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Question Number 30159 by naka3546 last updated on 17/Feb/18

Commented by mrW2 last updated on 18/Feb/18

sir, please check your question. I think the given configuration leads to no normal solution. If BD=5 cm, then AC must be between 7.5 and 9 cm to make a solution possible. E.g. if you set AC=8 cm, you′ll get DC=((40)/3) cm.

$${sir},\:{please}\:{check}\:{your}\:{question}. \\ $$$${I}\:{think}\:{the}\:{given}\:{configuration}\:{leads} \\ $$$${to}\:{no}\:{normal}\:{solution}. \\ $$$${If}\:{BD}=\mathrm{5}\:{cm},\:{then}\:{AC}\:{must}\:{be}\:{between} \\ $$$$\mathrm{7}.\mathrm{5}\:{and}\:\mathrm{9}\:{cm}\:{to}\:{make}\:{a}\:{solution} \\ $$$${possible}.\:{E}.{g}.\:{if}\:{you}\:{set}\:{AC}=\mathrm{8}\:{cm}, \\ $$$${you}'{ll}\:{get}\:{DC}=\frac{\mathrm{40}}{\mathrm{3}}\:{cm}. \\ $$

Answered by mrW2 last updated on 18/Feb/18

Commented by mrW2 last updated on 18/Feb/18

let λ=(a/b) (a/(sin 3θ))=((CB)/(sin 4θ)) ...(i) (b/(sin θ))=((CB)/(sin 2θ)) ...(ii) (i)/(ii): (a/(sin 3θ))×((sin θ)/b)=((sin 2θ)/(sin 4θ))=(1/(2 cos 2θ)) ⇒((λ sin θ)/(sin 3θ))=(1/(2 cos 2θ)) 2λ sin θ cos 2θ=sin 3θ=sin θ cos 2θ+cos θ sin 2θ (2λ−1) sin θ cos 2θ=cos θ sin 2θ (2λ−1) tan θ=tan 2θ=((2 tan θ)/(1−tan^2 θ)) 2λ−1=(2/(1−tan^2 θ)) tan^2 θ=1−(2/(2λ−1))>0 ⇒λ>(3/2) ⇒tan θ=(√(1−(2/(2λ−1)))) 3θ+4θ<180° ⇒θ<((180°)/7)≈2 tan^2 θ=1−(2/(2λ−1))<tan^2 (((180°)/7)) ⇒λ<(1/2)+(1/(1−tan^2 ((180°)/7)))=1.8 i.e. to make a solution possible, 1.5<λ=(a/b)<1.8 for b=5 cm: a>1.5×5=7.5 cm a<1.8×5=9 cm let′s say b=8 cm, ⇒λ=(8/5)=1.6 ⇒tan θ=(√(1−(2/(2×1.6−1))))=(1/(√(11))) ⇒θ=16.8° ⇒cos 2θ=2 cos^2 θ−1=(2/(1+tan^2 θ))−1 ⇒cos 2θ=(2/(1+(1/(11))))−1=((2×11)/(12))−1=(5/6) ((DC)/(sin 4θ))=(a/(sin 2θ)) ⇒DC=((a sin 4θ)/(sin 2θ))=2a cos 2θ=2×8×(5/6)=((40)/3)

$${let}\:\lambda=\frac{{a}}{{b}} \\ $$$$\frac{{a}}{\mathrm{sin}\:\mathrm{3}\theta}=\frac{{CB}}{\mathrm{sin}\:\mathrm{4}\theta}\:\:\:...\left({i}\right) \\ $$$$\frac{{b}}{\mathrm{sin}\:\theta}=\frac{{CB}}{\mathrm{sin}\:\mathrm{2}\theta}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\frac{{a}}{\mathrm{sin}\:\mathrm{3}\theta}×\frac{\mathrm{sin}\:\theta}{{b}}=\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{sin}\:\mathrm{4}\theta}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta} \\ $$$$\Rightarrow\frac{\lambda\:\mathrm{sin}\:\theta}{\mathrm{sin}\:\mathrm{3}\theta}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\theta} \\ $$$$\mathrm{2}\lambda\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\mathrm{2}\theta=\mathrm{sin}\:\mathrm{3}\theta=\mathrm{sin}\:\theta\:\mathrm{cos}\:\mathrm{2}\theta+\mathrm{cos}\:\theta\:\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\left(\mathrm{2}\lambda−\mathrm{1}\right)\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\mathrm{2}\theta=\mathrm{cos}\:\theta\:\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\left(\mathrm{2}\lambda−\mathrm{1}\right)\:\mathrm{tan}\:\theta=\mathrm{tan}\:\mathrm{2}\theta=\frac{\mathrm{2}\:\mathrm{tan}\:\theta}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\theta} \\ $$$$\mathrm{2}\lambda−\mathrm{1}=\frac{\mathrm{2}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\theta} \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\theta=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{2}\lambda−\mathrm{1}}>\mathrm{0}\:\Rightarrow\lambda>\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\sqrt{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{2}\lambda−\mathrm{1}}} \\ $$$$\mathrm{3}\theta+\mathrm{4}\theta<\mathrm{180}° \\ $$$$\Rightarrow\theta<\frac{\mathrm{180}°}{\mathrm{7}}\approx\mathrm{2} \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\theta=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{2}\lambda−\mathrm{1}}<\mathrm{tan}^{\mathrm{2}} \:\left(\frac{\mathrm{180}°}{\mathrm{7}}\right) \\ $$$$\Rightarrow\lambda<\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:\frac{\mathrm{180}°}{\mathrm{7}}}=\mathrm{1}.\mathrm{8} \\ $$$${i}.{e}.\:{to}\:{make}\:{a}\:{solution}\:{possible}, \\ $$$$\mathrm{1}.\mathrm{5}<\lambda=\frac{{a}}{{b}}<\mathrm{1}.\mathrm{8} \\ $$$${for}\:{b}=\mathrm{5}\:{cm}: \\ $$$${a}>\mathrm{1}.\mathrm{5}×\mathrm{5}=\mathrm{7}.\mathrm{5}\:{cm} \\ $$$${a}<\mathrm{1}.\mathrm{8}×\mathrm{5}=\mathrm{9}\:{cm} \\ $$$$ \\ $$$${let}'{s}\:{say}\:{b}=\mathrm{8}\:{cm},\:\Rightarrow\lambda=\frac{\mathrm{8}}{\mathrm{5}}=\mathrm{1}.\mathrm{6} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\sqrt{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{2}×\mathrm{1}.\mathrm{6}−\mathrm{1}}}=\frac{\mathrm{1}}{\sqrt{\mathrm{11}}}\:\Rightarrow\theta=\mathrm{16}.\mathrm{8}° \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{2}\theta=\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\theta−\mathrm{1}=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}−\mathrm{1} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{2}\theta=\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{11}}}−\mathrm{1}=\frac{\mathrm{2}×\mathrm{11}}{\mathrm{12}}−\mathrm{1}=\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$\frac{{DC}}{\mathrm{sin}\:\mathrm{4}\theta}=\frac{{a}}{\mathrm{sin}\:\mathrm{2}\theta} \\ $$$$\Rightarrow{DC}=\frac{{a}\:\mathrm{sin}\:\mathrm{4}\theta}{\mathrm{sin}\:\mathrm{2}\theta}=\mathrm{2}{a}\:\mathrm{cos}\:\mathrm{2}\theta=\mathrm{2}×\mathrm{8}×\frac{\mathrm{5}}{\mathrm{6}}=\frac{\mathrm{40}}{\mathrm{3}} \\ $$

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