Question-70277

Question Number 70277 by mr W last updated on 02/Oct/19

Answered by mind is power last updated on 02/Oct/19

(a/(sin(θ)))=((2a)/(Sin(π−3θ)))=((2a)/(sin(3θ))) ⇒sin(3θ)=2sin(θ) sin(3θ)=−4sin^3 (θ)+3sin(θ) ⇒−4sin^3 (θ)+sin(θ)=0 ⇒sin(θ)(1−4sin^2 (θ))=0⇒1−4sin^2 (θ)=0 ⇒(1−2sin(θ))(1+2sinθ))=0 sin(θ)=(1/2)⇒θ=30^°

$$\frac{{a}}{{sin}\left(\theta\right)}=\frac{\mathrm{2}{a}}{{Sin}\left(\pi−\mathrm{3}\theta\right)}=\frac{\mathrm{2}{a}}{{sin}\left(\mathrm{3}\theta\right)} \\ $$$$\Rightarrow{sin}\left(\mathrm{3}\theta\right)=\mathrm{2}{sin}\left(\theta\right) \\ $$$${sin}\left(\mathrm{3}\theta\right)=−\mathrm{4}{sin}^{\mathrm{3}} \left(\theta\right)+\mathrm{3}{sin}\left(\theta\right) \\ $$$$\Rightarrow−\mathrm{4}{sin}^{\mathrm{3}} \left(\theta\right)+{sin}\left(\theta\right)=\mathrm{0} \\ $$$$\Rightarrow{sin}\left(\theta\right)\left(\mathrm{1}−\mathrm{4}{sin}^{\mathrm{2}} \left(\theta\right)\right)=\mathrm{0}\Rightarrow\mathrm{1}−\mathrm{4}{sin}^{\mathrm{2}} \left(\theta\right)=\mathrm{0} \\ $$$$\left.\Rightarrow\left(\mathrm{1}−\mathrm{2}{sin}\left(\theta\right)\right)\left(\mathrm{1}+\mathrm{2}{sin}\theta\right)\right)=\mathrm{0} \\ $$$${sin}\left(\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\theta=\mathrm{30}^{°} \\ $$

Commented by mr W last updated on 02/Oct/19

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Commented by mr W last updated on 02/Oct/19

$${can}\:{we}\:{solve}\:{without}\:{trigonometry}? \\ $$

Commented by mind is power last updated on 03/Oct/19

$${i}\:{will}\:{Try}\:{sir} \\ $$

Answered by som(math1967) last updated on 03/Oct/19

Construct bisector of ∠B(2α) AD ∴((AD)/(DC))=(a/(2a))=(1/2) ∴AD=(1/3)AC again △ABC∼△ABD ∴((AB)/(AD))=((AC)/(AB)) ∴AB^2 =AD×AC=(1/3)AC^2 AC^2 =3AB^2 =3a^2 AB^2 +AC^2 =a^2 +3a^2 =4a^2 =BC^2 ∴∠A=90° ∴α+2α=90 α=30

$${Construct}\:{bisector}\:{of}\:\angle{B}\left(\mathrm{2}\alpha\right)\:{AD} \\ $$$$\therefore\frac{{AD}}{{DC}}=\frac{{a}}{\mathrm{2}{a}}=\frac{\mathrm{1}}{\mathrm{2}}\:\therefore{AD}=\frac{\mathrm{1}}{\mathrm{3}}{AC} \\ $$$${again}\:\bigtriangleup{ABC}\sim\bigtriangleup{ABD} \\ $$$$\therefore\frac{{AB}}{{AD}}=\frac{{AC}}{{AB}} \\ $$$$\therefore{AB}^{\mathrm{2}} ={AD}×{AC}=\frac{\mathrm{1}}{\mathrm{3}}{AC}^{\mathrm{2}} \\ $$$${AC}^{\mathrm{2}} =\mathrm{3}{AB}^{\mathrm{2}} =\mathrm{3}{a}^{\mathrm{2}} \\ $$$${AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} ={BC}^{\mathrm{2}} \\ $$$$\therefore\angle{A}=\mathrm{90}°\:\therefore\alpha+\mathrm{2}\alpha=\mathrm{90} \\ $$$$\:\alpha=\mathrm{30}\: \\ $$

Commented by som(math1967) last updated on 03/Oct/19