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Question Number 218267 by lmcp1203 last updated on 04/Apr/25

Commented by lmcp1203 last updated on 04/Apr/25

thanks

$${thanks} \\ $$

Answered by vnm last updated on 04/Apr/25

Nothing has changed with the drawings, they still won′t load.The dot on the left is A, on the top is B, on the right is C and at the bottom in the middle is D. let AC=1, AD=BC=a ∡ABC=120° ((AC)/(sin ∡ABC))=((BC)/(sin∡BAC )) a=BC=AC((sin ∡BAC)/(sin ∡ABC))=((sin (60°−2x))/(sin 120^° ))= (((√3)cos 2x−sin 2x)/( (√3))) ((BC)/(sin ∡BDC))=((CD)/(sin ∡CBD)) (a/(sin(180°−3x) ))=((1−a)/(sin x)) ((((√3)cos 2x−sin 2x)/( (√3)))/(sin 3x))=((1−(((√3)cos 2x−sin 2x)/( (√3))))/(sin x)) 2x=t (√3)(2cos^2 t−1)=2sin t∙(1+cos t) 16cos^4 t+8cos^3 t−12cos^2 t−8cos t−1=0 2cos t=u u^4 +u^3 −3u^2 −4u−1=0 u^3 (u+1)−(3u+1)(u+1)=0 (u^3 −3u−1)(u+1)=0 u^3 −3u−1=0 (1/2)=cos 60°=cos (3∙20°)= cos 40°cos 20°−sin 40^° sin 20°= (2cos^2 20°−1)cos 20°− 2cos 20^° (1−cos^2 20°)= 4cos^3 20°−3cos 20^° (2cos 20°)^3 −3(2cos 20°)−1=0 x=10°

$$ \\ $$$$\mathrm{Nothing}\:\mathrm{has}\:\mathrm{changed}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{drawings},\:\mathrm{they}\:\mathrm{still}\:\mathrm{won}'\mathrm{t} \\ $$$$\mathrm{load}.\mathrm{The}\:\mathrm{dot}\:\mathrm{on}\:\mathrm{the}\:\mathrm{left}\:\mathrm{is}\:\mathrm{A},\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{top}\:\mathrm{is}\:\mathrm{B},\:\mathrm{on}\:\mathrm{the}\:\mathrm{right}\:\mathrm{is}\:\mathrm{C}\: \\ $$$$\mathrm{and}\:\mathrm{at}\:\mathrm{the}\:\mathrm{bottom}\:\mathrm{in}\:\mathrm{the}\: \\ $$$$\mathrm{middle}\:\mathrm{is}\:\mathrm{D}. \\ $$$${let}\:{AC}=\mathrm{1},\:{AD}={BC}={a} \\ $$$$\measuredangle{ABC}=\mathrm{120}° \\ $$$$\frac{{AC}}{\mathrm{sin}\:\measuredangle{ABC}}=\frac{{BC}}{\mathrm{sin}\measuredangle{BAC}\:} \\ $$$${a}={BC}={AC}\frac{\mathrm{sin}\:\measuredangle{BAC}}{\mathrm{sin}\:\measuredangle{ABC}}=\frac{\mathrm{sin}\:\left(\mathrm{60}°−\mathrm{2}{x}\right)}{\mathrm{sin}\:\mathrm{120}^{°} }= \\ $$$$\frac{\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{2}{x}−\mathrm{sin}\:\mathrm{2}{x}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{{BC}}{\mathrm{sin}\:\measuredangle{BDC}}=\frac{{CD}}{\mathrm{sin}\:\measuredangle{CBD}} \\ $$$$\frac{{a}}{\mathrm{sin}\left(\mathrm{180}°−\mathrm{3}{x}\right)\:}=\frac{\mathrm{1}−{a}}{\mathrm{sin}\:{x}} \\ $$$$\frac{\frac{\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{2}{x}−\mathrm{sin}\:\mathrm{2}{x}}{\:\sqrt{\mathrm{3}}}}{\mathrm{sin}\:\mathrm{3}{x}}=\frac{\mathrm{1}−\frac{\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{2}{x}−\mathrm{sin}\:\mathrm{2}{x}}{\:\sqrt{\mathrm{3}}}}{\mathrm{sin}\:{x}} \\ $$$$ \\ $$$$\mathrm{2}{x}={t} \\ $$$$\sqrt{\mathrm{3}}\left(\mathrm{2cos}^{\mathrm{2}} {t}−\mathrm{1}\right)=\mathrm{2sin}\:{t}\centerdot\left(\mathrm{1}+\mathrm{cos}\:{t}\right) \\ $$$$\mathrm{16cos}^{\mathrm{4}} {t}+\mathrm{8cos}^{\mathrm{3}} {t}−\mathrm{12cos}^{\mathrm{2}} {t}−\mathrm{8cos}\:{t}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2cos}\:{t}={u} \\ $$$${u}^{\mathrm{4}} +{u}^{\mathrm{3}} −\mathrm{3}{u}^{\mathrm{2}} −\mathrm{4}{u}−\mathrm{1}=\mathrm{0} \\ $$$${u}^{\mathrm{3}} \left({u}+\mathrm{1}\right)−\left(\mathrm{3}{u}+\mathrm{1}\right)\left({u}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({u}^{\mathrm{3}} −\mathrm{3}{u}−\mathrm{1}\right)\left({u}+\mathrm{1}\right)=\mathrm{0} \\ $$$${u}^{\mathrm{3}} −\mathrm{3}{u}−\mathrm{1}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{cos}\:\mathrm{60}°=\mathrm{cos}\:\left(\mathrm{3}\centerdot\mathrm{20}°\right)= \\ $$$$\mathrm{cos}\:\mathrm{40}°\mathrm{cos}\:\mathrm{20}°−\mathrm{sin}\:\mathrm{40}^{°} \mathrm{sin}\:\mathrm{20}°= \\ $$$$\left(\mathrm{2cos}^{\mathrm{2}} \mathrm{20}°−\mathrm{1}\right)\mathrm{cos}\:\mathrm{20}°− \\ $$$$\mathrm{2cos}\:\mathrm{20}^{°} \left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{20}°\right)= \\ $$$$\mathrm{4cos}^{\mathrm{3}} \mathrm{20}°−\mathrm{3cos}\:\mathrm{20}^{°} \\ $$$$\left(\mathrm{2cos}\:\mathrm{20}°\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{2cos}\:\mathrm{20}°\right)−\mathrm{1}=\mathrm{0} \\ $$$${x}=\mathrm{10}° \\ $$

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