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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 θ.

FindABandACintermsofaandθ.

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))

ABsinθ=ACsin2θ=asin3θAB=asinθsin3θ=a34sin2θAC=asin2θsin3θ=2acosθ34sin2θ

Commented by ajfour last updated on 09/Sep/19

Thank you Sir. I shall find a  better question.

ThankyouSir.Ishallfindabetterquestion.

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))

letAC=apandAB=qapsin2θ=qsinθq=ap2cosθcos(π3θ)=(p24cos2θ+p21)(2p22cosθ)(34cos2θ)p2=p2+p24cos2θ11p2=14cos2θ+4cos2θ2If0<θ<π21p=12cosθ2cosθ=14cos2θ2cosθasAC=apandAB=ap2cosθsoAC=2acosθ14cos2θ;AB=a14cos2θ

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