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Question Number 215230 by Mingma last updated on 01/Jan/25

Answered by A5T last updated on 01/Jan/25

Commented by A5T last updated on 01/Jan/25

AC=BC=a; AB=c; ∠CAB=θ; IE∥AB ((a^2 c)/(4×1350))=((ac×sinθ)/2)⇒a=2700sinθ⇒sinθ=(a/(2700)) c^2 =2a^2 (1+cos(2θ)) cos2θ=cos^2 θ−sin^2 θ=1−2sin^2 θ=1−((2a^2 )/(2700^2 )) CF^2 =2×((a/2))^2 (1−cos2θ)=(a^4 /(2700^2 )) ⇒CF=(a^2 /(2700)) (CF varies since a is not constant)

$$\mathrm{AC}=\mathrm{BC}=\mathrm{a};\:\mathrm{AB}=\mathrm{c};\:\:\angle\mathrm{CAB}=\theta;\:\mathrm{IE}\parallel\mathrm{AB} \\ $$$$\frac{\mathrm{a}^{\mathrm{2}} \mathrm{c}}{\mathrm{4}×\mathrm{1350}}=\frac{\mathrm{ac}×\mathrm{sin}\theta}{\mathrm{2}}\Rightarrow\mathrm{a}=\mathrm{2700sin}\theta\Rightarrow\mathrm{sin}\theta=\frac{\mathrm{a}}{\mathrm{2700}} \\ $$$$\mathrm{c}^{\mathrm{2}} =\mathrm{2a}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\left(\mathrm{2}\theta\right)\right) \\ $$$$\mathrm{cos2}\theta=\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{sin}^{\mathrm{2}} \theta=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \theta=\mathrm{1}−\frac{\mathrm{2a}^{\mathrm{2}} }{\mathrm{2700}^{\mathrm{2}} } \\ $$$$\mathrm{CF}^{\mathrm{2}} =\mathrm{2}×\left(\frac{\mathrm{a}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos2}\theta\right)=\frac{\mathrm{a}^{\mathrm{4}} }{\mathrm{2700}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{CF}=\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2700}}\:\left(\mathrm{CF}\:\mathrm{varies}\:\mathrm{since}\:\mathrm{a}\:\mathrm{is}\:\mathrm{not}\:\mathrm{constant}\right) \\ $$

Commented by Mingma last updated on 02/Jan/25

Perfect ��

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