Question Number 40255 by ajfour last updated on 17/Jul/18 Commented by MJS last updated on 17/Jul/18 $$\mathrm{are}\:\mathrm{the}\:\mathrm{trapezoids}\:\mathrm{on}\:\mathrm{the}\:\mathrm{right}\:\mathrm{similar}? \\ $$ Commented by MJS last updated on…
Question Number 105769 by ajfour last updated on 31/Jul/20 Commented by ajfour last updated on 31/Jul/20 $${ACDB}\:{is}\:{a}\:{straight}\:{line}.\:{The} \\ $$$${three}\:{coloured}\:{regions}\:{have}\:{equal} \\ $$$${areas}.\:{Given}\:{OA}={p},\:{OB}={q}. \\ $$$${Find}\:{coordinates}\:{of}\:{A}\:{and}\:{B}. \\ $$…
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Question Number 171255 by mnjuly1970 last updated on 11/Jun/22 Commented by mr W last updated on 11/Jun/22 $${S}_{\Delta{APB}} =\frac{{AB}×{AP}\:\mathrm{cos}\:\theta}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{AD}×{AP}\:\mathrm{cos}\:\theta}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{AP}^{\mathrm{2}} }{\mathrm{4}} \\…
Question Number 40170 by ajfour last updated on 16/Jul/18 Commented by ajfour last updated on 16/Jul/18 Commented by ajfour last updated on 16/Jul/18 $${p}=\frac{\mathrm{1}}{\mathrm{2sin}\:\alpha}\:\:\:\:\:\:{q}=\frac{\mathrm{1}}{\mathrm{2sin}\:\beta} \\…
Question Number 105630 by I want to learn more last updated on 30/Jul/20 Commented by I want to learn more last updated on 30/Jul/20 $$\mathrm{Thanks}\:\mathrm{sir}…
Question Number 171132 by cortano1 last updated on 08/Jun/22 Answered by kapoorshah last updated on 08/Jun/22 $${PA}\:=\:\left(−\mathrm{3},\:−\mathrm{4}\right)\: \\ $$$${PT}\:\:=\:\:\left(−\mathrm{1},\:−\mathrm{4}\right) \\ $$$$\mathrm{cos}\:\theta\:=\:\frac{{PA}\:.\:{PT}}{\mid{PA}\mid\:.\:\mid{PT}\mid} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{3}\:+\:\mathrm{16}}{\mathrm{5}\:.\:\sqrt{\mathrm{17}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{19}}{\mathrm{5}\sqrt{\mathrm{17}}}…
Question Number 40063 by ajfour last updated on 15/Jul/18 Answered by ajfour last updated on 16/Jul/18 $${Q}\equiv\left({a},\sqrt{{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:\right) \\ $$$${TL}\:=\:{QL} \\ $$$$\Rightarrow\:\:{T}\left(−{a},\:\sqrt{{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:\right)…
Question Number 40031 by ajfour last updated on 15/Jul/18 Commented by ajfour last updated on 16/Jul/18 $${CE}\:{and}\:{DF}\:\bot\:{AF}\:.\:{Find}\:\theta. \\ $$ Answered by ajfour last updated on…
Question Number 40023 by ajfour last updated on 15/Jul/18 Answered by MrW3 last updated on 15/Jul/18 $${AB}=\mathrm{2}{R}\:\mathrm{cos}\:\theta \\ $$$${AD}=\frac{{R}}{\mathrm{cos}\:\theta} \\ $$$${BD}={AB}−{AD}={R}\left(\mathrm{2}\:\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\right)=\frac{{R}\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\theta} \\ $$$${Let}\:{E}={midpoint}\:{of}\:{BC}. \\ $$$${BE}={AB}\:\mathrm{sin}\:\theta={R}\:\mathrm{sin}\:\mathrm{2}\theta…