Question Number 203151 by sonukgindia last updated on 11/Jan/24
Answered by mr W last updated on 11/Jan/24
Commented by mr W last updated on 11/Jan/24
![((AB)/(sin (θ+60°)))=(1/(sin 60°)) ⇒AB=((sin θ)/( (√3)))+cos θ 2α+θ=60° ⇒α=30°−(θ/2) BD tan 30°=DA tan α (AB−DA) tan 30°=DA tan (30°−(θ/2)) ⇒DA=((AB)/(1+(√3) tan (30°−(θ/2)))) ⇒DA=((((sin θ)/( (√3)))+cos θ)/(1+(√3) tan (30°−(θ/2)))) AE=((DA)/(cos 2α))=((((sin θ)/( (√3)))+cos θ)/(cos (60°−θ)[1+(√3) tan (30°−(θ/2))])) AE=((2(sin θ+(√3) cos θ))/( (√3)(cos θ+(√3) sin θ)[1+(√3) tan (30°−(θ/2))])) AE=((2(tan θ+(√3)))/( (√3)(1+(√3) tan θ)[1+(√3) tan (30°−(θ/2))])) (AE)_(min) ≈0.9082 ⇒(EC)_(max) ≈0.0918 at θ≈0.4317 (24.73°)](https://www.tinkutara.com/question/Q203183.png)
$$\frac{{AB}}{\mathrm{sin}\:\left(\theta+\mathrm{60}°\right)}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{60}°} \\ $$$$\Rightarrow{AB}=\frac{\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{3}}}+\mathrm{cos}\:\theta \\ $$$$\mathrm{2}\alpha+\theta=\mathrm{60}° \\ $$$$\Rightarrow\alpha=\mathrm{30}°−\frac{\theta}{\mathrm{2}} \\ $$$${BD}\:\mathrm{tan}\:\mathrm{30}°={DA}\:\mathrm{tan}\:\alpha \\ $$$$\left({AB}−{DA}\right)\:\mathrm{tan}\:\mathrm{30}°={DA}\:\mathrm{tan}\:\left(\mathrm{30}°−\frac{\theta}{\mathrm{2}}\right) \\ $$$$\Rightarrow{DA}=\frac{{AB}}{\mathrm{1}+\sqrt{\mathrm{3}}\:\mathrm{tan}\:\left(\mathrm{30}°−\frac{\theta}{\mathrm{2}}\right)} \\ $$$$\Rightarrow{DA}=\frac{\frac{\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{3}}}+\mathrm{cos}\:\theta}{\mathrm{1}+\sqrt{\mathrm{3}}\:\mathrm{tan}\:\left(\mathrm{30}°−\frac{\theta}{\mathrm{2}}\right)} \\ $$$${AE}=\frac{{DA}}{\mathrm{cos}\:\mathrm{2}\alpha}=\frac{\frac{\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{3}}}+\mathrm{cos}\:\theta}{\mathrm{cos}\:\left(\mathrm{60}°−\theta\right)\left[\mathrm{1}+\sqrt{\mathrm{3}}\:\mathrm{tan}\:\left(\mathrm{30}°−\frac{\theta}{\mathrm{2}}\right)\right]} \\ $$$${AE}=\frac{\mathrm{2}\left(\mathrm{sin}\:\theta+\sqrt{\mathrm{3}}\:\mathrm{cos}\:\theta\right)}{\:\sqrt{\mathrm{3}}\left(\mathrm{cos}\:\theta+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\theta\right)\left[\mathrm{1}+\sqrt{\mathrm{3}}\:\mathrm{tan}\:\left(\mathrm{30}°−\frac{\theta}{\mathrm{2}}\right)\right]} \\ $$$${AE}=\frac{\mathrm{2}\left(\mathrm{tan}\:\theta+\sqrt{\mathrm{3}}\right)}{\:\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\mathrm{tan}\:\theta\right)\left[\mathrm{1}+\sqrt{\mathrm{3}}\:\mathrm{tan}\:\left(\mathrm{30}°−\frac{\theta}{\mathrm{2}}\right)\right]} \\ $$$$\left({AE}\right)_{{min}} \approx\mathrm{0}.\mathrm{9082}\:\Rightarrow\left({EC}\right)_{{max}} \approx\mathrm{0}.\mathrm{0918} \\ $$$${at}\:\theta\approx\mathrm{0}.\mathrm{4317}\:\left(\mathrm{24}.\mathrm{73}°\right) \\ $$
Commented by MathematicalUser2357 last updated on 20/Jan/24
$$=\frac{\mathrm{2}\left(\mathrm{tan}\:\mathrm{24}.\mathrm{73}°+\sqrt{\mathrm{3}}\right)}{\:\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{tan}\:\mathrm{24}.\mathrm{73}°\right)\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{tan}\left(\mathrm{30}°−\frac{\mathrm{24}.\mathrm{73}°}{\mathrm{2}}\right)\right)} \\ $$