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Question Number 187535 by ajfour last updated on 18/Feb/23

Commented by ajfour last updated on 18/Feb/23

$$Q.187482$$

Answered by ajfour last updated on 19/Feb/23

$$Wealraedyhaveobtained,$$$$s=\tan \alpha =\frac{R}{H}$$$$p\cos 30°=b\Rightarrow \frac{p}{b}=\frac{2}{\sqrt{3}}$$$$\frac{b\cos \alpha +p}{c\cos \alpha }=\frac{AE}{AG\cos \alpha }=\frac{2\sqrt{bc}}{c\cot \alpha }+1$$$$\Rightarrow \frac{b}{c}(1+\frac{2}{\sqrt{3}\cos \alpha })=\frac{2}{\cot \alpha }\sqrt{\frac{b}{c}}+1$$$$say\frac{b}{c}=\lambda \tan \alpha =s,\cos \alpha =\frac{1}{\sqrt{1+s^2}}$$$$\lambda (1+\frac{2\sqrt{1+s^2}}{\sqrt{3}})-2s\sqrt{\lambda }-1=0$$$$\frac{1}{\lambda }+\frac{2s}{\sqrt{\lambda }}-(1+\frac{2\sqrt{1+s^2}}{\sqrt{3}})=0$$$$\frac{1}{\sqrt{\lambda }}=\sqrt{\frac{c}{b}}=-s+\sqrt{s^2+1+2\sqrt{\frac{1+s^2}{3}}}$$$$----------------$$$$examples:$$$$fors=\frac{1}{\sqrt{2}}\Rightarrow \sin \alpha =\frac{1}{\sqrt{3}}$$$$\frac{1}{\sqrt{\lambda }}=\sqrt{\frac{c}{b}}=-\frac{1}{\sqrt{2}}+\sqrt{\frac{3}{2}+\sqrt{2}}=1$$$$for\alpha =60°$$$$\frac{1}{\sqrt{\lambda }}=-\sqrt{3}+\sqrt{3+1+\frac{4}{\sqrt{3}}}$$$$\sqrt{\frac{c}{b}}=-\sqrt{3}+2\sqrt{1+\frac{1}{\sqrt{3}}}$$$$\approx 0.7798$$$$\Rightarrow b\approx 1.6445$$$$for\alpha =30°$$$$\sqrt{\frac{c}{b}}=\frac{2\sqrt{2}-1}{\sqrt{3}}$$$$b\approx 0.8974$$

Commented by ajfour last updated on 19/Feb/23

$$ThanksSir,edited.$$

Commented by mr W last updated on 19/Feb/23

$$perfectsolutionsir!$$

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