Question Number 197095 by Mingma last updated on 07/Sep/23 Answered by mahdipoor last updated on 07/Sep/23 $$\frac{{AB}}{{sin}\mathrm{4}{a}}=\frac{{BM}}{{sina}}\:\:\: \\ $$$$\:\frac{{CD}}{{sin}\mathrm{4}{a}}=\frac{{MC}}{{sin}\mathrm{2}{a}}\Rightarrow\frac{\mathrm{2}.{CD}.{cosa}}{{sin}\mathrm{4}{a}}=\frac{{MC}}{{sina}}\Rightarrow \\ $$$$\frac{{MC}}{{sina}}+\frac{{BM}}{{sina}}=\frac{{CB}}{{sina}}=\frac{\mathrm{2}.{CD}.{cosa}}{{sin}\mathrm{4}{a}}+\frac{{AB}}{{sin}\mathrm{4}{a}} \\ $$$$\Rightarrow\frac{{sin}\mathrm{4}{a}}{{sina}}=\mathrm{4}{cos}\mathrm{2}{a}.{cosa}=\mathrm{2}{cosa}+\mathrm{1}\Rightarrow \\ $$$$\mathrm{8}{cos}^{\mathrm{3}}…
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Question Number 196836 by mathlove last updated on 01/Sep/23 Commented by mathlove last updated on 01/Sep/23 $${x}=? \\ $$ Answered by universe last updated on…
Question Number 196829 by cortano12 last updated on 01/Sep/23 Answered by universe last updated on 01/Sep/23 $${by}\:{apollonius}\:{theorem} \\ $$$${BC}^{\mathrm{2}} \:+\:{PC}^{\mathrm{2}\:} \:=\:\mathrm{2}\left({QC}^{\mathrm{2}} +{PQ}^{\mathrm{2}} \right)\:\:\:…..\left(\mathrm{1}\right) \\ $$$${AC}^{\mathrm{2}}…
Question Number 196766 by mr W last updated on 31/Aug/23 Commented by mr W last updated on 31/Aug/23 $${find}\:{the}\:{minimum}\:{and}\:{maximum} \\ $$$${of}\:{m}+{n}\:{with}\:{P}\:{on}\:{the}\:{circle}. \\ $$ Commented by…
Question Number 196774 by BaliramKumar last updated on 31/Aug/23 $$ \\ $$A man standing on top of Burj Khalifa. If the height of Burj Khalifa…
Question Number 196723 by mr W last updated on 30/Aug/23 $${prove}\:{that}\:{the}\:{curve}\: \\ $$$$\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }+\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{4}\: \\ $$$${is}\:{an}\:{ellipse}\:{and}\:{find}\:{its}\:{semi} \\ $$$${major}\:{axis}\:{and}\:{semi}\:{minor}\:{axis}. \\ $$ Answered by…
Question Number 196275 by AROUNAMoussa last updated on 21/Aug/23 Answered by MM42 last updated on 21/Aug/23 $$\langle{ABD}=\mathrm{35}\Rightarrow{AD}={AB} \\ $$$$\langle{BCD}=\mathrm{55} \\ $$$$\frac{{AB}}{{AC}}=\frac{{sinx}}{{sin}\mathrm{130}}\:\:\:\&\:\:\frac{{AD}}{{AC}}=\frac{{sin}\left(\mathrm{55}−{x}\right)}{{sin}\mathrm{65}} \\ $$$$\Rightarrow{sinx}×{sin}\mathrm{65}=\mathrm{2}{sin}\mathrm{65}×{cos}\mathrm{65}×{sin}\left(\mathrm{55}−{x}\right) \\ $$$$\Rightarrow{sinx}=\mathrm{2}{cos}\mathrm{65}{sin}\left(\mathrm{55}−{x}\right)={sin}\left(\mathrm{55}−{x}+\mathrm{65}\right)+{sin}\left(\mathrm{55}−{x}−\mathrm{65}\right)…
Question Number 196225 by mr W last updated on 20/Aug/23 Commented by mr W last updated on 20/Aug/23 $${find}\:{the}\:{area}\:{of}\:{red}\:{part}\:{in}\:{the} \\ $$$${parallelogram}. \\ $$ Answered by…
Question Number 196021 by universe last updated on 16/Aug/23 Answered by mr W last updated on 17/Aug/23 Commented by mr W last updated on 17/Aug/23…