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} \\…
Question Number 104309 by ajfour last updated on 20/Jul/20 Commented by ajfour last updated on 20/Jul/20 $${A},\:{B}\:{are}\:{two}\:{vertices}\:{of}\:\bigtriangleup{ABC} \\ $$$${while}\:{H}\:{is}\:{its}\:{orthocenter};\:{find} \\ $$$${coordinates}\:{of}\:{the}\:{vertex}\:{C} \\ $$$${of}\:{the}\:\bigtriangleup{ABC}. \\ $$…
Question Number 169821 by cortano1 last updated on 10/May/22 Answered by floor(10²Eta[1]) last updated on 10/May/22 $$\mathrm{let}\:\mathrm{AD}=\mathrm{a}=\mathrm{DC}\:\mathrm{and}\:\mathrm{BD}=\mathrm{b} \\ $$$$\mathrm{by}\:\mathrm{law}\:\mathrm{of}\:\mathrm{sines}: \\ $$$$\frac{\mathrm{a}}{\mathrm{sin30}°}=\frac{\mathrm{b}}{\mathrm{sinx}}\: \\ $$$$\frac{\mathrm{a}}{\mathrm{sin105}°}=\frac{\mathrm{b}}{\mathrm{sin}\left(\mathrm{45}°−\mathrm{x}\right)},\:\mathrm{45}°−\mathrm{x}=\angle\mathrm{BAC} \\ $$$$\Rightarrow\mathrm{a}=\frac{\mathrm{b}}{\mathrm{2sinx}}…
Question Number 38613 by rish@bh last updated on 27/Jun/18 $$\mathrm{The}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{circle}\:\mathrm{which} \\ $$$$\mathrm{passes}\:\mathrm{through}\:\left(\mathrm{1},\mathrm{2}\right)\:\mathrm{and}\:\left(\mathrm{3},\mathrm{4}\right)\:\mathrm{and}\:\mathrm{lies} \\ $$$$\mathrm{completely}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{quadrant}\:\mathrm{is} \\ $$$$\left.\mathrm{A}\right)\:\mathrm{3} \\ $$$$\left.\mathrm{B}\right)\:\mathrm{2} \\ $$$$\left.\mathrm{C}\right)\:\sqrt{\mathrm{6}} \\ $$$$\left.\mathrm{D}\right)\:\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\mathrm{I}\:\mathrm{got}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{as}\:\mathrm{2}\:\mathrm{but}\:\mathrm{the}\:\mathrm{answer}\: \\…
Question Number 103986 by ajfour last updated on 18/Jul/20 Commented by ajfour last updated on 18/Jul/20 $${If}\:{both}\:{coloured}\:{regions}\:{have}\:{equal} \\ $$$${area},\:{find}\:{x}_{{A}} ,\:{x}_{{B}} \:. \\ $$ Answered by…
Question Number 38289 by Tinkutara last updated on 23/Jun/18 Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18 $${centre}\left(−{g},−{f}\right)\:{lies}\:{on}\:{ax}+{by}+{d}=\mathrm{0} \\ $$$${a}\left(−{g}\right)+{b}\left(−{f}\right)+{d}=\mathrm{0}\: \\ $$$$\lambda\left(−{g}\right)−{a}\left(−{f}\right)+\mu{g}+{k}=\mathrm{0} \\ $$$${g}\left(−{a}\right)+{f}\left(−{b}\right)+{d}=\mathrm{0} \\ $$$${g}\left(−\lambda+\mu\right)+{f}\left({a}\right)+{k}=\mathrm{0}…