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 184557 by a.lgnaoui last updated on 08/Jan/23

Determiner  1•AB,  BC  AC en fonction de r  2•  ∡CBA ;  ∡BAC ;et ∡BCA

$${Determiner} \\ $$$$\mathrm{1}\bullet\mathrm{AB},\:\:\mathrm{BC}\:\:\mathrm{AC}\:\mathrm{en}\:\mathrm{fonction}\:\mathrm{de}\:\boldsymbol{\mathrm{r}} \\ $$$$\mathrm{2}\bullet\:\:\measuredangle\mathrm{CBA}\:;\:\:\measuredangle\mathrm{BAC}\:;{et}\:\measuredangle\mathrm{BCA} \\ $$

Commented by a.lgnaoui last updated on 08/Jan/23

Answered by a.lgnaoui last updated on 09/Jan/23

   Reponse  −−−−−−−−−−−−−−−−−  ∡AOB=60=2∡ACCB ⇒∡ACB=30  (Propriete de angle au centre)  1•AB^2 =2r^2 (1−cos 60)⇒AB=r(△AOB Equilaterale)     AB^2 =AC^2 +BC^2 −2AC×BCcos 30      AB=(5/2)BC   ⇒ BC=(2/5)r        r^2 =AC^2 +(4/(25))r^2 −(4/5)rACcos 30      r^2 =AC^2 +(4/(25))r^2 −(2/5)r(√3) ×AC      AC=x           x^2  −((2r(√3))/5)x−((21)/(25))r^2 =0         (x−(((√3) +2(√6) )/5)r)(x−(((√3) −2(√6)  )/5)r)=0              AC=(((√3) +2(√6))/5)r  3•calul des angles     ∡BCA=((60)/2)=30    ((sin (∡ABC))/(((((√3) +2(√6))/5) )r))=((sin 30)/r)=(1/(2r))    sin (∡ABC)=(((√3) +2(√6))/(10))=0,6631       ∡ABC=sin^(−1) ((((√3) +2(√6))/(10)))         ∡ABC=41,53    ∡BAC=(180−30−41,53)        ∡BAC=108,47

$$\: \\ $$$${Reponse} \\ $$$$−−−−−−−−−−−−−−−−− \\ $$$$\measuredangle{AOB}=\mathrm{60}=\mathrm{2}\measuredangle{ACCB}\:\Rightarrow\measuredangle{ACB}=\mathrm{30} \\ $$$$\left({Propriete}\:{de}\:{angle}\:{au}\:{centre}\right) \\ $$$$\mathrm{1}\bullet{AB}^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\mathrm{60}\right)\Rightarrow\boldsymbol{{AB}}=\boldsymbol{{r}}\left(\bigtriangleup{AOB}\:{Equilaterale}\right) \\ $$$$\:\:\:{AB}^{\mathrm{2}} ={AC}^{\mathrm{2}} +{BC}^{\mathrm{2}} −\mathrm{2}{AC}×{BC}\mathrm{cos}\:\mathrm{30} \\ $$$$\:\:\:\:{AB}=\frac{\mathrm{5}}{\mathrm{2}}{BC}\:\:\:\Rightarrow\:\boldsymbol{{BC}}=\frac{\mathrm{2}}{\mathrm{5}}\boldsymbol{{r}}\: \\ $$$$\:\:\:\:\:{r}^{\mathrm{2}} ={AC}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{25}}{r}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{5}}{rAC}\mathrm{cos}\:\mathrm{30} \\ $$$$\:\:\:\:{r}^{\mathrm{2}} ={AC}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{25}}{r}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{5}}{r}\sqrt{\mathrm{3}}\:×{AC} \\ $$$$\:\:\:\:{AC}={x}\:\: \\ $$$$\:\:\:\:\:\:\:{x}^{\mathrm{2}} \:−\frac{\mathrm{2}{r}\sqrt{\mathrm{3}}}{\mathrm{5}}{x}−\frac{\mathrm{21}}{\mathrm{25}}{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\left({x}−\frac{\sqrt{\mathrm{3}}\:+\mathrm{2}\sqrt{\mathrm{6}}\:}{\mathrm{5}}{r}\right)\left({x}−\frac{\sqrt{\mathrm{3}}\:−\mathrm{2}\sqrt{\mathrm{6}}\:\:}{\mathrm{5}}{r}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{AC}=\frac{\sqrt{\mathrm{3}}\:+\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{5}}\boldsymbol{{r}} \\ $$$$\mathrm{3}\bullet{calul}\:{des}\:{angles} \\ $$$$\:\:\:\measuredangle{BCA}=\frac{\mathrm{60}}{\mathrm{2}}=\mathrm{30} \\ $$$$\:\:\frac{\mathrm{sin}\:\left(\measuredangle{ABC}\right)}{\left(\frac{\sqrt{\mathrm{3}}\:+\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{5}}\:\right){r}}=\frac{\mathrm{sin}\:\mathrm{30}}{{r}}=\frac{\mathrm{1}}{\mathrm{2}{r}} \\ $$$$\:\:\mathrm{sin}\:\left(\measuredangle{ABC}\right)=\frac{\sqrt{\mathrm{3}}\:+\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{10}}=\mathrm{0},\mathrm{6631} \\ $$$$\:\:\:\:\:\measuredangle{ABC}=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}\:+\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{10}}\right)\: \\ $$$$\:\:\:\:\:\:\measuredangle\boldsymbol{{ABC}}=\mathrm{41},\mathrm{53} \\ $$$$ \\ $$$$\measuredangle{BAC}=\left(\mathrm{180}−\mathrm{30}−\mathrm{41},\mathrm{53}\right) \\ $$$$\:\:\:\:\:\:\measuredangle\boldsymbol{{BAC}}=\mathrm{108},\mathrm{47} \\ $$$$\:\:\:\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com