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Question Number 202967 by sonukgindia last updated on 06/Jan/24

Answered by mr W last updated on 06/Jan/24

Commented by mr W last updated on 07/Jan/24

$$sidelengthofhexagon=1$$$$\frac{1}{\sin (60°-\theta )}=\frac{c}{\sin \theta }$$$$\Rightarrow c=\frac{2\sin \theta }{\sqrt{3}\cos \theta -\sin \theta }$$$$\Rightarrow d=2-c=2-\frac{2\sin \theta }{\sqrt{3}\cos \theta -\sin \theta }$$$$\frac{1-x+1}{\sin (30°+\theta )}=\frac{d}{\sin (90°+\theta )}$$$$\frac{2(2-x)}{\cos \theta +\sqrt{3}\sin \theta }=\frac{1}{\cos \theta }\times (2-\frac{2\sin \theta }{\sqrt{3}\cos \theta -\sin \theta })$$$$x=2-\frac{(1+\sqrt{3}\tan \theta )(\sqrt{3}-2\tan \theta )}{\sqrt{3}-\tan \theta }$$$$lett=\tan \theta$$$$x=3-\frac{2\sqrt{3}(1-t^2)}{\sqrt{3}-t}$$$$\frac{dx}{dt}=0$$$$\Rightarrow \frac{-2t}{\sqrt{3}-t}+\frac{1-t^2}{{(\sqrt{3}-t)}^2}=0$$$$\Rightarrow t^2-2\sqrt{3}t+1=0$$$$\Rightarrow t=\sqrt{3}-\sqrt{2}\Rightarrow \theta ={\tan }^{-1}(\sqrt{3}-\sqrt{2})\approx 17.63°$$$$x_{min}=4\sqrt{6}-9\approx 0.798$$$${\left(\frac{yellow}{red}\right)}_{min}=\frac{x_{min}}{1-x_{min}}=\frac{4\sqrt{6}-9}{10-4\sqrt{6}}$$$$=\frac{3}{2}+\sqrt{6}\approx 3.9495$$

Commented by mr W last updated on 06/Jan/24

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