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 \\ $$$${AD}=\frac{{R}}{\mathrm{cos}\:\theta} \\ $$$${BD}={AB}−{AD}={R}\left(\mathrm{2}\:\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\right)=\frac{{R}\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\theta} \\ $$$${Let}\:{E}={midpoint}\:{of}\:{BC}. \\ $$$${BE}={AB}\:\mathrm{sin}\:\theta={R}\:\mathrm{sin}\:\mathrm{2}\theta…
Question Number 39993 by ajfour last updated on 14/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 14/Jul/18 $${let}\:\angle{AOT}=\theta \\ $$$${OT}\parallel\:{toBC}\:{so}\:\frac{{AO}}{{AC}}=\frac{{AT}}{{AB}}=\frac{{OT}}{{BC}} \\ $$$${point}\:{T}\left({cos}\theta,{sin}\theta\right)\:\:{andpoint}\:{C}\left(−\mathrm{1},\mathrm{0}\right) \\ $$$${eqn}\:\:\:{of}\:{OT}\:{is}\:{y}={xtan}\theta \\ $$$${eqn}\:{ofCB}\:{is}\:{y}=\left({x}+\mathrm{1}\right){tan}\theta…
Question Number 39985 by ajfour last updated on 14/Jul/18 Answered by ajfour last updated on 15/Jul/18 $${eq}.\:{of}\:{line}\:\:{OBC}\::\:\:{y}={x} \\ $$$${eq}.\:{of}\:{AEC}:\:\:\:\:{y}\mathrm{tan}\:\theta={x}−\mathrm{1} \\ $$$${For}\:{point}\:{C}\:: \\ $$$$\:\:\:\:\:{x}_{{C}} ={y}_{{C}} =\frac{\mathrm{1}}{\mathrm{1}−\mathrm{tan}\:\theta}\:\:…
Question Number 105474 by Don08q last updated on 29/Jul/20 Answered by ajfour last updated on 29/Jul/20 Commented by ajfour last updated on 29/Jul/20 $${BD}=\sqrt{\mathrm{25}+\mathrm{36}−\mathrm{2}×\mathrm{5}×\mathrm{6cos}\:\mathrm{30}°} \\…
Question Number 39854 by ajfour last updated on 12/Jul/18 Commented by ajfour last updated on 12/Jul/18 $${Choose}\:{the}\:{correct}\:{options}: \\ $$$$\left({i}\right)\:{a}\approx\mathrm{1}.\mathrm{6}\:\:\:\:\:\:\:\:\:\:\left({ii}\right)\:\:{b}\approx\mathrm{2}.\mathrm{7} \\ $$$$\left({iii}\right)\:\:{a}\:\approx\:\mathrm{1}.\mathrm{8}\:\:\:\left({iv}\right)\:\:{b}\:\approx\:\mathrm{2}.\mathrm{4} \\ $$$$ \\ $$…
Question Number 39848 by ajfour last updated on 12/Jul/18 Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jul/18 $${tan}\theta=\frac{{a}}{\mathrm{1}} \\ $$$${sin}\theta=\frac{\mathrm{1}}{\mathrm{1}+{a}} \\ $$$${tan}\theta=\frac{{a}}{\mathrm{1}}\:\:{so}\:{sin}\theta=\frac{{a}}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }} \\ $$$$\frac{{a}}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}=\frac{\mathrm{1}}{{a}+\mathrm{1}}…