Question Number 19709 by NECC last updated on 14/Aug/17 $${In}\:{the}\:{cyclic}\:{quadrilateral}\:{ABCD} \\ $$$${AB}=\mathrm{7},{BC}=\mathrm{8},{CD}=\mathrm{8},{DA}=\mathrm{15}. \\ $$$${Calculate}\:{the}\:{angle}\:{ADC}\:{and} \\ $$$${the}\:{length}\:{ofAC}. \\ $$ Answered by Tinkutara last updated on 15/Aug/17…
Question Number 85241 by Power last updated on 20/Mar/20 Answered by jagoll last updated on 20/Mar/20 $$\mathrm{minimum}\:\frac{\mathrm{4s}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }{\mathrm{5st}}\:=\:\frac{\mathrm{4s}^{\mathrm{2}} }{\mathrm{5st}}\:+\:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{5st}} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{5}}\left(\frac{\mathrm{s}}{\mathrm{t}}\right)\:+\:\frac{\mathrm{1}}{\mathrm{5}}\left(\frac{\mathrm{t}}{\mathrm{s}}\right). \\ $$$$\mathrm{let}\:\frac{\mathrm{s}}{\mathrm{t}}\:=\:\mathrm{u}\:\Rightarrow\mathrm{f}\left(\mathrm{u}\right)\:=\:\frac{\mathrm{4}}{\mathrm{5}}\mathrm{u}+\frac{\mathrm{1}}{\mathrm{5}}\mathrm{u}^{−\mathrm{1}}…
Question Number 19699 by Tinkutara last updated on 14/Aug/17 $$\mathrm{Let}\:{S}\:\mathrm{be}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{centre}\:{O}.\:\mathrm{A}\:\mathrm{chord} \\ $$$${AB},\:\mathrm{not}\:\mathrm{a}\:\mathrm{diameter},\:\mathrm{divides}\:{S}\:\mathrm{into}\:\mathrm{two} \\ $$$$\mathrm{regions}\:{R}_{\mathrm{1}} \:\mathrm{and}\:{R}_{\mathrm{2}} \:\mathrm{such}\:\mathrm{that}\:{O}\:\mathrm{belongs} \\ $$$$\mathrm{to}\:{R}_{\mathrm{2}} .\:\mathrm{Let}\:{S}_{\mathrm{1}} \:\mathrm{be}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{centre}\:\mathrm{in} \\ $$$${R}_{\mathrm{1}} ,\:\mathrm{touching}\:{AB}\:\mathrm{at}\:{X}\:\mathrm{and}\:{S}\:\mathrm{internally}. \\ $$$$\mathrm{Let}\:{S}_{\mathrm{2}}…
Question Number 19668 by Joel577 last updated on 14/Aug/17 Commented by Joel577 last updated on 15/Aug/17 $$\mathrm{An}\:\mathrm{equateral}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{side}\:\mathrm{lenght} \\ $$$${d}.\:\mathrm{Circle}\:{L}_{\mathrm{1}} \:\mathrm{touching}\:\Delta{ABC}\:\mathrm{at}\:{A}\:\mathrm{and}\:{B}. \\ $$$$\mathrm{Circle}\:{L}_{\mathrm{2}} \:\mathrm{touching}\:{AC},\:{BC},\:\mathrm{and}\:{L}_{\mathrm{1}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{circle}\:{L}_{\mathrm{2}}…
Question Number 19659 by thukada last updated on 14/Aug/17 $$\frac{\mathrm{1}+{sec}\theta}{{sec}\theta}=\frac{{sin}^{\mathrm{2}\:} \theta}{\mathrm{1}−{cos}\theta} \\ $$$$ \\ $$ Answered by prakash jain last updated on 14/Aug/17 $$\frac{\mathrm{1}+\mathrm{sec}\:\theta}{\mathrm{sec}\:\theta}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{sec}\:\theta} \\…
Question Number 150688 by ajfour last updated on 14/Aug/21 Commented by ajfour last updated on 14/Aug/21 $${Find}\:{maximum}\:{tetrahedron} \\ $$$${volume}. \\ $$ Answered by mr W…
Question Number 150683 by Jonathanwaweh last updated on 14/Aug/21 Commented by Tawa11 last updated on 14/Aug/21 $$\mathrm{Great} \\ $$ Answered by ajfour last updated on…
Question Number 85074 by ajfour last updated on 18/Mar/20 Commented by ajfour last updated on 19/Mar/20 $${If}\:{a}\:{cylindrical}\:{cavity}\:{be}\:{drilled} \\ $$$${out}\:{of}\:{a}\:{hemisphere}\:{of}\:{radius}\:{R}, \\ $$$${in}\:{the}\:{place}\:{shown};\:{with}\:{a}\:{driller} \\ $$$$\:{of}\:{radius}\:{r},\:{find}\:{fraction}\:{of} \\ $$$${hemisphere}\:{material}\:{drilled}\:{out}.…
Question Number 150517 by ajfour last updated on 13/Aug/21 Commented by ajfour last updated on 13/Aug/21 $${Cannot}\:{OP}\:{be}\:{a}\:{bit}\:{greater} \\ $$$${than}\:{OB}={R}\:\left({radius}\:{of}\:{hemi}-\right. \\ $$$$\left.{sphere}\right)?\:{If}\:{yes}\:{then}\:{for}\:{what} \\ $$$${range}\:{of}\:{semi}-{vertical}\:\angle\:\theta \\ $$$${of}\:{cone}\:\left({right}\right),\:{is}\:{that}\:{so}?…
Question Number 84970 by Power last updated on 18/Mar/20 Answered by mr W last updated on 18/Mar/20 $$\mathrm{2}{x}+\mathrm{3}{y}=\pi \\ $$$$\Rightarrow{x}=\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}{y}}{\mathrm{2}} \\ $$$$\frac{{AB}}{\mathrm{sin}\:\left({x}+{y}\right)}=\frac{{BD}}{\mathrm{sin}\:{x}} \\ $$$$\Rightarrow\frac{\mathrm{11}}{\mathrm{cos}\:\frac{{y}}{\mathrm{2}}}=\frac{\mathrm{2}}{\mathrm{cos}\:\frac{\mathrm{3}{y}}{\mathrm{2}}} \\…