Question Number 39860 by ajfour last updated on 12/Jul/18 Commented by ajfour last updated on 12/Jul/18 $${OA}\:=\:{BP}\:\:\:,\:{OT}\:=\mathrm{1},\:\:\:\&\: \\ $$$$\angle{OTB}\:=\:\angle{BTP}\:\:\:\left({given}\:{conditions}\right). \\ $$ Answered by ajfour last…
Question Number 170890 by nadovic last updated on 02/Jun/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{which} \\ $$$$\mathrm{touches}\:\mathrm{the}\:\mathrm{line}\:{x}−\mathrm{3}{y}+\mathrm{13}\:=\:\mathrm{0}\: \\ $$$$\mathrm{and}\:\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{points}\:\left(\mathrm{6},\:\mathrm{3}\right) \\ $$$$\mathrm{and}\:\left(\mathrm{4},\:−\mathrm{1}\right). \\ $$ Answered by aleks041103 last updated on 02/Jun/22…
Question Number 105273 by ajfour last updated on 27/Jul/20 Commented by ajfour last updated on 27/Jul/20 $${Given}\:{P}\left({h},{k}\right)\:\:\:{and}\:\:{B}\left({p},{q}\right) \\ $$$${To}\:{find}\:{image}\:{I}\left({r},{t}\right)\:{of}\:{B}\left({p},{q}\right) \\ $$$${in}\:{the}\:{mirror}\:{line}\:{AT}. \\ $$ Answered by…
Question Number 105001 by ajfour last updated on 25/Jul/20 Commented by ajfour last updated on 25/Jul/20 $${If}\:{both}\:{circles}\:{have}\:{unit}\:{radius},\:{and} \\ $$$${regions}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5}\:{have}\:{equal}\:{areas}, \\ $$$${find}\:{eq}.\:{of}\:{both}\:{circles}. \\ $$ Answered by…
Question Number 170323 by cortano1 last updated on 20/May/22 Answered by mr W last updated on 20/May/22 $$\frac{{AC}}{\mathrm{sin}\:\mathrm{18}°}=\frac{{AD}}{\mathrm{sin}\:\mathrm{12}°}\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{sin}\:\left(\alpha+\mathrm{18}°\right)}{{DB}}=\frac{\mathrm{sin}\:\alpha}{{AD}}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\frac{\mathrm{sin}\:\left(\alpha+\mathrm{18}°\right)}{\mathrm{sin}\:\mathrm{18}°}=\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\mathrm{12}°} \\…
Question Number 39175 by ajfour last updated on 03/Jul/18 Answered by MJS last updated on 03/Jul/18 $$\mathrm{let}\:{d}>{R}+{r} \\ $$$$\mathrm{2}\:\mathrm{rectangular}\:\mathrm{triangles}: \\ $$$${d}^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} \\ $$$${d}^{\mathrm{2}}…
Question Number 170181 by cortano1 last updated on 17/May/22 Answered by som(math1967) last updated on 18/May/22 $$\frac{\mathrm{1}}{\mathrm{2}}×{AB}×\mathrm{4}.\mathrm{5}=\frac{\mathrm{1}}{\mathrm{2}}×{BC}×\mathrm{5}.\mathrm{5} \\ $$$$\:{AB}:{BC}=\mathrm{11}:\mathrm{9} \\ $$$${let}\:{AB}=\mathrm{11}{k},\:{BC}=\mathrm{9}{k} \\ $$$$\:\therefore{BD}=\frac{{BC}}{\mathrm{2}}=\mathrm{4}.\mathrm{5}{k} \\ $$$$\therefore\left(\mathrm{11}{k}\right)^{\mathrm{2}}…
Question Number 170012 by cortano1 last updated on 14/May/22 Answered by som(math1967) last updated on 14/May/22 $${From}\:\bigtriangleup{ABC} \\ $$$$\:\frac{{AC}}{{sin}\mathrm{50}}=\frac{{BC}}{{sin}\mathrm{30}}\:\left[\because\measuredangle{ABC}=\mathrm{180}−\mathrm{100}−\mathrm{30}=\mathrm{50}\right] \\ $$$${from}\:\bigtriangleup{DBC} \\ $$$$\:\frac{{BD}}{{sin}\mathrm{100}}=\frac{{BC}}{{sinx}} \\ $$$$\:\frac{{BD}}{{sin}\mathrm{50}}=\frac{{BC}}{{sinxsin}\mathrm{50}}×{sin}\left(\mathrm{180}−\mathrm{80}\right)…
Question Number 38936 by ajfour last updated on 01/Jul/18 Commented by ajfour last updated on 01/Jul/18 $${Find}\:{the}\:{locus}\:{of}\:{point}\:{P}\:\left({the}\right. \\ $$$${point}\:{of}\:{intersection}\:{of}\:{tangents} \\ $$$${to}\:{ellipse}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\:.\:{The}\:{tangents}…
Question Number 169947 by cortano1 last updated on 13/May/22 Commented by cortano1 last updated on 13/May/22 Answered by Rasheed.Sindhi last updated on 13/May/22 $${BD}=\mathrm{1}\left({say}\right)\Rightarrow{AC}=\mathrm{2} \\…