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 $${centre}\left(−{g},−{f}\right)\:{lies}\:{on}\:{ax}+{by}+{d}=\mathrm{0} \\ $$$${a}\left(−{g}\right)+{b}\left(−{f}\right)+{d}=\mathrm{0}\: \\ $$$$\lambda\left(−{g}\right)−{a}\left(−{f}\right)+\mu{g}+{k}=\mathrm{0} \\ $$$${g}\left(−{a}\right)+{f}\left(−{b}\right)+{d}=\mathrm{0} \\ $$$${g}\left(−\lambda+\mu\right)+{f}\left({a}\right)+{k}=\mathrm{0}…
Question Number 38281 by Tinkutara last updated on 23/Jun/18 Answered by ajfour last updated on 23/Jun/18 $$\frac{{ap}_{{A}\rightarrow{BC}} }{\mathrm{2}}\:=\:\frac{{bp}_{{B}\rightarrow{CA}} }{\mathrm{2}}\:=\:\frac{{cp}_{{C}\rightarrow{AB}} }{\mathrm{2}}\:=\bigtriangleup \\ $$$$\frac{{ap}_{{P}\rightarrow{BC}} }{\mathrm{2}}\:+\:\frac{{bp}_{{P}\rightarrow{CA}} }{\mathrm{2}}\:+\:\frac{{cp}_{{P}\rightarrow{AB}} }{\mathrm{2}}\:=\:\Delta…
Question Number 103756 by ajfour last updated on 17/Jul/20 Commented by ajfour last updated on 17/Jul/20 $${Find}\:{radius}\:{r}\:{of}\:{shown}\:{circle}. \\ $$ Answered by mr W last updated…
Question Number 38006 by ajfour last updated on 20/Jun/18 Commented by ajfour last updated on 20/Jun/18 $${For}\:{each}\:{of}\:{the}\:{four}\:{single} \\ $$$${coloured}\:{areas}\:{to}\:{be}\:{equal},\:{locate} \\ $$$${P}\left({r},\theta\right)\:\:{and}\:{C}\left(\mathrm{0},{y}\right)\:{in}\:{terms}\:{of} \\ $$$${a}\:{and}\:{b}.\:\:\:\left(\:{r}=\:{OP}\:\:,\:\:{y}={OC}\:\right) \\ $$…
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Question Number 168980 by BHOOPENDRA last updated on 22/Apr/22 Commented by mr W last updated on 22/Apr/22 $${not}\:{clear}\:{what}\:{you}\:{want}\:{to}\:{show}… \\ $$ Commented by BHOOPENDRA last updated…