Question Number 166928 by amin96 last updated on 02/Mar/22 Answered by mr W last updated on 02/Mar/22 $${a}\:{picture}\:{says}\:{more}\:{than}\:{a}\:{thousant} \\ $$$${words}… \\ $$ Commented by mr…
Question Number 166917 by ajfour last updated on 02/Mar/22 Commented by mr W last updated on 02/Mar/22 Commented by ajfour last updated on 02/Mar/22 $$\left\{\left(\mathrm{2}{b}−{c}\right)\mathrm{cos}\:\theta−{b}\right\}^{\mathrm{2}}…
Question Number 101375 by ajfour last updated on 02/Jul/20 Commented by ajfour last updated on 02/Jul/20 $${In}\:{terms}\:{of}\:{R},\:{find}\:{radii}\:{a},{b},{c}. \\ $$ Commented by ajfour last updated on…
Question Number 166888 by cherokeesay last updated on 01/Mar/22 Answered by mr W last updated on 01/Mar/22 Commented by mr W last updated on 01/Mar/22…
Question Number 35798 by ajfour last updated on 23/May/18 Commented by ajfour last updated on 23/May/18 $$\bigtriangleup{ABC}\:{and}\:\bigtriangleup{CDE}\:{are}\:{congruent}. \\ $$$${Find}\:{the}\:{area}\:{of}\:{quadrilateral} \\ $$$${APQC}\:{in}\:{terms}\:{of}\:{a},\:{b},\:{and}\:{c}\:;\:{the} \\ $$$${sides}\:{of}\:\bigtriangleup{ABC}\:. \\ $$…
Question Number 166875 by amin96 last updated on 01/Mar/22 Commented by amin96 last updated on 01/Mar/22 $$\boldsymbol{\mathrm{circle}}\:\boldsymbol{\mathrm{radius}}=??? \\ $$ Answered by MJS_new last updated on…
Question Number 166861 by Tawa11 last updated on 01/Mar/22 Commented by cortano1 last updated on 01/Mar/22 $$\Rightarrow\mathrm{3}.\mathrm{x}=\mathrm{6}.\mathrm{2}\Rightarrow\mathrm{x}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{r}=\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }\:=\sqrt{\mathrm{49}+\mathrm{16}}=\sqrt{\mathrm{65}} \\ $$…
Question Number 35763 by $@ty@m last updated on 23/May/18 Answered by ajfour last updated on 23/May/18 $${let}\:\angle{BDF}\:=\:\alpha \\ $$$${AF}=\mathrm{30cos}\:{A}\:\:;\:\:\:{AE}=\mathrm{36cos}\:{A} \\ $$$$\mathrm{cot}\:\alpha\:=\:\frac{{AF}\mathrm{cos}\:{B}+{AE}\mathrm{cos}\:{C}}{{BF}\mathrm{sin}\:{B}−{CE}\mathrm{sin}\:{C}} \\ $$$$\:\:\:=\frac{\mathrm{30cos}\:{A}\mathrm{cos}\:{B}+\mathrm{36cos}\:{A}\mathrm{cos}\:{C}}{\left(\mathrm{18}−\mathrm{30cos}\:{A}\right)\mathrm{sin}\:{B}−\left(\mathrm{15}−\mathrm{36cos}\:{A}\right)\mathrm{sin}\:{C}} \\ $$$${BD}={BF}\mathrm{cos}\:{B}+{BF}\mathrm{sin}\:{B}\mathrm{cot}\:\alpha…
Question Number 35750 by ajfour last updated on 23/May/18 Commented by ajfour last updated on 23/May/18 $${Find}\:{proportion}\:{of}\:{blue},\:{brown}, \\ $$$${yellow},\:{and}\:{red}\:{coloured}\:{areas}. \\ $$$${Given}:\:{chord}\:{makes}\:{angle}\:\theta\:{with} \\ $$$${the}\:{diameter}\:{shown}\:{and}\:{that}\:{the} \\ $$$${point}\:{of}\:{intersection}\:{of}\:{the}\:{chord}…
Question Number 166822 by ajfour last updated on 28/Feb/22 Answered by mr W last updated on 28/Feb/22 Commented by mr W last updated on 28/Feb/22…