Category: Trigonometry

Question Number 212384 by a.lgnaoui last updated on 14/Oct/24 $$\mathrm{Reponse}\:\mathrm{a}\:\mathrm{la}\:\mathrm{question}\:\mathrm{N}° \\ $$$$\mathrm{Q212291}\:\:\:\:\:\:\mathrm{S}\left(\mathrm{ABCD}\right)=\:\:\mathrm{66},\mathrm{69} \\ $$ Commented by hardmath last updated on 12/Oct/24 $$\mathrm{dear}\:\mathrm{professor},\:\mathrm{answer}:\:\mathrm{90} \\ $$ Terms…

Question Number 212011 by a.lgnaoui last updated on 28/Sep/24 $$\boldsymbol{\mathrm{D}}\mathrm{eterminer}:\:\:\boldsymbol{\mathrm{R}}\mathrm{1}\:\:\:\boldsymbol{\mathrm{R}}\mathrm{2}\:\:\:\boldsymbol{\mathrm{R}}\mathrm{3} \\ $$$$\boldsymbol{\mathrm{pour}}\:\:\:\boldsymbol{\mathrm{b}}=\mathrm{12}\boldsymbol{\mathrm{cm}}\:\:\:\: \\ $$$$\boldsymbol{\mathrm{EF}}\://\:\boldsymbol{\mathrm{MN}}\:;\:\:\boldsymbol{\mathrm{EF}}\:\boldsymbol{\mathrm{Tangent}}\:\boldsymbol{\mathrm{aux}}\:\boldsymbol{\mathrm{cercles}}:\: \\ $$$$\:\:\boldsymbol{\mathrm{C}}\mathrm{1}\left(\boldsymbol{\mathrm{R}}\mathrm{1}\right)\:\:\:\boldsymbol{\mathrm{C}}\mathrm{2}\left(\boldsymbol{\mathrm{R}}\mathrm{2}\right)\:\:;\:\:\:\boldsymbol{\mathrm{EF}}=\boldsymbol{\mathrm{a}}\:\:\:\:\:\:\boldsymbol{\mathrm{MN}}=\boldsymbol{\mathrm{b}} \\ $$$$\boldsymbol{\mathrm{MN}}:\:\boldsymbol{\mathrm{tangent}}\:\boldsymbol{\mathrm{au}}\:\boldsymbol{\mathrm{cercle}}\:\boldsymbol{\mathrm{C}}\mathrm{2} \\ $$$$\boldsymbol{\mathrm{OM}}=\boldsymbol{\mathrm{ON}}=\frac{\mathrm{3}\boldsymbol{\mathrm{a}}}{\mathrm{2}}\:\:\:\:\:\:\:\:\measuredangle\mathrm{MON}=\mathrm{2}\boldsymbol{\mathrm{x}}\:\:\: \\ $$$$ \\ $$ Commented…

Question Number 211956 by Durganand last updated on 25/Sep/24 Answered by Frix last updated on 25/Sep/24 $$\mathrm{tan}\:\alpha\:={t}\:\:\:\:\:\mathrm{tan}\:\mathrm{2}\alpha\:=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }\:\:\:\:\:\mathrm{sin}\:\mathrm{2}\alpha\:=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}=\frac{\mathrm{2}−\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{2}{t}}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}\alpha} \\…

Question Number 211672 by MATHEMATICSAM last updated on 15/Sep/24 $$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{the}\:\mathrm{side}\:{c}\:\mathrm{is} \\ $$$$\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{side}\:{b}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{2tanC}\:+\:\mathrm{tanA}\:=\:\mathrm{0}. \\ $$ Commented by Frix last updated on 16/Sep/24 $$\mathrm{If}\:\mathrm{the}\:\mathrm{bisector}\:\mathrm{of}\:\mathrm{a}\:{side}\:\mathrm{is}\:\mathrm{perpendicular}\:\mathrm{to} \\…

Question Number 211609 by MATHEMATICSAM last updated on 14/Sep/24 $$\mathrm{Prove}\:\mathrm{that},\:\mathrm{in}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{the}\:\mathrm{ratios}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{sides}\:\mathrm{and}\:\mathrm{the}\:\mathrm{sine}\:\mathrm{of}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{angles} \\ $$$$\mathrm{are}\:\mathrm{equal}.\:\mathrm{Also}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{each}\:\mathrm{ratio}\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{diameter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circum}\:\mathrm{circle} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}. \\ $$ Answered by Frix last updated…