Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 71993 by TawaTawa last updated on 23/Oct/19

Answered by mind is power last updated on 23/Oct/19

let∠ AC^ E=∠ADE=x  ⇒∠ECD=3x−x=2x⇒EC=ED  ECD isocel Triangl  ∠ADC=∠EDC−∠EDA=2x−x=x  △ADE=AD.ED.((sin(x))/2)  △ACE=((AC.CEsin(x))/2)  ((△ADE)/(△ACE))=((AD.DEsin(x))/(AC.CE.sin(x)))=((AD)/(AC))   {≪ED=EC≫}  we have In ADE  ((AD)/(sin(3x)))=((AC)/(sin(x)))  ⇒((AD)/(AC))=((sin(3x))/(sin(x)))=((△ADE)/(△ACE))  lim_(x→0) ((sin(3x))/(sin(x)))=lim_(x→0) ((3.sin(3x))/(3x)).(((sin(x))/x))^(−1)   lim ((sint)/t)=1⇒lim_(x→0)  ((3sin(3x))/(3x)).(((sin(x))/x))^(−1) =3  we can also  use sin(3x)=−4sin^3 (x)+3sin(x)  ⇒((sin(3x))/(sin(x)))=−4sin^2 (x)+3→3

$$\mathrm{let}\angle\:\mathrm{A}\overset{} {\mathrm{C}E}=\angle\mathrm{ADE}=\mathrm{x} \\ $$$$\Rightarrow\angle\mathrm{ECD}=\mathrm{3x}−\mathrm{x}=\mathrm{2x}\Rightarrow\mathrm{EC}=\mathrm{ED}\:\:\mathrm{ECD}\:\mathrm{isocel}\:\mathrm{Triangl} \\ $$$$\angle\mathrm{ADC}=\angle\mathrm{EDC}−\angle\mathrm{EDA}=\mathrm{2x}−\mathrm{x}=\mathrm{x} \\ $$$$\bigtriangleup\mathrm{ADE}=\mathrm{AD}.\mathrm{ED}.\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{2}} \\ $$$$\bigtriangleup\mathrm{ACE}=\frac{\mathrm{AC}.\mathrm{CEsin}\left(\mathrm{x}\right)}{\mathrm{2}} \\ $$$$\frac{\bigtriangleup\mathrm{ADE}}{\bigtriangleup\mathrm{ACE}}=\frac{\mathrm{AD}.\mathrm{DEsin}\left(\mathrm{x}\right)}{\mathrm{AC}.\mathrm{CE}.\mathrm{sin}\left(\mathrm{x}\right)}=\frac{\mathrm{AD}}{\mathrm{AC}}\:\:\:\left\{\ll\mathrm{ED}=\mathrm{EC}\gg\right\} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{In}\:\mathrm{ADE}\:\:\frac{\mathrm{AD}}{\mathrm{sin}\left(\mathrm{3x}\right)}=\frac{\mathrm{AC}}{\mathrm{sin}\left(\mathrm{x}\right)} \\ $$$$\Rightarrow\frac{\mathrm{AD}}{\mathrm{AC}}=\frac{\mathrm{sin}\left(\mathrm{3x}\right)}{\mathrm{sin}\left(\mathrm{x}\right)}=\frac{\bigtriangleup\mathrm{ADE}}{\bigtriangleup\mathrm{ACE}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\left(\mathrm{3x}\right)}{\mathrm{sin}\left(\mathrm{x}\right)}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3}.\mathrm{sin}\left(\mathrm{3x}\right)}{\mathrm{3x}}.\left(\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{x}}\right)^{−\mathrm{1}} \\ $$$$\mathrm{lim}\:\frac{\mathrm{sint}}{\mathrm{t}}=\mathrm{1}\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3sin}\left(\mathrm{3x}\right)}{\mathrm{3x}}.\left(\frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{x}}\right)^{−\mathrm{1}} =\mathrm{3} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{also} \\ $$$$\mathrm{use}\:\mathrm{sin}\left(\mathrm{3x}\right)=−\mathrm{4sin}^{\mathrm{3}} \left(\mathrm{x}\right)+\mathrm{3sin}\left(\mathrm{x}\right) \\ $$$$\Rightarrow\frac{\mathrm{sin}\left(\mathrm{3x}\right)}{\mathrm{sin}\left(\mathrm{x}\right)}=−\mathrm{4sin}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{3}\rightarrow\mathrm{3} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by TawaTawa last updated on 23/Oct/19

God bless you sir, thanks for your time

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time} \\ $$

Commented by mind is power last updated on 23/Oct/19

happy this help you sir have a nice day

$$\mathrm{happy}\:\mathrm{this}\:\mathrm{help}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{have}\:\mathrm{a}\:\mathrm{nice}\:\mathrm{day} \\ $$$$ \\ $$

Commented by TawaTawa last updated on 23/Oct/19

Yes sir,  help me look at question  71761

$$\mathrm{Yes}\:\mathrm{sir},\:\:\mathrm{help}\:\mathrm{me}\:\mathrm{look}\:\mathrm{at}\:\mathrm{question}\:\:\mathrm{71761} \\ $$

Commented by mind is power last updated on 24/Oct/19

3002!!=[2003] ?

$$\mathrm{3002}!!=\left[\mathrm{2003}\right]\:? \\ $$$$ \\ $$

Commented by TawaTawa last updated on 24/Oct/19

No sir

$$\mathrm{No}\:\mathrm{sir} \\ $$

Commented by TawaTawa last updated on 24/Oct/19

Terms of Service

Privacy Policy

Contact: info@tinkutara.com