $$$${sin}\mathrm{2}\alpha=\mathrm{2}×\frac{\sqrt{\mathrm{11}}}{\mathrm{6}}×\frac{\mathrm{5}}{\mathrm{6}}=\frac{\mathrm{5}\sqrt{\mathrm{11}}}{\mathrm{18}} \\ $$$${again}\:{sin}\mathrm{2}\alpha=\frac{{LM}}{{AL}} \\ $$$$\:\:\frac{\mathrm{10}}{{AL}}=\frac{\mathrm{5}\sqrt{\mathrm{11}}}{\mathrm{18}}\Rightarrow{AL}=\frac{\mathrm{36}}{\:\sqrt{\mathrm{11}}}=\frac{\mathrm{36}\sqrt{\mathrm{11}}}{\mathrm{11}} \\ $$$${DL}^{\mathrm{2}} =\left(\frac{\mathrm{36}\sqrt{\mathrm{11}}}{\mathrm{11}}\right)^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} \\ $$$${DL}=\frac{\mathrm{14}}{\:\sqrt{\mathrm{11}}}=\frac{\mathrm{14}\sqrt{\mathrm{11}}}{\mathrm{11}} \\ $$$${LC}=\left(\mathrm{10}−\frac{\mathrm{14}\sqrt{\mathrm{11}}}{\mathrm{11}}\right) \\ $$$${BN}=\sqrt{\mathrm{12}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{11}} \\ $$$${CN}=\mathrm{10}−\mathrm{2}\sqrt{\mathrm{11}} \\ $$$${perimeter}={AL}+{LC}+{CN}+{AN} \\ $$$$=\frac{\mathrm{36}\sqrt{\mathrm{11}}}{\mathrm{11}}\:+\left(\mathrm{10}−\frac{\mathrm{14}\sqrt{\mathrm{11}}}{\mathrm{11}}\right)+\mathrm{10}−\mathrm{2}\sqrt{\mathrm{11}}+\mathrm{12} \\ $$$$=\mathrm{32}+\frac{\mathrm{36}\sqrt{\mathrm{11}}}{\mathrm{11}}\:−\frac{\mathrm{36}\sqrt{\mathrm{11}}}{\mathrm{11}}=\mathrm{32}{unit} \\ $$
Commented by cortano last updated on 14/Oct/21
oo yes gave kudos
$$\mathrm{oo}\:\mathrm{yes}\:\mathrm{gave}\:\mathrm{kudos} \\ $$
Commented by otchereabdullai@gmail.com last updated on 14/Oct/21
wow
$$\mathrm{wow} \\ $$
Commented by Tawa11 last updated on 14/Oct/21
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$
Answered by john_santu last updated on 15/Oct/21
Commented by cortano last updated on 15/Oct/21
nice
$$\mathrm{nice} \\ $$

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