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Question-49725

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Question Number 49725 by behi83417@gmail.com last updated on 09/Dec/18
Commented by behi83417@gmail.com last updated on 09/Dec/18
ABC,is equilaterl.((AD)/(DB))=(1/2) ⇒∡ADC=?
$${ABC},{is}\:{equilaterl}.\frac{{AD}}{{DB}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\measuredangle{ADC}=? \\ $$
Answered by ajfour last updated on 09/Dec/18
((AD)/(sin ∠ACD)) = ((AC)/(sin ∠ADC)) ⇒ ((a/3)/(sin α)) = (a/(sin θ)) ⇒ sin θ = 3sin α ((BD)/(sin ∠BCD)) = ((BC)/(sin ∠BDC)) ⇒ ((2a/3)/(sin (π/3−α))) = (a/(sin θ)) ⇒ sin θ = (3/2)sin ((π/3)−α) ⇒ 2sin θ=3(((√3)/2)cos α−(1/2)sin α) 4sin θ = 3(√3)(√(1−((sin^2 θ)/9)))−sin θ ⇒ 25sin^2 θ = 3(9−sin^2 θ) sin θ = (√((27)/(28))) = ((3(√3))/(2(√7))) ⇒ θ = ∠ADC =sin^(−1) (((3(√3))/(2(√7)))) .
$$\frac{{AD}}{\mathrm{sin}\:\angle{ACD}}\:=\:\frac{{AC}}{\mathrm{sin}\:\angle{ADC}} \\ $$$$\Rightarrow\:\:\frac{{a}/\mathrm{3}}{\mathrm{sin}\:\alpha}\:=\:\frac{{a}}{\mathrm{sin}\:\theta}\:\:\Rightarrow\:\:\mathrm{sin}\:\theta\:=\:\mathrm{3sin}\:\alpha \\ $$$$\frac{{BD}}{\mathrm{sin}\:\angle{BCD}}\:=\:\frac{{BC}}{\mathrm{sin}\:\angle{BDC}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{2}{a}/\mathrm{3}}{\mathrm{sin}\:\left(\pi/\mathrm{3}−\alpha\right)}\:=\:\frac{{a}}{\mathrm{sin}\:\theta} \\ $$$$\Rightarrow\:\:\mathrm{sin}\:\theta\:=\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\alpha\right) \\ $$$$\Rightarrow\:\:\mathrm{2sin}\:\theta=\mathrm{3}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\alpha−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\alpha\right) \\ $$$$\:\:\mathrm{4sin}\:\theta\:=\:\mathrm{3}\sqrt{\mathrm{3}}\sqrt{\mathrm{1}−\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{9}}}−\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\:\mathrm{25sin}\:^{\mathrm{2}} \theta\:=\:\mathrm{3}\left(\mathrm{9}−\mathrm{sin}\:^{\mathrm{2}} \theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{sin}\:\theta\:=\:\sqrt{\frac{\mathrm{27}}{\mathrm{28}}}\:=\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{7}}} \\ $$$$\:\Rightarrow\:\:\:\:\theta\:=\:\angle{ADC}\:=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{7}}}\right)\:. \\ $$
Answered by ajfour last updated on 09/Dec/18
sin ∠ADC =((Altitude(CE))/(CD))= ((((((√3)a)/2)))/( (√(((a/2)−(a/3))^2 +((((√3)a)/2))^2 )))) = ((3(√3))/(2(√7))) ∠ADC = sin^(−1) (((3(√3))/(2(√7)))) .
$$\mathrm{sin}\:\angle{ADC}\:=\frac{{Altitude}\left({CE}\right)}{{CD}}=\:\frac{\left(\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}\right)}{\:\sqrt{\left(\frac{{a}}{\mathrm{2}}−\frac{{a}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}\right)^{\mathrm{2}} }}\:=\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{7}}} \\ $$$$\:\:\angle{ADC}\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{7}}}\right)\:. \\ $$
Commented by behi83417@gmail.com last updated on 09/Dec/18
nice sir!thanks. sir Ajfour!good questions are uploaded above that you may intrested in.
$${nice}\:{sir}!{thanks}. \\ $$$${sir}\:{Ajfour}!{good}\:{questions}\:{are}\:{uploaded} \\ $$$${above}\:{that}\:{you}\:{may}\:{intrested}\:{in}. \\ $$

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