Question Number 172195 by cherokeesay last updated on 24/Jun/22 Answered by som(math1967) last updated on 24/Jun/22 $${let}\:{side}\:{of}\:{biggest}\:{square}={rad}.\:{of} \\ $$$${quarter}\:{circle}\:={r}\:{unit} \\ $$$$\therefore\:{AB}={AL}={r} \\ $$$$\therefore\:{side}\:{of}\:{S}_{\mathrm{2}} =\frac{{r}}{\:\sqrt{\mathrm{2}}} \\…
Question Number 41096 by ajfour last updated on 02/Aug/18 Commented by ajfour last updated on 02/Aug/18 $${AB}={c}\:\:{and}\:\:{CE}={FD}=\mathrm{1} \\ $$$${Find}\:{GC}={DH}\:{in}\:{terms}\:{of}\:{c}\:; \\ $$$$\left({geometrically}\right). \\ $$ Terms of…
Question Number 172144 by mnjuly1970 last updated on 23/Jun/22 Answered by infinityaction last updated on 23/Jun/22 Commented by infinityaction last updated on 23/Jun/22 $$\:\:\:\mathrm{cos}\alpha\:=\:\frac{\mathrm{2}}{{r}}\:\:\:{and}\:\:\:\:\mathrm{sin}\alpha\:\:=\:\:\frac{{r}}{\mathrm{6}}\: \\…
Question Number 41062 by Tawa1 last updated on 01/Aug/18 Commented by tanmay.chaudhury50@gmail.com last updated on 01/Aug/18 Commented by tanmay.chaudhury50@gmail.com last updated on 01/Aug/18 $${pls}\:{wait}\:{destination}\:{yet}\:{to}\:{reach}… \\…
Question Number 41063 by Tawa1 last updated on 01/Aug/18 Commented by alex041103 last updated on 07/Aug/18 $$\mathrm{72} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 41038 by Tawa1 last updated on 31/Jul/18 Commented by tanmay.chaudhury50@gmail.com last updated on 31/Jul/18 $$\left.{about}\:{question}\:{number}\:\mathrm{25}\right) \\ $$$${there}\:{is}\:{eight}\:{corner}…{at}\:{each}\:{corner}\:{three}\:{face} \\ $$$${merge}…{number}\:{at}\:{any}\:{corner}\:{share}\:{by}\:{three} \\ $$$${face}..{target}\:{to}\:{find}\:{sum}\:{of}\:{four}\:{number}\:{such} \\ $$$${that}\:{each}\:{face}\:{give}\:{same}\:{value}…{the}\:{sumtotal}…
Question Number 41027 by behi83417@gmail.com last updated on 31/Jul/18 Answered by MJS last updated on 01/Aug/18 $${A}={B}\:\wedge\:{a}={b} \\ $$$$\mathrm{2}{a}\mathrm{tan}\:{A}={c}\mathrm{tan}\left(\pi−\mathrm{2}{A}\right) \\ $$$$\mathrm{2}{a}\mathrm{tan}\:{A}=−{c}\mathrm{tan}\:\mathrm{2}{A} \\ $$$$\mathrm{let}\:\mathrm{c}=\mathrm{1} \\ $$$$−\mathrm{2}{a}=\frac{\mathrm{tan}\:\mathrm{2}{A}}{\mathrm{tan}\:{A}}…
Question Number 41023 by Tawa1 last updated on 31/Jul/18 Commented by maxmathsup by imad last updated on 31/Jul/18 $${drive}\:{before}\:{calculating}… \\ $$ Terms of Service Privacy…
Question Number 172062 by cortano1 last updated on 23/Jun/22 Answered by kapoorshah last updated on 23/Jun/22 $${s}\:=\:\frac{\mathrm{10}\:+\:\mathrm{14}\:+\:\mathrm{16}}{\mathrm{2}} \\ $$$$\:\:\:=\:\mathrm{20} \\ $$$$ \\ $$$${r}\:=\:\sqrt{\frac{\left(\mathrm{20}\:−\:\mathrm{10}\right)\left(\mathrm{20}\:−\:\mathrm{14}\right)\left(\mathrm{20}\:−\:\mathrm{16}\right)}{\mathrm{20}}} \\ $$$$\:\:\:=\:\mathrm{2}\sqrt{\mathrm{3}}…
Question Number 40948 by behi83417@gmail.com last updated on 30/Jul/18 Commented by MJS last updated on 30/Jul/18 $${B}−{C}=\mathrm{2}{A}\:\Rightarrow\:{C}={B}−\mathrm{2}{A} \\ $$$${A}+{B}+{C}=\mathrm{180}°\:\Rightarrow\:{C}=\mathrm{180}°−{A}−{B} \\ $$$${B}−\mathrm{2}{A}=\mathrm{180}°−{A}−{B}\:\Rightarrow\:{A}=\mathrm{2}{B}−\mathrm{180}° \\ $$$$\Rightarrow\:{C}=\mathrm{360}°−\mathrm{3}{B} \\ $$$$…