Question Number 187925 by Rupesh123 last updated on 23/Feb/23 Answered by HeferH last updated on 23/Feb/23 Commented by HeferH last updated on 23/Feb/23 $${Trapezoid}\Rightarrow\: \\…
Question Number 56838 by necx1 last updated on 25/Mar/19 $${What}\:{would}\:{be}\:{the}\:{diameter}\:{of}\:{a}\:{circle} \\ $$$${having}\:{a}\:{heptagon}\:{of}\:{sides}\:\mathrm{45}{m},\mathrm{60}{m}, \\ $$$$\mathrm{60}{m},\mathrm{50}{m},\mathrm{40}{m},\mathrm{45}{m}\:{and}\:\mathrm{50}{m}\:{inscribed} \\ $$$${in}\:{it}? \\ $$ Answered by MJS last updated on 25/Mar/19…
Question Number 187893 by Rupesh123 last updated on 23/Feb/23 Answered by a.lgnaoui last updated on 23/Feb/23 $$\bigtriangleup{ABD}\:\:\:\:\frac{\mathrm{sin}\:\mathrm{53}}{{AD}}=\frac{\mathrm{sin}\:\left(\mathrm{14}+{x}\right)}{{AB}}=\frac{\mathrm{sin}\:\left(\mathrm{14}+\mathrm{53}+{x}\right)}{\mathrm{2}{k}} \\ $$$$\:\:\:\mathrm{2}{k}\mathrm{sin}\:\mathrm{53}={AD}\mathrm{sin}\:\left(\mathrm{67}+{x}\right)\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\bigtriangleup{ACD}\:\:\:\frac{\mathrm{sin}\:\mathrm{14}}{{k}}=\frac{\mathrm{sin}\:{x}}{{AD}} \\ $$$$\:\:\:\:\:{k}\mathrm{sin}\:{x}={AD}\mathrm{sin}\:\mathrm{14}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$…
Question Number 187828 by Rupesh123 last updated on 22/Feb/23 Answered by HeferH last updated on 22/Feb/23 Commented by HeferH last updated on 22/Feb/23 $${i}.\:\mathrm{3}\alpha\:+\:\beta\:+\omega\:=\:\mathrm{180}° \\…
Question Number 187830 by normans last updated on 22/Feb/23 Commented by normans last updated on 22/Feb/23 $$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{find}}\:\boldsymbol{{r}}\:=??? \\ $$$$\: \\ $$ Answered by HeferH last…
Question Number 187817 by normans last updated on 22/Feb/23 Answered by mr W last updated on 22/Feb/23 Commented by mr W last updated on 22/Feb/23…
Question Number 187819 by normans last updated on 22/Feb/23 Answered by mr W last updated on 22/Feb/23 Commented by mr W last updated on 22/Feb/23…
Question Number 187808 by LowLevelLump last updated on 22/Feb/23 Answered by a.lgnaoui last updated on 22/Feb/23 $$\bigtriangleup{ABC}'\:\:\begin{cases}{{AB}={AC}={BC}={a}}\\{{S}_{\mathrm{1}} =\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}}}\end{cases} \\ $$$$\bigtriangleup{ACB}'\:\:\:\begin{cases}{{AC}={AB}'={CB}'={b}}\\{{S}_{\mathrm{2}} =\frac{{b}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}}}\end{cases} \\ $$$$\bigtriangleup{BCC}'\:\:\:\begin{cases}{{BC}={BC}'={CC}^{'}…
Question Number 187790 by cherokeesay last updated on 21/Feb/23 Answered by mr W last updated on 21/Feb/23 Commented by mr W last updated on 22/Feb/23…
Question Number 187774 by normans last updated on 21/Feb/23 Answered by mr W last updated on 21/Feb/23 Commented by mr W last updated on 21/Feb/23…