Question Number 200357 by ajfour last updated on 17/Nov/23 Commented by ajfour last updated on 17/Nov/23 $$\:\:\:\:\:\:{Find}\:{r},\:{if}\:{square}\:{side}\:{is}\:{unity}. \\ $$ Commented by Frix last updated on…
Question Number 200325 by cherokeesay last updated on 17/Nov/23 $$\mathrm{A}\:\mathrm{point}\:\mathrm{P}\:\mathrm{is}\:\mathrm{taken}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{rectangleC} \\ $$$$\mathrm{ABD}.\:\mathrm{This}\:\mathrm{point}\:\mathrm{joins}\:\mathrm{the}\:\mathrm{four}\:\mathrm{vertices}\:\mathrm{ofh} \\ $$$$\mathrm{te}\:\mathrm{rectangle}.\:\:\mathrm{Knowing}\:\mathrm{that}\:\mathrm{PA}\:\mathrm{is}\:\mathrm{15}\:\mathrm{cm}.\:\mathrm{B} \\ $$$$\mathrm{P}\:\:\mathrm{24}\:\mathrm{cm}\:\mathrm{and}\:\mathrm{PC}\:\:\mathrm{20}\:\mathrm{cm}\:\mathrm{find}\:\mathrm{the}\:\mathrm{distancef} \\ $$$$\mathrm{o}\:\mathrm{point}\:\mathrm{P}\:\mathrm{from}\:\mathrm{point}\:\mathrm{D}. \\ $$ Answered by som(math1967) last updated…
Question Number 200270 by ajfour last updated on 16/Nov/23 Commented by mr W last updated on 17/Nov/23 $${i}\:{think}\:{one}\:{vertex}\:{must}\:{lie}\:{at}\:{the} \\ $$$${center}\:{of}\:{a}\:{circle}\:{as}\:{in}\:{diagram}.\:{so} \\ $$$${s}_{{max}} ={radius}=\mathrm{1}. \\ $$…
Question Number 200204 by cherokeesay last updated on 15/Nov/23 Commented by MM42 last updated on 15/Nov/23 $${is}\:,\:{ABCD},\:{is}\:{square}? \\ $$ Commented by ajfour last updated on…
Question Number 200120 by Mingma last updated on 14/Nov/23 Answered by ajfour last updated on 14/Nov/23 Commented by ajfour last updated on 14/Nov/23 $$\frac{{s}}{{t}+{p}}=\frac{{t}}{{s}+{q}}=\frac{{q}}{{p}}\:\:\:\:\left({similarity}\:{of}\:\bigtriangleup{s}\right) \\…
Question Number 200092 by ajfour last updated on 13/Nov/23 Commented by ajfour last updated on 13/Nov/23 $${If}\:{semicircle}\:{has}\:{radius}\:\mathrm{1},\:{find} \\ $$$${radii}\:{of}\:{blue}\:{and}\:{green}\:{circles}. \\ $$$${One}\:{end}\:{of}\:{each}\:{semicircle}\:{is}\:{at} \\ $$$${center}\:{of}\:{other}. \\ $$…
Question Number 199922 by Mingma last updated on 11/Nov/23 Answered by AST last updated on 11/Nov/23 $${WLOG},{let}\:{C}\:{be}\:{the}\:{origin}\:{and}\:{let}\:{AD}\:{coincide} \\ $$$${with}\:{the}\:{real}\:{axis};{then}\:{a}=\overset{−} {{a}};{b}=−\overset{−} {{b}};{d}=\overset{−} {{d}};{e}=−\overset{−} {{e}} \\ $$$$\frac{{b}−{a}}{−{b}−{a}}=\frac{{a}−{e}}{−{e}−{a}}\Rightarrow−{be}+\mathrm{2}{a}^{\mathrm{2}}…
Question Number 199911 by mr W last updated on 11/Nov/23 Commented by mr W last updated on 11/Nov/23 $${three}\:{mirrors}\:{build}\:{the}\:{sides}\:{of}\:{an} \\ $$$${equilateral}\:{triangle}.\:{a}\:{laser}\:{ray} \\ $$$${emitted}\:{from}\:{the}\:{midpoint}\:{of}\:{a}\:{side} \\ $$$${should}\:{reach}\:{the}\:{opposite}\:{vertex}\:{after}…
Question Number 199706 by cherokeesay last updated on 08/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 199728 by Mingma last updated on 08/Nov/23 Answered by AST last updated on 08/Nov/23 $$\frac{{sin}\mathrm{120}°}{{BC}}=\frac{{sin}\mathrm{20}°}{{AB}}\Rightarrow{AB}=\frac{\mathrm{2}\sqrt{\mathrm{3}}{BCsin}\mathrm{20}°}{\mathrm{3}} \\ $$$$\frac{{sin}\mathrm{20}°}{{AB}}=\frac{{sin}\mathrm{40}°}{{AC}}\Rightarrow{AC}=\mathrm{2}{ABcos}\mathrm{20}°=\frac{\mathrm{2}\sqrt{\mathrm{3}}{BCsin}\mathrm{40}}{\mathrm{3}} \\ $$$$\frac{{sin}\left({x}\right)}{{DC}}=\frac{{sinADC}}{{AC}}=\frac{{sin}\left(\mathrm{140}−{x}\right)}{{AC}}\Rightarrow\frac{{AC}}{{DC}}=\frac{{sin}\left(\mathrm{140}−{x}\right)}{{sin}\left({x}\right)} \\ $$$$=\frac{{sin}\mathrm{40}}{{sin}\mathrm{20}}=\mathrm{2}{cos}\mathrm{20}=\frac{{sin}\mathrm{140}{cosx}−{sin}\left({x}\right){cos}\left(\mathrm{140}\right)}{{sinx}} \\ $$$$={sin}\mathrm{40}{tan}\left({x}\right)−{cos}\left(\mathrm{90}+\mathrm{50}\right)={sin}\mathrm{40}{tanx}+{sin}\mathrm{50}…