Question Number 28464 by ajfour last updated on 26/Jan/18 Answered by mrW2 last updated on 26/Jan/18 $${eqn}.\:{of}\:{ellipse}: \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{\mathrm{2}{x}}{{a}^{\mathrm{2}}…
Question Number 93986 by $@ty@m123 last updated on 16/May/20 Answered by mr W last updated on 16/May/20 Commented by mr W last updated on 16/May/20…
Question Number 159392 by ajfour last updated on 16/Nov/21 Commented by ajfour last updated on 16/Nov/21 $$\:\:{Find}\:{radius}\:{of}\:{the}\:{circles}\left({equal}\right). \\ $$ Answered by mr W last updated…
Question Number 28240 by $@ty@m last updated on 22/Jan/18 Answered by ajfour last updated on 22/Jan/18 $${DE}×{EA}={CE}×{EF} \\ $$$$\mathrm{6}×\mathrm{2}=\mathrm{6}\sqrt{\mathrm{2}}×{EF} \\ $$$$\Rightarrow\:\:{EF}=\sqrt{\mathrm{2}}\:\:{so}\:\:{CF}=\mathrm{7}\sqrt{\mathrm{2}} \\ $$$${let}\:{foot}\:{of}\:\bot\:{from}\:{F}\:{on}\:{BC}\:{be}\:{G}. \\ $$$${CG}=\mathrm{7}\:,\:{GB}=\mathrm{1},\:{FG}=\mathrm{7}…
Question Number 27959 by ajfour last updated on 17/Jan/18 Commented by ajfour last updated on 17/Jan/18 $${Inspired}\:{with}\:{Q}.#\mathrm{27942} \\ $$ Commented by beh.i83417@gmail.com last updated on…
Question Number 27942 by beh.i83417@gmail.com last updated on 17/Jan/18 Commented by beh.i83417@gmail.com last updated on 17/Jan/18 $$\angle{A}=\mathrm{60}^{°} ,{AB}={c},{AC}={b},{BC}={a} \\ $$$${draw}\:{tangent}\:{lines}\:{from}:{B}\:{and}\:{C},{to}\:{circle}, \\ $$$${such}\:{that}:\angle{FBC}=\angle{FCB}. \\ $$$$\left.\mathrm{1}\right){find}\:{radius}\:\boldsymbol{{r}}\:{in}\:{terms}\:{of}:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}. \\…
Question Number 159005 by Tawa11 last updated on 11/Nov/21 Answered by ajfour last updated on 12/Nov/21 $${s}=\mathrm{2}{r}\mathrm{sin}\:\alpha \\ $$$${p}={r}\mathrm{sin}\:\mathrm{2}\alpha \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{35}/\mathrm{2}}{{r}}\:=\frac{\mathrm{7}}{\mathrm{8}};\:\:\mathrm{sin}\:\beta=\frac{\sqrt{\mathrm{15}}}{\mathrm{8}} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{90}°−\alpha\right)}{\mathrm{27}}=\frac{\mathrm{sin}\:\left(\mathrm{3}\alpha+\beta\right)}{{p}} \\ $$$$\mathrm{20sin}\:\mathrm{2}\alpha\mathrm{cos}\:\alpha=\mathrm{27sin}\:\left(\mathrm{3}\alpha+\beta\right)…
Question Number 158964 by amin96 last updated on 11/Nov/21 Commented by mr W last updated on 19/Nov/21 $${see}\:{Q}\mathrm{159606} \\ $$ Terms of Service Privacy Policy…
Question Number 158924 by ajfour last updated on 10/Nov/21 Commented by ajfour last updated on 10/Nov/21 $${a},\:{b},\:{c}\:{are}\:{slant}\:{edges}\:{of}\:{outer} \\ $$$${tetrahedron}.\:{Find}\:{minimum} \\ $$$${volume}\:{of}\:{inner}\:{tetrahedron} \\ $$$${whose}\:{vertices}\:{lie}\:{on}\:{faces}\:{of} \\ $$$${outer}\:{one}\:{when}\:{the}\:{outer}\:{one}…
Question Number 158893 by ajfour last updated on 10/Nov/21 Answered by ajfour last updated on 10/Nov/21 $$\frac{{r}^{\mathrm{4}} −\left({r}+{c}\right)^{\mathrm{2}} }{{c}}=\frac{{r}^{\mathrm{2}} }{\:{r}^{\mathrm{2}} −\mathrm{1}} \\ $$$${r}^{\mathrm{3}} −{r}={c}\:\:\:\:\Rightarrow\:\:{r}^{\mathrm{6}} =\left({r}+{c}\right)^{\mathrm{2}}…