Question Number 181329 by manxsol last updated on 24/Nov/22 Answered by som(math1967) last updated on 24/Nov/22 Commented by som(math1967) last updated on 24/Nov/22 $$\bigtriangleup{AOD}\sim\bigtriangleup{COB} \\…
Question Number 181327 by HeferH last updated on 24/Nov/22 Answered by a.lgnaoui last updated on 24/Nov/22 $$\measuredangle{BED}=\mathrm{45}° \\ $$ Answered by a.lgnaoui last updated on…
Question Number 181232 by mr W last updated on 23/Nov/22 Commented by mr W last updated on 23/Nov/22 $${a}\:{train}\:{drives}\:{from}\:{A}\:{in}\:{direction}\:{to} \\ $$$${D}\:{with}\:{speed}\:\mathrm{160}\:{km}/{h}.\:{at}\:{the}\:{same} \\ $$$${time}\:{a}\:{car}\:{drives}\:{from}\:{B}\:{in}\:{direction} \\ $$$${to}\:{E}\:{with}\:{speed}\:\mathrm{100}\:{km}/{h}.…
Question Number 181201 by HeferH last updated on 22/Nov/22 Answered by ARUNG_Brandon_MBU last updated on 23/Nov/22 $$\mathrm{tan}\theta=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{tan2}\theta=\frac{\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\centerdot\frac{\mathrm{3}}{\mathrm{4}}}=\frac{\mathrm{24}}{\mathrm{7}}=\frac{{AC}}{{AE}}=\frac{{AC}}{\mathrm{4}} \\ $$$$\Rightarrow{AC}=\frac{\mathrm{96}}{\mathrm{7}} \\ $$$$\mathrm{area}\left({ABC}\right)=\frac{\mathrm{1}}{\mathrm{2}}×{AB}×{AC} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{9}×\frac{\mathrm{96}}{\mathrm{7}}=\frac{\mathrm{432}}{\mathrm{7}}…
Question Number 115642 by 24224 Opiyo Kamuki last updated on 27/Sep/20 $${A}\:{vertical}\:{post}\:{of}\:{height}\:{h}\:{m}\:{rises}\:{from}\:{a}\:{plane}\:{which}\: \\ $$$${slopes}\:{down}\:{towards}\:{the}\:{South}\:{at}\:{an}\:{angle} \\ $$$$\alpha\:{to}\:{the}\:{horizontal}.\:{Prove}\:{that}\:{the}\:{length} \\ $$$${of}\:{its}\:{shadow}\:{when}\:{the}\:{sun}\:{is}\:\boldsymbol{{S}\theta{W}}\:\: \\ $$$${at}\:{an}\:{elevation}\:\beta\:{is} \\ $$$$ \\ $$$$\frac{{h}\sqrt{\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\:{cos}^{\mathrm{2}}…
Question Number 181151 by mr W last updated on 22/Nov/22 Answered by SEKRET last updated on 22/Nov/22 $$\:\sqrt{\mathrm{2}} \\ $$ Answered by SEKRET last updated…
Question Number 181128 by Acem last updated on 21/Nov/22 Commented by Acem last updated on 21/Nov/22 $$\:{Find}\:{the}\:{length}\:{of}\:{the}\:{crease}\:{L}\left({w},\:\theta\right) \\ $$ Commented by Acem last updated on…
Question Number 181125 by depressiveshrek last updated on 21/Nov/22 $${Let}\:{the}\:{acute}\:{triangle}\:\Delta{ABC}\:\:{have} \\ $$$${an}\:{outer}\:{circumscribed}\:{circle}, \\ $$$${whose}\:{tangents}\:{at}\:{the}\:{points}\:{B}\:{and}\:{C} \\ $$$${intersect}\:{at}\:{point}\:{P}.\:{Let}\:{D}\:{and}\:{E}\:{be} \\ $$$${the}\:{projections}\:{of}\:{perpendicular} \\ $$$${lines}\:{from}\:{point}\:{P}\:{on}\:{AC}\:{and}\:{AB}. \\ $$$${Prove}\:{that}\:{the}\:{interdection}\:{point}\:{of} \\ $$$${the}\:{heights}\:{of}\:\Delta{ADE}\:{is}\:{the}\:{midpoint} \\…
Question Number 181085 by mr W last updated on 21/Nov/22 Commented by mr W last updated on 21/Nov/22 $${find}\:{the}\:{diameter}\:{of}\:{semicircle}. \\ $$ Commented by HeferH last…
Question Number 181070 by mr W last updated on 21/Nov/22 Commented by mr W last updated on 21/Nov/22 $$\left[{Q}\mathrm{181055}\:{modified}\right] \\ $$$${find}\:{the}\:{area}\:{of}\:{trapazoid}\:{ABCD}. \\ $$ Commented by…