Question Number 206430 by cortano21 last updated on 14/Apr/24 Answered by mr W last updated on 14/Apr/24 $${AB}={a},\:{say} \\ $$$$\mathrm{7}^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} −\mathrm{2}×\mathrm{8}{a}\:\mathrm{cos}\:\mathrm{60}° \\ $$$${a}^{\mathrm{2}}…
Question Number 206396 by cortano21 last updated on 13/Apr/24 Answered by mr W last updated on 13/Apr/24 Commented by mr W last updated on 13/Apr/24…
Question Number 206321 by cortano21 last updated on 12/Apr/24 Answered by TonyCWX08 last updated on 12/Apr/24 $${By}\:\Delta{BCD}\:\backsim\:\Delta{CDE} \\ $$$$\frac{{x}}{\:\sqrt{\mathrm{14}}}=\frac{\mathrm{7}\sqrt{\mathrm{14}}}{{x}} \\ $$$${x}^{\mathrm{2}} =\mathrm{98} \\ $$$$ \\…
Question Number 206292 by cortano21 last updated on 11/Apr/24 Answered by A5T last updated on 11/Apr/24 $${Let}\:{AE}={x};{BE}={y};{BF}={v};{FC}={w} \\ $$$${S}+\mathrm{39}=\frac{\left(\mathrm{2}{v}+{w}\right)\left({x}+{y}\right)}{\mathrm{2}}=\frac{\mathrm{2}{v}\left({x}+{y}\right)}{\mathrm{2}}+\mathrm{15}\Rightarrow{S}={vx} \\ $$$${wx}=\mathrm{54}−{S}=\mathrm{30}−{yw}\Rightarrow{yw}={S}−\mathrm{24} \\ $$$$\frac{\left[{BFD}\right]}{\left[{DFC}\right]}=\frac{{v}}{{w}}\Rightarrow\left[{BFD}\right]=\frac{\mathrm{15}{v}}{{w}} \\ $$$$\frac{\left[{EDB}\right]}{\left[{ADE}\left[\right.\right.}=\frac{{y}}{{x}}\Rightarrow{EDB}=\frac{\mathrm{27}{y}}{{x}}…
Question Number 206275 by cortano21 last updated on 11/Apr/24 Answered by HeferH24 last updated on 11/Apr/24 $$\:{CDEF}\:=\:{m} \\ $$$$\:{ABFE}\:=\:\mathrm{3}{m} \\ $$$$\:\left(\frac{\mathrm{4}}{\mathrm{6}}\right)^{\mathrm{2}} =\:\frac{\mathrm{4}}{\mathrm{9}}=\frac{\mathrm{4}{k}}{\mathrm{9}{k}} \\ $$$$\:\mathrm{5}{k}\:=\:\mathrm{4}{m} \\…
Question Number 206269 by cortano21 last updated on 10/Apr/24 Commented by A5T last updated on 11/Apr/24 Commented by A5T last updated on 11/Apr/24 $${General}\:{formula}.\:{When}\:{k}_{\mathrm{1}} {k}_{\mathrm{2}}…
Question Number 206267 by mr W last updated on 10/Apr/24 Commented by mr W last updated on 10/Apr/24 $${if}\:{S}_{\mathrm{1}} +{S}_{\mathrm{2}} +{S}_{\mathrm{3}} =\mathrm{10},\:{find}\:{S}_{\mathrm{7}} +{S}_{\mathrm{8}} +{S}_{\mathrm{9}} =?…
Question Number 206232 by mr W last updated on 09/Apr/24 Answered by A5T last updated on 09/Apr/24 $${S}_{\mathrm{2}} ={S}_{\mathrm{1}} +{S}_{\mathrm{3}} −\mathrm{2}{AB}×{BCcos}\left(\theta\right) \\ $$$${S}_{\mathrm{1}} ={S}_{\mathrm{2}} +{S}_{\mathrm{3}}…
Question Number 206216 by cortano21 last updated on 09/Apr/24 Answered by A5T last updated on 09/Apr/24 $${HC}=\sqrt{\mathrm{108}+\mathrm{4}}=\sqrt{\mathrm{112}}=\mathrm{4}\sqrt{\mathrm{7}} \\ $$$${Let}\:{HC}\:{meet}\:{the}\:{circumcircle}\:{of}\:{the}\:{hexagon} \\ $$$${at}\:{L}\:{then},\:{HC}×{HL}={FH}×{HE}=\mathrm{8}\Rightarrow{HL}=\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{7}} \\ $$$${sin}\angle{EHC}=\frac{\sqrt{\mathrm{108}}}{\:\sqrt{\mathrm{112}}}=\frac{\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{14}}\Rightarrow{cos}\angle{EHC}=\frac{\sqrt{\mathrm{7}}}{\mathrm{14}} \\ $$$$\Rightarrow\sqrt{{x}^{\mathrm{2}}…
Question Number 206198 by lmcp1203 last updated on 09/Apr/24 Answered by A5T last updated on 09/Apr/24 $$\frac{{sin}\mathrm{4}\theta}{{AD}}=\frac{{sin}\left(\mathrm{90}−\mathrm{3}\theta\right)}{{AB}};\frac{{sin}\left(\theta\right)}{{AD}}=\frac{{sin}\left(\mathrm{90}−\mathrm{4}\theta\right)}{{CD}={AB}} \\ $$$$\Rightarrow\frac{{sin}\left(\mathrm{4}\theta\right)}{{sin}\left(\mathrm{90}−\mathrm{3}\theta\right)}=\frac{{sin}\left(\theta\right)}{{sin}\left(\mathrm{90}−\mathrm{4}\theta\right)}\Rightarrow{x}=\mathrm{20}° \\ $$ Answered by lmcp1203 last…