Question-68309

Question Number 68309 by ajfour last updated on 08/Sep/19

Commented by ajfour last updated on 08/Sep/19

$${Find}\:{AB}\:{and}\:{AC}\:{in}\:{terms}\:{of}\:{a}\: \\ $$$${and}\:\theta. \\ $$

Commented by mr W last updated on 08/Sep/19

((AB)/(sin θ))=((AC)/(sin 2θ))=(a/(sin 3θ)) ⇒AB=((a sin θ)/(sin 3θ))=(a/(3−4 sin^2 θ)) ⇒AC=((a sin 2θ)/(sin 3θ))=((2a cos θ)/(3−4 sin^2 θ))

$$\frac{{AB}}{\mathrm{sin}\:\theta}=\frac{{AC}}{\mathrm{sin}\:\mathrm{2}\theta}=\frac{{a}}{\mathrm{sin}\:\mathrm{3}\theta} \\ $$$$\Rightarrow{AB}=\frac{{a}\:\mathrm{sin}\:\theta}{\mathrm{sin}\:\mathrm{3}\theta}=\frac{{a}}{\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\Rightarrow{AC}=\frac{{a}\:\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{sin}\:\mathrm{3}\theta}=\frac{\mathrm{2}{a}\:\mathrm{cos}\:\theta}{\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$

Commented by ajfour last updated on 09/Sep/19

$${Thank}\:{you}\:{Sir}.\:{I}\:{shall}\:{find}\:{a} \\ $$$${better}\:{question}. \\ $$

Answered by ajfour last updated on 09/Sep/19

let AC=ap and AB=q ((ap)/(sin 2θ))=(q/(sin θ)) ⇒ q=((ap)/(2cos θ)) cos (π−3θ)=((((p^2 /(4cos^2 θ))+p^2 −1))/((((2p^2 )/(2cos θ))))) ⇒ (3−4cos^2 θ)p^2 =p^2 +(p^2 /(4cos^2 θ))−1 ⇒ (1/p^2 )=(1/(4cos^2 θ))+4cos^2 θ−2 If 0< θ < (π/2) ⇒ (1/p)=(1/(2cos θ))−2cos θ =((1−4cos^2 θ)/(2cos θ)) as AC=ap and AB=((ap)/(2cos θ)) so AC= ((2acos θ)/(1−4cos^2 θ)) ; AB=(a/(1−4cos^2 θ))

$${let}\:\:{AC}={ap}\:\:{and}\:{AB}={q} \\ $$$$\frac{{ap}}{\mathrm{sin}\:\mathrm{2}\theta}=\frac{{q}}{\mathrm{sin}\:\theta}\:\:\Rightarrow\:\:{q}=\frac{{ap}}{\mathrm{2cos}\:\theta} \\ $$$$\mathrm{cos}\:\left(\pi−\mathrm{3}\theta\right)=\frac{\left(\frac{{p}^{\mathrm{2}} }{\mathrm{4cos}\:^{\mathrm{2}} \theta}+{p}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\frac{\mathrm{2}{p}^{\mathrm{2}} }{\mathrm{2cos}\:\theta}\right)} \\ $$$$\Rightarrow\:\left(\mathrm{3}−\mathrm{4cos}\:^{\mathrm{2}} \theta\right){p}^{\mathrm{2}} ={p}^{\mathrm{2}} +\frac{{p}^{\mathrm{2}} }{\mathrm{4cos}\:^{\mathrm{2}} \theta}−\mathrm{1} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{p}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4cos}\:^{\mathrm{2}} \theta}+\mathrm{4cos}\:^{\mathrm{2}} \theta−\mathrm{2} \\ $$$$\:\:\:\:{If}\:\:\:\:\mathrm{0}<\:\theta\:<\:\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{p}}=\frac{\mathrm{1}}{\mathrm{2cos}\:\theta}−\mathrm{2cos}\:\theta\:=\frac{\mathrm{1}−\mathrm{4cos}\:^{\mathrm{2}} \theta}{\mathrm{2cos}\:\theta} \\ $$$${as}\:\:{AC}={ap}\:\:{and}\:{AB}=\frac{{ap}}{\mathrm{2cos}\:\theta} \\ $$$${so}\:\:\boldsymbol{{AC}}=\:\frac{\mathrm{2}{a}\mathrm{cos}\:\theta}{\mathrm{1}−\mathrm{4cos}\:^{\mathrm{2}} \theta}\:\:\:;\:\:\boldsymbol{{AB}}=\frac{{a}}{\mathrm{1}−\mathrm{4cos}\:^{\mathrm{2}} \theta}\: \\ $$

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