Question Number 177047 by Ar Brandon last updated on 30/Sep/22 Answered by som(math1967) last updated on 30/Sep/22 $${cosD}=\frac{\mathrm{7}^{\mathrm{2}} +{DC}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }{\mathrm{2}×\mathrm{7}{DC}}=\frac{\mathrm{24}+{DC}^{\mathrm{2}} }{\mathrm{14}{DC}} \\ $$$${again}\:\:{cosD}=\frac{\mathrm{7}^{\mathrm{2}} +{DB}^{\mathrm{2}}…
Question Number 177046 by Ar Brandon last updated on 30/Sep/22 Answered by som(math1967) last updated on 30/Sep/22 $${Area}\left(\bigtriangleup{ADE}+\bigtriangleup{BEC}\right)=\frac{\mathrm{1}}{\mathrm{2}}{AreaABCD} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{x}}{\mathrm{2}}\:\left({let}\right) \\ $$$${Area}\bigtriangleup{AEB}=\frac{\mathrm{1}}{\mathrm{2}}{ABCD}=\frac{{x}}{\mathrm{2}} \\ $$$${Area}\left({ADEP}+\bigtriangleup{ADE}\right) \\…
Question Number 177043 by Ar Brandon last updated on 30/Sep/22 Answered by a.lgnaoui last updated on 30/Sep/22 $$\left.\mathrm{A}\right){only}\:\mathrm{I} \\ $$ Commented by Rasheed.Sindhi last updated…
Question Number 176973 by Ar Brandon last updated on 28/Sep/22 Commented by Ar Brandon last updated on 28/Sep/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{triangle}. \\ $$ Terms of Service Privacy…
Question Number 176946 by Ar Brandon last updated on 28/Sep/22 Answered by mr W last updated on 29/Sep/22 Commented by mr W last updated on…
Question Number 176933 by Ar Brandon last updated on 28/Sep/22 Answered by mr W last updated on 28/Sep/22 $${AC}+{BC}>{AB}=\mathrm{7} \\ $$$$\left({AC}+{BC}\right)_{{min}} =\mathrm{8} \\ $$$${p}_{{min}} =\left({AC}+{BC}+{AB}\right)_{{min}}…
Question Number 176936 by Ar Brandon last updated on 28/Sep/22 Answered by mr W last updated on 28/Sep/22 $$\frac{{x}}{\mathrm{sin}\:\left(\mathrm{15}+\mathrm{30}\right)}=\frac{\mathrm{4}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\mathrm{sin}\:\mathrm{15}} \\ $$$${x}=\frac{\mathrm{4}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)×\mathrm{4}}{\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}\right)}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{8}\:{cm}\:\checkmark \\ $$ Commented by…
Question Number 176942 by Ar Brandon last updated on 28/Sep/22 Answered by som(math1967) last updated on 28/Sep/22 $${HE}={GF}=\frac{\mathrm{1}}{\mathrm{2}}×{AC}=\frac{\mathrm{17}}{\mathrm{2}}{cm} \\ $$$${EF}={HG}=\frac{\mathrm{1}}{\mathrm{2}}×{DB}=\frac{\mathrm{13}}{\mathrm{2}}{cm} \\ $$$${perimeter}\:{EFGH}=\mathrm{2}\left(\frac{\mathrm{17}}{\mathrm{2}}\:+\frac{\mathrm{13}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{30}{cm} \\…
Question Number 176853 by Ar Brandon last updated on 27/Sep/22 Answered by som(math1967) last updated on 27/Sep/22 $$\:\frac{{PT}}{{TA}}=\frac{{PR}}{{RK}}=\frac{{PS}}{{SE}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${let}\:{PT}={PR}={PS}=\mathrm{2}{x}\: \\ $$$${TA}=\mathrm{3}{x} \\ $$$$\therefore\:{each}\:{side}\:{of}\:{cube}={PT}+{TA}=\mathrm{5}{x} \\…
Question Number 176848 by Ar Brandon last updated on 27/Sep/22 Answered by mr W last updated on 27/Sep/22 $$\beta=\frac{\pi}{\mathrm{2}}−\alpha \\ $$$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}\beta\right)=\mathrm{sin}\:\mathrm{2}\beta=\mathrm{sin}\:\left(\pi−\mathrm{2}\alpha\right) \\ $$$$=\mathrm{sin}\:\mathrm{2}\alpha=\mathrm{2}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha \\ $$$$=\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}}…