Question Number 176913 by a.lgnaoui last updated on 28/Sep/22 $${Determiner}\:{la}\:{hauteur}\:\mathrm{D}{E}\left({r}+{x}\right)\:{en}\:{fonction}\:{de}\:{r} \\ $$$${r}=\mathrm{OOC}=\mathrm{BH}\:\:\:\:\:\mathrm{BF}=\mathrm{20} \\ $$$$\mathrm{pour}\:\mathrm{que}\:\mathrm{distance}\left(\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DE}+\mathrm{EF}\:\:\mathrm{soit}\:\right. \\ $$$$\mathrm{sgale}\:\mathrm{AC}+\mathrm{arcCDF} \\ $$ Commented by a.lgnaoui last updated on 28/Sep/22…
Question Number 176906 by HeferH last updated on 27/Sep/22 Commented by ajfour last updated on 27/Sep/22 Commented by ajfour last updated on 27/Sep/22 $${Let}\:{radius}\:{of}\:{circle}\:{be}\:{r}\:\&\:{side}\:{of} \\…
Question Number 45785 by Tawa1 last updated on 16/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 16/Oct/18 $${let}\:{centre}\:{of}\:{circle}\:{is}\:\left(\alpha,\beta\right)\:{and}\:{radius}\:{r} \\ $$$${so}\:{eqn}\:{of}\:{circle}\:\left({x}−\alpha\right)^{\mathrm{2}} +\left({y}−\beta\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${now}\:{as}\:{per}\:{condition} \\ $$$$\left.\mathrm{1}\right){x}\:{axis}\:{is}\left[{the}\:{tangent}\:{of}\:{circle}\:{so}\:{distance}\right.…
Question Number 45753 by Tawa1 last updated on 16/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 16/Oct/18 $${for}\:{cylnder}\:{volume}\:=\pi{R}^{\mathrm{2}} {H}\:\:{when}\:{base}\:{is}\:{circle} \\ $$$${of}\:{area}\:\pi{R}^{\mathrm{2}} \\ $$$${but}\:{here}\:{volume}\:{is}\:{sector}\:{aresfor}\:\mathrm{30}^{{o}} ×{height} \\ $$$${requiref}\:{volume}\:{is}…
Question Number 111284 by Aina Samuel Temidayo last updated on 03/Sep/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{whose}\:\mathrm{vertices}\:\mathrm{lie}\:\mathrm{on}\:\mathrm{a}\:\mathrm{regular} \\ $$$$\mathrm{hexagon}\:\mathrm{of}\:\mathrm{unit}\:\mathrm{area}. \\ $$ Commented by mr W last updated on…
Question Number 111278 by Aina Samuel Temidayo last updated on 03/Sep/20 $$\mathrm{Towns}\:\mathrm{A},\mathrm{B},\mathrm{C}\:\mathrm{and}\:\mathrm{D}\:\mathrm{are}\:\mathrm{located}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:\mathrm{whose}\:\mathrm{area}\:\mathrm{is} \\ $$$$\mathrm{1000km}^{\mathrm{2}} .\:\mathrm{There}\:\mathrm{is}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\mathrm{highway}\:\mathrm{passing}\:\mathrm{through}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{square}\:\mathrm{but}\:\mathrm{not}\:\mathrm{through}\:\mathrm{any}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{towns}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{squares} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{distances}\:\mathrm{of}\:\mathrm{the}\:\mathrm{towns}\:\mathrm{to}\:\mathrm{the}…
Question Number 111279 by Aina Samuel Temidayo last updated on 03/Sep/20 $$\mathrm{Triangle}\:\mathrm{ABC}\:\mathrm{has}\:\mathrm{AB}=\mathrm{2}\centerdot\mathrm{AC}.\:\mathrm{Let} \\ $$$$\mathrm{D}\:\mathrm{and}\:\mathrm{E}\:\mathrm{be}\:\mathrm{on}\:\mathrm{AB}\:\mathrm{and}\:\mathrm{BC} \\ $$$$\mathrm{respectively}\:\mathrm{such}\:\mathrm{that}\:\angle\mathrm{BAE} \\ $$$$=\angle\mathrm{ACD}.\:\mathrm{Let}\:\mathrm{F}\:\mathrm{be}\:\mathrm{the}\:\mathrm{intersections}\:\mathrm{of} \\ $$$$\mathrm{segments}\:\mathrm{AE}\:\mathrm{and}\:\mathrm{CD},\:\mathrm{and}\:\mathrm{suppose} \\ $$$$\mathrm{that}\:\bigtriangleup\mathrm{CFE}\:\mathrm{is}\:\mathrm{equilateral}.\:\mathrm{What}\:\mathrm{is} \\ $$$$\angle\mathrm{ACB}? \\…
Question Number 111277 by Aina Samuel Temidayo last updated on 03/Sep/20 $$\mathrm{In}\:\mathrm{a}\:\mathrm{quadrilateral}\:\mathrm{ABCD},\:\angle\mathrm{B}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{right}\:\mathrm{angle},\:\mathrm{diagonal}\:\mathrm{AC}\:\mathrm{is} \\ $$$$\mathrm{perpendicular}\:\mathrm{to} \\ $$$$\mathrm{CD},\mathrm{BC}=\mathrm{21cm},\mathrm{CD}=\mathrm{14cm}\:\mathrm{and} \\ $$$$\mathrm{AD}=\mathrm{31cm}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{ABCD}. \\ $$ Answered by 1549442205PVT…
Question Number 111267 by I want to learn more last updated on 03/Sep/20 Answered by Her_Majesty last updated on 03/Sep/20 $${tan}\alpha−{tan}\beta=\frac{{sin}\left(\alpha−\beta\right)}{{cos}\alpha{cos}\beta};\:{cos}\alpha{cos}\beta=\frac{{cos}\left(\alpha−\beta\right)+{cos}\left(\alpha+\beta\right)}{\mathrm{2}} \\ $$$$\frac{{a}}{{b}}=\frac{{tan}\mathrm{10}°}{{tan}\mathrm{50}°−{tan}\mathrm{40}°}=\frac{\frac{{sin}\mathrm{10}°}{{cos}\mathrm{10}°}}{\frac{{sin}\mathrm{10}°}{{cos}\mathrm{50}°{cos}\mathrm{40}°}}=\frac{{cos}\mathrm{50}°{cos}\mathrm{40}°}{{cos}\mathrm{10}°}= \\ $$$$=\frac{\frac{{cos}\mathrm{10}°+{cos}\mathrm{90}°}{\mathrm{2}}}{{cos}\mathrm{10}°}=\frac{\frac{{cos}\mathrm{10}°}{\mathrm{2}}}{{cos}\mathrm{10}°}=\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 176776 by cherokeesay last updated on 26/Sep/22 Answered by BaliramKumar last updated on 27/Sep/22 $$\frac{\mathrm{3}−{a}}{{a}}\:=\:\frac{{a}}{\mathrm{4}−{a}} \\ $$$${a}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} −\mathrm{7}{a}+\mathrm{12} \\ $$$${a}\:=\:\frac{\mathrm{12}}{\mathrm{7}} \\ $$$$\frac{{a}^{\mathrm{2}}…