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Category: Geometry

Question-194410

Question Number 194410 by necx122 last updated on 05/Jul/23 Commented by necx122 last updated on 05/Jul/23 $${I}'{ve}\:{dedicated}\:{sometime}\:{to}\:{solving}\:{the} \\ $$$${question}\:{but}\:{it}\:{seems}\:{as}\:{though}\:{some} \\ $$$${parameters}\:{are}\:{missing}.{Please}\:{help} \\ $$$${me}\:{out}.\: \\ $$…

Question-194056

Question Number 194056 by ajfour last updated on 26/Jun/23 Answered by a.lgnaoui last updated on 27/Jun/23 $$\mathrm{ABC}\:\:\:\mathrm{triangle}\:\mathrm{equilaterale}\:\: \\ $$$$\mathrm{AB}=\mathrm{BC}=\mathrm{AC}\:\:\:\mathrm{BC}=\mathrm{2acos}\:\mathrm{30}=\mathrm{a}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{AH}=\mathrm{BCsin}\:\mathrm{60}=\frac{\mathrm{3a}}{\mathrm{2}} \\ $$$$\:\:\mathrm{AI}=\mathrm{AM}+\mathrm{MI}=\mathrm{2a}\:\:\:\:\:\left(\mathrm{i}\right) \\ $$$$\:\:\measuredangle\mathrm{MAN}=\mathrm{30}\:;\mathrm{sin}\:\mathrm{30}=\frac{\mathrm{c}}{\mathrm{AM}}\Rightarrow\:\mathrm{AM}=\mathrm{2}\boldsymbol{\mathrm{c}}…