Menu Close

Question-40023

Tinku Tara 0 Comments
Pin




Question Number 40023 by ajfour last updated on 15/Jul/18
Answered by MrW3 last updated on 15/Jul/18
AB=2R cos θ AD=(R/(cos θ)) BD=AB−AD=R(2 cos θ−(1/(cos θ)))=((R cos 2θ)/(cos θ)) Let E=midpoint of BC. BE=AB sin θ=R sin 2θ BC=2R sin 2θ ((AB)/(BE))=((BC)/(BD)) ⇒((2R cos θ)/(R sin 2θ))=((2R sin 2θ cos θ)/(R cos 2θ)) ⇒(1/(sin 2θ))=((sin 2θ)/(cos 2θ)) cos 2θ=sin^2 2θ=1−cos^2 2θ cos^2 2θ+cos 2θ−1=0 4 cos^4 θ−2 cos^2 θ−1=0 ⇒cos^2 θ=((1+(√5))/4) ⇒cos θ=((√(1+(√5)))/2) ⇒θ=cos^(−1) ((√(1+(√5)))/2)≈25.9°
$${AB}=\mathrm{2}{R}\:\mathrm{cos}\:\theta \\ $$$${AD}=\frac{{R}}{\mathrm{cos}\:\theta} \\ $$$${BD}={AB}−{AD}={R}\left(\mathrm{2}\:\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\right)=\frac{{R}\:\mathrm{cos}\:\mathrm{2}\theta}{\mathrm{cos}\:\theta} \\ $$$${Let}\:{E}={midpoint}\:{of}\:{BC}. \\ $$$${BE}={AB}\:\mathrm{sin}\:\theta={R}\:\mathrm{sin}\:\mathrm{2}\theta \\ $$$${BC}=\mathrm{2}{R}\:\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\frac{{AB}}{{BE}}=\frac{{BC}}{{BD}} \\ $$$$\Rightarrow\frac{\mathrm{2}{R}\:\mathrm{cos}\:\theta}{{R}\:\mathrm{sin}\:\mathrm{2}\theta}=\frac{\mathrm{2}{R}\:\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{cos}\:\theta}{{R}\:\mathrm{cos}\:\mathrm{2}\theta} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{2}\theta}=\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{cos}\:\mathrm{2}\theta} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\theta=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\theta \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\theta+\mathrm{cos}\:\mathrm{2}\theta−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{4}\:\mathrm{cos}^{\mathrm{4}} \:\theta−\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{cos}^{\mathrm{2}} \:\theta=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{5}}}}{\mathrm{2}} \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{5}}}}{\mathrm{2}}\approx\mathrm{25}.\mathrm{9}° \\ $$
Commented by ajfour last updated on 15/Jul/18
Thank you Sir, very correct!
$${Thank}\:{you}\:{Sir},\:{very}\:{correct}! \\ $$
Answered by ajfour last updated on 15/Jul/18
AB=2cos θ BC = 4cos θsin θ BD = 4cos θsin^2 θ BD+AD =AB ⇒ 4cos θsin^2 θ+(1/(cos θ)) =2cos θ let cos θ =t ⇒ 4t(1−t^2 )+(1/t)=2t or 4t^2 (1−t^2 )+1−2t^2 =0 ⇒ 4t^4 −2t^2 −1=0 or t^2 =((1+(√5))/4) =cos^2 θ cos θ = ((√(1+(√5)))/2) .
$${AB}=\mathrm{2cos}\:\theta \\ $$$${BC}\:=\:\mathrm{4cos}\:\theta\mathrm{sin}\:\theta \\ $$$${BD}\:=\:\mathrm{4cos}\:\theta\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$${BD}+{AD}\:={AB} \\ $$$$\Rightarrow\:\mathrm{4cos}\:\theta\mathrm{sin}\:^{\mathrm{2}} \theta+\frac{\mathrm{1}}{\mathrm{cos}\:\theta}\:=\mathrm{2cos}\:\theta \\ $$$${let}\:\mathrm{cos}\:\theta\:={t} \\ $$$$\Rightarrow\:\:\:\mathrm{4}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{{t}}=\mathrm{2}{t} \\ $$$${or}\:\:\:\:\mathrm{4}{t}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)+\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\mathrm{4}{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${or}\:\:\:{t}^{\mathrm{2}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\:=\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\mathrm{cos}\:\theta\:=\:\frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{5}}}}{\mathrm{2}}\:\:. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *