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Question Number 72666 by aliesam last updated on 31/Oct/19

	Answered by mind is power last updated on 31/Oct/19
	$let α=∠CDA=BAE ,OCA=2α E=(AD)∩BC (3/5)=((BE)/(EC)),BE+EC=4 ⇒BE=(3/2) AE^2 =(9/4)+9−2.(3/2).3cos(θ)=((45)/4)−9cos(θ)⇒AE=(3/2)(√(5−4cos(θ))) AC^2 =9+16−2.3.4cos(θ)=25−24cos(θ) ((AE)/(sin(θ)))=((BE)/(sin(α)))⇒sin(α)=((BE)/(AE)).sin(θ)=((sin(θ))/(√(5−4cos(θ)))) AC^2 =oA^2 +OC^2 −2OA.OCcos(2α)=2R^2 (1−cos(2α))=52sin^2 (α) ⇔ { ((52sin^2 (α)=25−24cos(θ))),((sin(α)=((sin(θ))/(√(5−4cos(θ)))))) :} ⇒((52sin^2 (θ))/(5−4cos(θ)))=25−24cos(θ) ⇒52(1−cos^2 (θ))=125+96cos^2 (θ)−220cos(θ) ⇔148cos^2 (θ)−220cos(θ)+73=0 Δ=220^2 −4.73.148=5184 cos(θ)=((220−72)/(296))=((148)/(296))=(1/2) cos(θ)=((292)/(296))=((73)/(74))....impossibl cos(DOC)=((2.13−5^2 )/(26))=(1/(26))⇒DOC≤90⇒DCO≥45 ⇒cos(θ)=cos(DcO)≤((√2)/2)⇒cos(θ)=(1/2)⇒θ=(π/3)=60^° BO DOC=cos^(−1) ((1/(26))) OCD=((π−DOC)/2) OCB=θ−OCD OC=(√(13)),CB=4 OB=(√(OC^2 +CB^2 −2OC.CBcos(OCB)))$
	$let α = ∠ CDA = BAE, OCA = 2 α$ $E = (AD) \cap BC$ $\frac{3}{5} = \frac{BE}{EC}, BE + EC = 4$ $\Rightarrow BE = \frac{3}{2}$ ${AE}^{2} = \frac{9}{4} + 9 - 2 . \frac{3}{2} . 3 \cos (θ) = \frac{45}{4} - 9 \cos (θ) \Rightarrow AE = \frac{3}{2} \sqrt{5 - 4 \cos (θ)}$ ${AC}^{2} = 9 + 16 - 2.3 . 4 \cos (θ) = 25 - 24 \cos (θ)$ $\frac{AE}{\sin (θ)} = \frac{BE}{\sin (α)} \Rightarrow \sin (α) = \frac{BE}{AE} . \sin (θ) = \frac{\sin (θ)}{\sqrt{5 - 4 \cos (θ)}}$ ${AC}^{2} = {oA}^{2} + {OC}^{2} - 2 OA . OCcos (2 α) = {2 R}^{2} (1 - \cos (2 α)) = {52 \sin}^{2} (α)$ $\Leftrightarrow {\begin{cases} {52 \sin}^{2} (α) = 25 - 24 \cos (θ) \\ \sin (α) = \frac{\sin (θ)}{\sqrt{5 - 4 \cos (θ)}} \end{cases}$ $\Rightarrow \frac{{52 \sin}^{2} (θ)}{5 - 4 \cos (θ)} = 25 - 24 \cos (θ)$ $\Rightarrow 52 (1 - \cos^{2} (θ)) = 125 + {96 \cos}^{2} (θ) - 220 \cos (θ)$ $\Leftrightarrow {148 \cos}^{2} (θ) - 220 \cos (θ) + 73 = 0$ $Δ = 220^{2} - 4.73.148 = 5184$ $\cos (θ) = \frac{220 - 72}{296} = \frac{148}{296} = \frac{1}{2}$ $\cos (θ) = \frac{292}{296} = \frac{73}{74} . . . . impossibl$ $\cos (DOC) = \frac{2.13 - 5^{2}}{26} = \frac{1}{26} \Rightarrow DOC ⩽ 90 \Rightarrow DCO ⩾ 45$ $\Rightarrow \cos (θ) = \cos (DcO) ⩽ \frac{\sqrt{2}}{2} \Rightarrow \cos (θ) = \frac{1}{2} \Rightarrow θ = \frac{π}{3} = 60^{°}$ $BO$ $DOC = \cos^{- 1} (\frac{1}{26})$ $OCD = \frac{π - DOC}{2}$ $OCB = θ - OCD$ $OC = \sqrt{13}, CB = 4$ $OB = \sqrt{{OC}^{2} + {CB}^{2} - 2 OC . CBcos (OCB)}$
	Commented by aliesam last updated on 31/Oct/19

	$n i c e s o l u t i o n s i r g o d b l e s s y o u$
	Commented by mind is power last updated on 01/Nov/19

	$Thank you Sir$
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