Question Number 88933 by TawaTawa1 last updated on 13/Apr/20 Commented by john santu last updated on 14/Apr/20 $$\mathrm{30}^{{o}} \\ $$ Commented by mr W last…
Question Number 88894 by TawaTawa1 last updated on 13/Apr/20 Commented by john santu last updated on 13/Apr/20 $$\mathrm{120}^{{o}} \:=\:{x} \\ $$ Terms of Service Privacy…
Question Number 154424 by amin96 last updated on 18/Sep/21 Commented by amin96 last updated on 18/Sep/21 $${prove} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 88886 by I want to learn more last updated on 13/Apr/20 Commented by I want to learn more last updated on 13/Apr/20 $$\mathrm{ABCD}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{square},\:\:\mathrm{find}\:\:\mathrm{AB}.…
Question Number 154415 by mr W last updated on 18/Sep/21 Commented by mr W last updated on 18/Sep/21 $$\left[{Q}\mathrm{154210}\right] \\ $$ Answered by mr W…
Question Number 88865 by tw000001 last updated on 13/Apr/20 Commented by tw000001 last updated on 13/Apr/20 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{who}\:\mathrm{the}\:\mathrm{question}\:\mathrm{can}\:\mathrm{solve}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 23317 by tapan das last updated on 28/Oct/17 $$\int\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}\:\mathrm{cos}\:\mathrm{x}\:\mathrm{dx} \\ $$ Answered by mrW1 last updated on 28/Oct/17 $$=\int\mathrm{sin}^{\mathrm{3}} \:\mathrm{x}\:\mathrm{dsin}\:\mathrm{x} \\ $$$$=\frac{\mathrm{sin}^{\mathrm{4}}…
Question Number 23312 by tawa tawa last updated on 28/Oct/17 Commented by tawa tawa last updated on 28/Oct/17 $$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$ Commented by Tinkutara last…
Question Number 154343 by Lekhraj last updated on 17/Sep/21 Answered by qaz last updated on 17/Sep/21 $$\frac{\mathrm{sin}\:\mathrm{x}}{\bullet\mathrm{O}}=\frac{\mathrm{sin}\:\angle\mathrm{M}\bullet\mathrm{C}}{\mathrm{OC}},\:\:\:\:\:\:\frac{\mathrm{sin}\:\mathrm{30}°}{\bullet\mathrm{O}}=\frac{\mathrm{sin}\:\mathrm{105}°}{\mathrm{OA}} \\ $$$$\because\:\:\mathrm{OA}=\mathrm{OC}\:\:\:\:\:\: \\ $$$$\therefore\:\:\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{sin}\:\mathrm{30}°}{\mathrm{sin}\:\mathrm{105}°}\centerdot\mathrm{sin}\:\angle\mathrm{M}\bullet\mathrm{C}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}+\mathrm{1}}\mathrm{sin}\:\angle\mathrm{M}\bullet\mathrm{C} \\ $$$$\because\:\:\:\angle\mathrm{M}\bullet\mathrm{C}=\mathrm{180}°−\mathrm{135}°−\mathrm{x}=\mathrm{45}°−\mathrm{x} \\ $$$$\therefore\:\:\:\mathrm{sin}\:\mathrm{x}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}+\mathrm{1}}\mathrm{sin}\:\left(\mathrm{45}°−\mathrm{x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+\mathrm{1}}\left(\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\right)…
Question Number 23262 by ajfour last updated on 28/Oct/17 Commented by ajfour last updated on 28/Oct/17 $$\frac{{CN}}{{AC}}=\frac{{x}}{{x}+{y}}\:\:\:;\:\:\:\frac{{AN}}{{AC}}=\frac{{y}}{{x}+{y}} \\ $$$$\frac{{MB}}{{AB}}\:=\frac{{y}}{{x}+{y}}\:;\:\:\frac{{AM}}{{AB}}=\frac{{x}}{{x}+{y}}\:. \\ $$$${based}\:{on}\:{similarity}\:\:{of}\:{triangles}. \\ $$$$\bigtriangleup{CNP}\:\sim\:\bigtriangleup{CAB} \\ $$$$\Rightarrow\:\:\frac{{CN}}{{AC}}=\frac{{NP}\left(={AM}\right)}{{AB}}=\frac{{CP}}{{BC}}…